---
abstract: 'The union of an ascending chain of prime ideals is not always prime. The union of an ascending chain of semi-prime ideals is not always semi-prime. We show that these two properties are independent. We also show that the number of non-prime unions of subchains in a chain of primes in a PI-algebra does not exceed the PI-class minus one, and this bound is tight.'
address: 'Department of Mathematics, Bar Ilan University, Ramat Gan 5290002, Israel'
author:
- 'Be’eri Greenfeld'
- 'Louis H. Rowen'
- Uzi Vishne
title: Unions of chains of primes 
---

Introduction {#sec:intro}
============

In a commutative ring, the union of a chain of prime ideals is prime, and the union of a chain of semiprime ideals is semiprime. This paper demonstrates and measures the failure of these chain conditions in general.

A ring has the [**[(semi)prime chain property]{}**]{} (denoted  and , respectively) if the union of any countable chain of (semi)prime ideals is always (semi)prime.[^1]

The property  was recognized by Fisher and Snider [@FS] as the missing hypothesis for Kaplansky’s conjecture on regular rings, and they gave an example of a ring without .

Our focus is on . The class of rings satisfying  is quite large. An easy exercise shows that every commutative ring satisfies , and the same argument yields that the union of strongly prime ideals is strongly prime ($P \normali R$ is strongly prime if $R/P$ is a domain). In fact, we have the following result:

\[first\] Every ring $R$ which is a finite module over a central subring, satisfies .

Write $R = \sum_{i=1}^t Cr_i$ where $C \sub
\operatorname{Cent}(R)$. Suppose $P_1 \subset P_2 \subset \cdots$ is a chain of prime ideals, with $P = \cup P_i$. If $a,b \in R$ with $$\sum C ar_i b = \sum aCr_i b = aRb \subseteq P,$$ then there is $n$ such that $ar_i b \in P_n$ for $1 \le i \le t,$ implying $aRb =
\sum Car_i b \subseteq P_n$, and thus $a \in P_n$ or $b \in P_n$.

(For a recent treatment of the correspondence of infinite chains of primes between a ring $R$ and a central subring, see [@Shai]).

The class of rings satisfying  also contains every ring that satisfies ACC (ascending chain condition) on primes, and is closed under homomorphic images and central localizations. This led some mathematicians to believe that it holds in general. On the other hand, Bergman produced an example lacking  (see [[Example \[Ex1\]]{}]{} below), implying that the free algebra on two generators does not have .

Obviously, the property  follows from the maximum property on families of primes. On the other hand,  implies (by Zorn’s lemma) the following maximum property: for every prime $Q$ contained in any ideal $I$, there is a prime $P$ maximal with respect to $Q \sub P \sub I$.

In [[Section \[sec:mat\]]{}]{} we show that  and  are independent, by presenting an example (due to Kaplansky and Lanski) of a ring satisfying  and not , and an example of a ring satisfying  but not .

We say that an ideal is [****]{} if it is a union of a chain of primes, but is not itself prime. (If ${{\{P_\lam\}}}$ is an ascending chain of primes, then $R/\bigcup P_{\lambda} =
\lim_{\rightarrow} R/P_{\lam}$ is a direct limit of prime rings). The [**[$\PP$-index]{}**]{} of the ring $R$ is the maximal number of non-prime unions of subchains of a chain of prime ideals in $R$ (or infinity if the number is unbounded, see [[Proposition \[PPindex\]]{}]{}). [[Section \[sec:mon\]]{}]{} extends Bergman’s example by showing that the -index of the free (countable) algebra is infinity. A variation of this construction, based on free products, is presented in [[Section \[sec:example2.2\]]{}]{}. After defining the -index in [[Section \[sec:PP\]]{}]{}, in [[Section \[sec:PI\]]{}]{} we discuss PI-rings, showing that the -index does not exceed the PI-class minus one, and this bound is tight. We thank the anonymous referee for careful comments on a previous version of this paper.

Monomial algebras {#sec:mon}
=================

Fix a field $F$. We show that  and  fail in the free algebra (over $F$) by constructing an (ascending) chain of primitive ideals whose union is not semiprime. Let us start with a simpler theme, whose variations have extra properties.

\[Ex1\] Let $R$ be the free algebra in the (noncommuting) variables $x,y$. For each $n$, let $$P_n = {{\left<xx,xyx, xy^2x, \dots, xy^{n-1}x\right>}}.$$ As a monomial ideal, it is enough to check primality on monomials. If $uRu' \sub P_n$ for some words $u,u'$, then in particular $uy^{n}u' \in P_n$, which forces a subword of the form $xy^ix$ (with $i<n$) in $u$ or in $u'$; hence either $u\in P_n$ or $u'\in
P_n$.

On the other hand $\bigcup P_n = (RxR)^2$ which is not semiprime.

This example, due to G. Bergman, appears in [@P Exmpl. 4.2]. Interestingly, primeness is always maintained in the following sense ([@P Lem. 4.1], also due to Bergman): for every countable chain of primes $P_1 \subset P_2 \subset \cdots$ in a ring $R$, the union $\bigcup (P_n[[\zeta]])$ is a prime ideal of the power series ring $R[[\zeta]]$.

Since in [[Example \[Ex1\]]{}]{} $\bigcup P_n = (RxR)^2$, if $Q \normali R$ is a prime containing the union then $x \in Q$ so $R/Q$ is commutative. In particular, a chain of prime ideals starting from the chain $P_1
\subset P_2 \subset \cdots$ has only one . Let us exhibit a (countable) chain providing infinitely many s.

Let $R$ be the free algebra generated by $x,y,z$. For a monomial $w$ we denote by $\deg_yw$ the degree of $w$ with respect to $y$. For $i,n \geq 1$, consider the monomial ideals $$I_{i,n} = RxxR+RxzxR+\cdots+Rxz^{i-1}xR+\sum_{\deg_y w < n} R xz^ixwxz^ix R,$$ which form an ascending chain with respect to the lexicographic order on the indices $(i,n)$, since $xz^ix \in I_{i',n}$ for every $i'>i$. To show that $I_{i,n}$ are prime, suppose that $u,u'$ are monomials such that $u,u' \not \in I_{i,n}$ but $u R u' \sub
I_{i,n}$. Then $u z^i y^n z^i u' \in I_{i,n}$. Since none of the monomials $xz^{i'}x$ ($i'<i$) is a subword of $u$ or $u'$, they are not subwords of $u z^i y^n z^i u'$, forcing $u z^i y^n z^i u'$ to have a subword of the form $xz^ixwxz^ix$ where $\deg_yw < n$. It follows that $z^iy^nz^i$ is a subword of $z^ixwxz^i$, contrary to the degree assumption. Now, for every $i$, $$\bigcup_{n} I_{i,n} =
RxxR+RxzxR+\cdots+Rxz^{i-1}xR+(Rxz^ixR)^2,$$ which contains $(Rxz^ixR)^2$ but not $Rxz^ixR$, so it is not semiprime.

In particular the -index of $R$ (see [[Proposition \[PPindex\]]{}]{}) is infinity. In Section \[sec:PI\] we show that this phenomenon is impossible in PI algebras: there, the number of s in a prime chain is bounded by the PI-class.

The ideals $P_n$ in [[Example \[Ex1\]]{}]{} are in fact primitive. Indeed, Bell and Colak [@Bell] proved that any finitely presented prime monomial algebra is either primitive or PI (also see [@Ok]), and $R/P_n$ contains a free subalgebra, e.g. $k{{\left<xy^n,xy^{n+1},\dots\right>}}$.

The same effect can be achieved by using idempotents. Let $R$ be a binomial algebra, namely a quotient of a free algebra with respect to relations of the form $w_1 = w_2$ (or $w_1 = 0$) where $w_1,w_2$ are monomials. We say that an ideal $I$ in $R$ is monomially prime if $uRu' \sub I$ forces $u \in I$ or $u' \in I$ for every two monomials $u,u'$.

Let $R$ be the free algebra generated by $e,y$ subject to the relation $e^2 = e$. Then every monomial ideal which is monomially prime is prime.

Let $I$ be a monomial ideal which is monomially prime. Let $f,g \in R$ be any two elements such that $fRg \sub I$.

\[Ex3\] Let $R$ be the free algebra in the variables $e,y$, modulo the relation $e^2 = e$. Every monomial has a unique shortest presentation as a word (replacing $e^2$ by $e$ throughout). Ordering monomials first by length and then lexicographically, every element $f$ has an upper monomial $\bar{f}$. Notice that $\overline{f y^n g}
= \bar{f}y^n \bar{g}$.

For each $n$, let $$P_n = {{\left<eye, ey^2e, \dots, ey^{n-1}e\right>}}.$$ To show that $P_n$ is a prime ideal, assume that $f y^n g \in
P_n$. Then $\bar{f}y^n \bar{g} = \overline{f y^n g} \in P_n$, forcing $\bar{f} \in P_n$ or $\bar{g} \in P_n$ as in [[Example \[Ex1\]]{}]{}. The claim follows by induction on the number of monomials.

To show that the ideal $P_n$ is primitive, it is enough by [@LRS] to prove that $e(R/P_n)e$ is a primitive ring. We construct an isomorphism between $e(R/P_n)e$ and the countably generated free algebra $F{{\left<z_0,z_1,\dots\right>}}$ by sending $ey^me$ for $m \geq n$ (which clearly generate a free algebra) to $z_{m-n}$. But the free algebra is primitive (see [@Lam Prop. 11.23]).

On the other hand $\bigcup P_n = ReyReR = ReRyeR$, which contains $(ReyR)^2$ but not $ey$, so is not semiprime.

We say that a ring is [**[uniquely-]{}**]{} if it has a unique minimal prime over every chain of prime ideals. Since the intersection of a descending chain of primes is prime, Zorn’s lemma shows that there are minimal primes over every ideal, in particular over every .

In the topology of the spectrum, a net ${{\{P_{\lambda}\}}}_{\lambda
\in \Lambda}$ of primes converges to a prime $Q$ if and only if $\bigcap_{\lambda \in \Lambda}\bigcup_{\lambda' \geq \lambda}
P_{\lambda'} \subseteq Q$; in particular when ${{\{P_{\lambda}\}}}$ is an ascending chain, $\lim P_\lambda = Q$ if and only if $\bigcup
P_\lambda \sub Q$. Therefore, the spectrum can identify minimal primes over s. It seems that the spectrum cannot distinguish  from uniquely-.

In the examples of this section, there is a unique minimal prime over every . In [[Example \[Ex4\]]{}]{} below the situation is different: the  ideal constructed there is the intersection of two primes containing it.

Prime ideals in free products {#sec:example2.2}
=============================

In [[Example \[Ex1\]]{}]{} there are infinitely many (incomparable) prime ideals lying over the chain. We modify this example, in order to obtain a chain over which there is unique prime. In [[Example \[Ex1\]]{}]{} we considered ideals of the free algebra, which can be written as a free product $F[x] *_F F[y]$. The quotient over the radical of the union over the chain is the “second” component $F[y]$, which we would like to replace by the field $F(y)$. The proof that the ideals are prime is somewhat delicate; we thank the referee for pointing this out.

\[Ex1div\] Let $D$ be the quotient division ring of the free algebra $F{{\left<x,y\right>}}$. Let $R$ be the subalgebra generated by $x$ and the subfield $F(y)$. Extend $\deg_y \co F[y] \ra \N$ to $\deg_y \co {{{F(y)}^{\times}}} \ra \Z$ in the obvious manner. Similarly to the previous example, take the prime ideals $$P_n = {{\left<xax {{\,:\ }}a \in F(y), \deg_y(a) < n\right>}}.$$ Again $P = \bigcup P_n = (RxR)^2$, which is not semiprime. But now $R/\sqrt P = R/{{\left<x\right>}} \cong F(y)$.

Let $F$ be a field and let $A,B$ be $F$-algebras, with given vector space decompositions $A = F \oplus A_0$ and $B = F \oplus B_0$. The free product $A *_F B$ can be viewed as the tensor algebra $T(A_0
\oplus B_0) = F \oplus \bigoplus_{n\geq 1} (A_0\oplus B_0)^{{{\otimes_{}}}n}$, modulo the relations $a {{\otimes_{}}}a' = aa'$ and $b
{{\otimes_{}}}b' = bb'$ for every $a,a' \in A$ and $b,b' \in B$. We will omit the tensor symbol.

Fixing the decomposition $F[x] = F \oplus xF[x]$ and an arbitrary decomposition $B = F \oplus B_0$, we consider ideals of the free product $R = F[x] *_F B$. The tensor algebra is graded by $x$, once we declare that $\deg(b) = 0$ for every $b \in B_0$, and this grading induces a grading on $R$.

Let $W \sub B$ be a vector space containing $F$. We say that $W$ is [**[restricted]{}**]{} if for every finite dimensional subspace $V \sub B$ there is an element $b \in B$ such that $Vb \sub B_0$ and $Vb \not \sub W$; and an element $b' \in B$ such that $b'V \sub B_0$ and $b'V \not \sub W$.

\[main\] The ideal $P = RxWxR$ of $R$ is prime whenever $W \sub B$ is a restricted subspace.

Write $W = F \oplus W_0$ where $W_0 = W \cap B_0$. Let $L'$ be the ideal of $R$ generated by $x^2$. For $n \geq 0$ let us denote the vector spaces $$L_n = B x B_0 x B_0 \cdots x B_0 x B,$$ where the degree with respect to $x$ is $n$; so that $L_0 = B$ and $L_1 = B x B$. Setting $L = \sum_{n\geq 0} L_n$, we have that $R = L' \oplus L$.

Let $P_n = L_n \cap P$; so $P_0 = P_1 = 0$, and for $n \geq 2$, $$P_n = \sum B x B_0 x \cdots x B_0 x W_0 x B_0 x \cdots x B_0 x B$$ where in each summand one of the intermediate entries is $W_0$ and all the others are equal to $B_0$. For example $P_2 = B x W_0 x B$ and $P_3 = B x W_0 x B_0 x B + B x B_0 x W_0 x B$. Now, since $F \sub W$, we have that $L' = RxxR = RxFxR \sub RxWxR = P$, and we can compute: $$\begin{aligned}
P & = & L' + P \\
& = & L' + (L'+L)x W x(L'+L) \\
& = & L' + Lx W x L \\
& = & L' + Lx W_0 x L \\
& = & L ' + \sum_{d,d' \geq 0} L_d x W_0 x L_{d'} \\
& = & L ' + \sum_{n \geq 0} \(\sum_{d+d' = n} L_d x W_0 x L_{d'}\) \\
& = & L' + \sum_{n \geq 2} P_n,\end{aligned}$$ since modulo $L'$, $xBx \equiv xB_0x$ and $xWx \equiv xW_0 x$.

Let $m \geq 1$. As a vector space, $L_m {{\,\cong\,}}B {{\otimes_{}}}B_0 {{\otimes_{}}}\cdots {{\otimes_{}}}B_0 {{\otimes_{}}}B$ with $m+1$ factors. This isomorphism carries $P_m$ to $\sum B {{\otimes_{}}}B_0 {{\otimes_{}}}\cdots {{\otimes_{}}}B_0 {{\otimes_{}}}W_0 {{\otimes_{}}}B_0 {{\otimes_{}}}\cdots {{\otimes_{}}}B_0 {{\otimes_{}}}B$ as above, and there is an isomorphism $$\psi_m \co L_m / P_m \,\longrightarrow\, B {{\otimes_{}}}\overline{B_0} {{\otimes_{}}}\cdots {{\otimes_{}}}\overline{B_0} {{\otimes_{}}}B$$ with $m-1$ factors of the form $\overline{B_0} = B_0/W_0$. The image of $g \in L_m$ in $\overline{L_m} = L_m/P_m$ will be denoted by $\overline{g}$, hoping that no confusion is incurred by the double usage of the over-line.

We need to show that $P$ is prime. Since $L' \sub P$, it suffices to show that if $f,f' \in L$ and $f,f' \not \in P$, then $fBf' \not \sub P$. Furthermore since $R$ is graded with respect to $x$, and $P$ is a homogeneous ideal with respect to this grading, we may assume that $f,f'$ are homogeneous with respect to $x$, so we can write $$f = \sum_i a_{0,i} x a_{1,i} x \cdots x a_{n,i} \in L_n$$ and $$f' = \sum_j a'_{0,j} x a'_{1,j} x \cdots x a'_{n',j} \in
L_{n'}$$ where $a_{0,i},a_{n,i}, a'_{0,j}, a'_{n',j} \in B$ and $a_{t,i}, a'_{t',j} \in B_0$ for $0 < t < n$ and $0 < t' < n'$. Let $V$ be the vector space spanned by all the $a_{n,i}$ and $V'$ the vector space spanned by all the $a'_{0,j}$. We say that $f$ “ends in $V$” and $f'$ “begins in $V'$”. By assumption, there are elements $b,b' \in B$ such that $Vb, b'V' \sub B_0$, while $Vb \not \sub W_0$ and $b'V' \not \sub W_0$.

Since $Vb, b'V' \sub B_0$, we have that $f b x b' f' \in L_{n+n'+1}$. Consider the commutative diagram $$\xymatrix@C=32pt{L_n {{\otimes_{}}}L_{n'} \ar@{->}[d]^{m} \ar@{->}[r]^{\theta} &
     \overline{L_n} {{\otimes_{}}}\overline{L_{n'}} \ar@{->}[r]^(0.24){\psi_n {{\otimes_{}}}\psi_{n'}} \ar@{->}[d]^{\bar{m}} &
     (B {{\otimes_{}}}\overline{B_0} {{\otimes_{}}}\cdots {{\otimes_{}}}\overline{B_0}
     {{\otimes_{}}}B) {{\otimes_{}}}(B {{\otimes_{}}}\overline{B_0} {{\otimes_{}}}\cdots {{\otimes_{}}}\overline{B_0}
     {{\otimes_{}}}B) \ar@{->}[d]
    \\
L_{n+n'+1} \ar@{->}[r] &  \overline{L_{n+n'+1}}  \ar@{->}[r]^(0.4){\psi_{n+n'+1}} & B {{\otimes_{}}}\overline{B_0}  {{\otimes_{}}}\cdots {{\otimes_{}}}\overline{B_0}
     {{\otimes_{}}}B
     }$$ $$\xymatrix{L_n {{\otimes_{}}}L_{n'} \ar@{->}[r]^{m} \ar@{->}[d]^{\theta} & L_{n+n'+1} \ar@{->}[d] \\
     \overline{L_n} {{\otimes_{}}}\overline{L_{n'}} \ar@{->}[d]^{\psi_n {{\otimes_{}}}\psi_{n'}} \ar@{->}[r]^{\bar{m}} & \overline{L_{n+n'+1}}  \ar@{->}[d]^{\psi_{n+n'+1}}
     \\
     (B {{\otimes_{}}}\cdots      {{\otimes_{}}}B) {{\otimes_{}}}(B {{\otimes_{}}}\cdots      {{\otimes_{}}}B) \ar@{->}[r] & B {{\otimes_{}}}\cdots      {{\otimes_{}}}B
     }$$ where the domain of definition of the top-to-bottom maps is the elements in $L_n {{\otimes_{}}}L_{n'}$ such that the left factor ends in $B_0$ and the right factor begins in $B_0$ (and not merely in $B$), and their image. Here, $m(g {{\otimes_{}}}g') = g x g'$ and $\bar{m}(\overline{g} {{\otimes_{}}}\overline{g'}) = \overline{g x g'}$ which is easily checked to be well-defined. The right-most arrow is reduction of the two intermediate factors along $B_0 \ra B_0/W_0$.

$$\xymatrix{L_n {{\otimes_{}}}L_{n'} \ar@{->}[r]^(0.4){m} \ar@{->}[d]^{\theta} & (L'+L_{n+n'+1})/L' \ar@{->}[d]\ar@{->}[r]^(0.61)\cong & L_{n+n'+1} \ar@{->}[d] \\
     \overline{L_n} {{\otimes_{}}}\overline{L_{n'}} \ar@{->}[d]^{\psi_n {{\otimes_{}}}\psi_{n'}} \ar@{->}[r]^(0.3){\bar{m}} & (L'+L_{n+n'+1})/(L'+P_{n+n'+1}) \ar@{->}[r]^(0.7)\cong & \overline{L_{n+n'+1}}  \ar@{->}[d]^{\psi_{n+n'+1}}
     \\
     (B {{\otimes_{}}}\cdots      {{\otimes_{}}}B) {{\otimes_{}}}(B {{\otimes_{}}}\cdots      {{\otimes_{}}}B) \ar@{-->}[rr] & & B {{\otimes_{}}}\cdots      {{\otimes_{}}}B
     }$$ where: $m(g {{\otimes_{}}}g') = g x g' + L'$; $\bar{m}(\overline{g}
{{\otimes_{}}}\overline{g'}) = g x g' + P_{n+n'+1}+L'$ which is easily checked to be well-defined. The isomorphisms to the right are the natural ones, induced by the fact that $L' \cap L_{n+n'+1} = 0$; the bottom arrow is reduction of the two intermediate factors along $B_0 \ra B_0/W_0$, and is defined on the subspace of elements whose two intermediate factors are indeed in $B_0$.

Now consider the element $f b{{\otimes_{}}}b'f' \in L_n {{\otimes_{}}}L_{n'}$, for the given $f \in L_n$ and $f' \in L_{n'}$. By assumption $\theta (f b{{\otimes_{}}}b'f') = \overline{fb} {{\otimes_{}}}\overline{b'f'}$ is non-zero, because $\overline{fb}, \overline{bf'} \neq 0$. Furthermore $\psi_n(\overline{fb})$ ends in $B_0$ and $\psi_{n'}(\overline{b'f'})$ begins in $B_0$, so $\psi_n(\overline{fb}) {{\otimes_{}}}\psi_{n'}(\overline{b'f'})$ is in the domain of definition of the right-most arrow, which takes this element to $\psi_{n+n'+1}(\overline{fbxbf'})$. It remains to show that this element is nonzero. But $\psi_n(\overline{fb})$ does not end in $W_0$, and $\psi_{n'}(\overline{b'f'})$ does not begin in $W_0$; hence their images in $B {{\otimes_{}}}\overline{B_0} {{\otimes_{}}}\cdots {{\otimes_{}}}\overline{B_0} {{\otimes_{}}}B$ are nonzero, and their tensor product, equal to $\psi_{n+n'+1}(\overline{fbxbf'})$, is nonzero as well.

\[3.2\] Let $(K,\nu)$ be a valued field containing $F$ as a field of scalars. For any $m \geq 0$, $W = {{\{k \in K {{\,:\ }}\nu(k) \geq
-m\}}}$ is a restricted subspace of $K$ (where the decomposition $K = F \oplus K_0$ is arbitrary).

We first claim that if $U \sub K$ is an $F$-vector subspace of finite codimension, then ${{\{\nu(u) {{\,:\ }}u \in U\}}}$ is unbounded from below. Indeed, choose any finite dimensional complement $U'$, and notice that ${{\{\nu(u'){{\,:\ }}u' \in U'\}}}$ is bounded from below; so if $\nu(u)$ were bounded for $u \in U$, then $\nu(k)$ would be bounded over the set of $k \in K$. Let $V\neq 0$ be a finite dimensional space. The space of elements $y$ such that $Vy \sub K_0$ has finite codimension, so by the previous argument contains elements of arbitrarily small value, for which $Vy \not \sub W$.

\[almostexample\] Let $(K,\nu)$ be a valued field containing $F$ as a field of scalars. For fixed $m \geq 0$, let $W_m = {{\{k \in K {{\,:\ }}\nu(k) \geq -m\}}}$. Then the ideal generated by $xW_mx$ in $R = F[x] *_F K$ is prime.

With the notation of [[Corollary \[almostexample\]]{}]{}, we now formulate the promised counterexample:

\[6.4\] Let $F$, $K$, $R$ and the $W_m$ be as above. By definition $\bigcup_{m\geq 0} W_m = K$. Let $P_m$ be the (prime) ideal generated by $xW_mx$. Then $P_1 \sub P_2 \sub \cdots$ is a chain of prime ideals in $R$, and $\bigcup_{m\geq 0} RxW_mxR = RxKxR = (RxR)^2$. The radical $RxR$ is thus maximal, as $R/RxR \,{{\,\cong\,}}\, K$.

Matrix constructions {#sec:mat}
====================

This section shows that  and  are independent: the algebra in [[Example \[ExKL\]]{}]{} satisfies  but not , and the algebra in [[Example \[Ex4\]]{}]{} satisfies  but not .

 does not imply 
----------------

As mentioned in the introduction, Kaplansky conjectured that a semiprime ring all of whose prime quotients are von Neumann regular, is regular. Fisher and Snider  [@FS] proved that this is the case if the ring satisfies  (also see [@G Thm. 1.17]), and gave a counterexample which lacks this property, due to Kaplansky and Lanski [@G Example 1.19]. We repeat the example and exhibit, in this ring, an explicit ascending chain of semiprime ideals whose union is not semiprime.

\[ExKL\] (A ring whose prime ideals are maximal, but without ). Let $R$ be the ring of sequences of $2$-by-$2$ matrices which eventually have the form $\begin{pmatrix}
\alpha & \beta_n \\
0 & \alpha
\end{pmatrix}$ in the $n$th place, clearly a semiprime ring. Let $I_n$ be the set of sequences in $R$, which are zero from the $n$th place onward. Clearly $R/I_n {{\,\cong\,}}R$, so the ideals are semiprime. However $\bigcup I_n$ is composed of sequences of matrices which are eventually zero, and $aRa$ is eventually zero for $a = \left(\begin{pmatrix}
0 & 1  \\
0 & 0
\end{pmatrix}, \begin{pmatrix}
0 & 1  \\
0 & 0
\end{pmatrix}, \dots \right)$; hence $R/\bigcup I_n$ is not semiprime. On the other hand by the argument in [@FS], every prime ideal of $R$ is maximal, so there are no infinite chains of primes and  holds trivially.

 does not imply 
----------------

In the rest of this section we investigate  and  for rings of the form $\hat{A} = {{\left(\begin{array}{cc} {A} & {M} \\ {M} & {A} \end{array}\right)}}$ where $A$ is an integral domain and $M \normali A$ is a nonzero ideal. We show that they always satisfy , and give an example which does not have . Clearly $\hat{A}$ is a prime ring. Let us describe the ideals of this ring.

\[whoissp\]

1.  The ideals of $\hat{A}$ have the form $\hat{I} = {{\left(\begin{array}{cc} {I_{11}} & {I_{12}} \\ {I_{21}} & {I_{22}} \end{array}\right)}}$, where for $1 \le i,j\le 2$, $I_{ij} \normal A$ (not necessarily proper), $I_{ii'} \sub M$, and $MI_{ij} \sub I_{i'j} \cap I_{ij'}$ (where $1'
    = 2$ and $2' = 1$).

2.  The semiprime ideals of $\hat{A}$ are of the form $${{\left(\begin{array}{cc} {I} & {M\cap I} \\ {M \cap I'} & {I'} \end{array}\right)}},$$ where $I,I'$ are semiprime ideals of $A$, and $M \cap I' = M \cap I$.

<!-- -->

1.  This is well known and easy.

2.  Write $A_{ij} = A$ if $i = j$ and $A_{ij} = M$ otherwise. We are given an ideal $\hat{I} \normali \hat{A}$ which thus can be written as $\hat{I} = {{\left(\begin{array}{cc} {I_{11}} & {I_{12}} \\ {I_{21}} & {I_{22}} \end{array}\right)}}$, satisfying the conditions of (1). Clearly $\hat{I}$ is semiprime if and only if for every $a_{ij} \in A_{ij}$,

    ( $\sum_{j,k} A_{jk} a_{ij}a_{k\ell} \sub I_{i\ell}$ for every $i,\ell$) implies ( $a_{i\ell} \in I_{i\ell}$ for every $i,\ell$).

    Assuming that this condition holds, fix $i,j$ and choose $a_{k\ell}
    = 0$ for every $(k,\ell) \neq (i,j)$; then

    $\COND$ for $a_{ij} \in A_{ij}$, $A_{ji} a_{ij}^2 \sub I_{ij}$ implies $a_{ij} \in I_{ij}$.

    On the other hand if Condition $\COND$ holds and $\sum_{j,k} A_{jk}
    a_{ij}a_{k\ell} \sub I_{i\ell}$ for every $i,\ell$, then in particular $A_{ji} a_{ij}^2 \sub I_{ij}$, so each $a_{i\ell} \in
    I_{i\ell}$. We conclude that $\hat{I}$ is semiprime if and only if $\COND$ holds for every $i,j$.

    We claim that $\COND$ is equivalent to $I_{ii}$ being semiprime in $A$ with $$M \cap I_{11} \sub I_{ij},\qquad \forall i\neq j.$$ Indeed, for $i = j$, condition $\COND$ requires that the $I_{ii}$ are semiprime in $A$. Assuming this, the condition is “for $a_{ij} \in A_{ij}$, $Ma_{ij}^2
    \sub I_{ij}$ implies $a_{ij} \in I_{ij}$ for $i \neq j$.” In light of the standing assumption that $a_{ij} \in A_{ij} = M$, we claim that this is equivalent to $M \cap I_{11} \sub I_{ij}$. Indeed, for every $b \in M$, $Mb^2 \sub I_{ij}$ iff $b \in I_{11}$ (Proof: If $b^2 M \sub I_{ij}$ then $b^4 \in (bM)^2 = b^2M\cdot M \sub I_{ij}M \sub I_{11}$, so $b \in I_{11}$. On the other hand if $b \in M \cap I_{11}$ then $b^2 M \sub MI_{11} \sub I_{ij}$), so the condition becomes “for $b \in M$, $b \in I_{11}$ implies $b \in I_{ij}$ for $i\neq j$”, as claimed.

    We have shown that $\hat{I}$ is semiprime if and only if $I_{11},I_{22}$ are semiprime in $A$ and $M \cap I_{11} \sub I_{12}
    \cap I_{21}$.

    Now assume that $I_{ii}$ are semiprime, and that $M \cap I_{11} \sub
    I_{12} \cap I_{21}$. Denote the idealizer of an ideal $I$ by $(I:M)
    = {{\{x \in A {{\,:\ }}xM \sub I\}}}$, and notice that $M \cap (I:M)
    = M \cap I$ when $I$ is semiprime. But $I_{12} M \sub I_{11}$, implying $$I_{12}\sub M \cap (I_{11}:M) = M \cap I_{11} \sub I_{12},$$ so $I_{12} = M \cap I_{11}$ and likewise $I_{21} = M\cap I_{11}$. By symmetry $I_{12} = M \cap I_{22}$ as well, so $M \cap I_{11} = M
    \cap I_{22}$.

Recall that for ideals $I,M$ in a commutative ring $A$, $(I:M) =
{{\{a \in A{{\,:\ }}aM \sub I\}}}$. The properties of $(I:M)$ are well known, and we review the ones that we need.

\[compactness\] Let $A$ be a commutative ring, with an ideal $M \normali A$. For any semiprime ideal $I \normali A$, $$M \cap (I:M) = M \cap I.$$

The inclusion $I \sub (I:M)$ is trivial. In the other direction let $x \in M \cap (I:M)$, then $x^2 \in xM \sub I$, so $x \in I$ by assumption.

\[prep\] Let $A$ be a commutative ring with an ideal $M \normali A$. Let $I,J \sub M$ be ideals of $A$ such that $I$ is semiprime, $M I\sub J$ and $M J\sub I$.

1.  \[x1\] For $b \in M$, $b^2 M \sub J$ iff $b \in I$.

2.  \[x4\] The condition “For every $b \in M$, if $b^2M \sub
    J$, then $b \in I$" is equivalent to $M \cap I \sub J$.

3.  \[x3\] Let $I'$ be another semiprime ideal such that $MI' \sub J$ and $MJ \sub I'$. Then $M \cap I = M \cap I'$.

<!-- -->

1.  If $b^2 M \sub J$ then $(bM)^2 \sub JM \sub I$ so $bM \sub I$ and $b \in M \cap (I:M) \sub I$ by [[Proposition \[compactness\]]{}]{}. On the other hand if $b \in M \cap I$ then $b^2 \in MI \sub J$ and $b^2M \sub J$.

2.  This is [[(\[x1\])]{}]{}.

3.  Indeed, by [[Proposition \[compactness\]]{}]{} and [[(\[x1\])]{}]{}, $b \in M \cap I$ iff $b \in M \cap (I:M)$ iff ($b \in M$ and $b^2M \sub J$) iff $b \in M \cap (I':M)$ iff $b \in M \cap I'$.

\[main0\] The ring $\hat{A}$ satisfies .

By [[Proposition \[whoissp\]]{}]{} every chain of semiprime ideals $T_1 \sub T_2 \sub
\cdots$ in $\hat{A}$ has the form $T_n =
{{\left(\begin{array}{cc} {I_n} & {J_n} \\ {J_n} & {I'_n} \end{array}\right)}}$, $I_n$ and $I'_n$ are ascending chains of semiprime ideals of $A$, and $J_n = M \cap I_n = M \cap I_n'$. The union of this chain is ${{\left(\begin{array}{cc} {\bigcup I_n} & {L} \\ {L} & {\bigcup I_n'} \end{array}\right)}}$ where $L = M \cap \bigcup I_n = M \cap \bigcup I_n'$, which is semiprime.

Using the description of the semiprime ideals, it is not difficult to obtain the following.

\[main0.2\]

1.  \[Y5\] The prime ideals of $\hat{A}$ are ${{\left(\begin{array}{cc} {J} & {M} \\ {M} & {A} \end{array}\right)}}$ and ${{\left(\begin{array}{cc} {A} & {M} \\ {M} & {J} \end{array}\right)}}$ for prime ideals $J \normali A$ containing $M$, and $I^{0} = {{\left(\begin{array}{cc} {I} & {M \cap I} \\ {M \cap I} & {I} \end{array}\right)}}$ for prime ideals $I \normali A$ not containing $M$.

2.  The  ideals of $\hat{A}$ are of the form ${{\left(\begin{array}{cc} {M'} & {M} \\ {M} & {M'} \end{array}\right)}}$ where $M' \normali A$ is a prime ideal containing $M$, which can be written as a union of an ascending chain of primes not containing $M$.

Notation as in [[Proposition \[whoissp\]]{}]{}(1), we are done by [[Proposition \[whoissp\]]{}]{}(2) unless some ${I_{11}} = A$, in which case we must have (1). If the chain of primes includes an ideal with $A$ in one of the corners, then every higher term has the same form, and the union is determined by the union of the ideals in the other corner, which is prime since $A$ is commutative. We thus assume that the chain has the form $I_1^0 \sub I_2^0 \sub \cdots$ where $I_1 \sub I_2 \sub \cdots$ are primes in $A$, not containing $M$. The union is clearly $\hat I^0$ where $\hat I = \bigcup I_n$ is prime, and $\hat I^0$ is not a prime iff $M \sub \hat I$.

Suppose $A$ is a prime PI-ring, integral over its center $C$. In [@BV] it is shown that $A$ satisfies the properties Lying Over and Going Up over $C$, which gives a correspondence of chains of primes between the two rings. The next example shows that the union of chains in not preserved.

\[Ex4\] Let $F$ be a field. Let $\hat{A} = {{\left(\begin{array}{cc} {A} & {M} \\ {M} & {A} \end{array}\right)}}$, where $A =
F[\lam_1,\dots]$ is the ring of polynomials in countably many variables $\lam_1,\lam_2,\dots$, and $M = {{\left<\lam_1,\dots\right>}}$. Clearly $\hat{A} \subset M_2(A)$ is integral over $A$. Choose $I_n
= {{\left<\lam_1,\dots,\lam_n\right>}}$. Then $T_n =
{{\left(\begin{array}{cc} {I_n} & {I_n} \\ {I_n} & {I_n} \end{array}\right)}} \normali \hat{A}$ form an ascending chain of primes by [[Proposition \[main0.2\]]{}]{}.[[(\[Y5\])]{}]{}, but their union $\bigcup T_n = {{\left(\begin{array}{cc} {M} & {M} \\ {M} & {M} \end{array}\right)}}$ is obviously not prime. Therefore $\hat{A}$ satisfies  ([[Proposition \[main0\]]{}]{}) but not . Furthermore $\hat{A}/\bigcup T_n\cong A/M\times A/M$ which has two minimal ideals, so uniquely- also fails.

The $\PP$-index {#sec:PP}
===============

Let $\tilde P = {{\{P_{\alpha}\}}}$ be a chain of prime ideals in a ring $R$. The number of non-prime unions of subchains of $\tilde P$ is called the [**[index]{}**]{} of $\tilde P$ (either finite or infinite). The [**[-index]{}**]{} of $R$, denoted by $\PP(R)$, is the supremum of the indices of all chains of primes in $R$.

For any ring $R$, $\PP(R) = \sup \PP(R/P)$ where $P\normali R$ ranges over the prime ideals.

By definition $\PP(R/P)$ is the supremum of indices of chains of primes containing $P$. Therefore, the supremum on the right-hand side is the supremum of indices of chains of primes containing some prime $P$, but any chain contains its own intersection, so this supremum is by definition $\PP(R)$.

If $n = \PP(R)$ then $R$ has a chain of $n$ . It is not clear if the converse holds. For example if ${{\{P_\lam\}}}$ and ${{\{P_{\lam}'\}}}$ are ascending chains of primes such that $\bigcup P_{\lam} \subset \bigcup P_{\lam}'$ are not primes, does it follow that there is a chain of primes with at least two non-prime unions?

We claim that:

\[PPindex\] For any ring $R$, $$\PPind(R) = \left\{ \begin{array}{cl} 0 & i\!f\mbox{\, $R$\, has the property\, \PP} \\ \sup_I \PPind(R/I)+1 & \mbox{otherwise} \end{array} \right.$$ where the supremum is taken over the  ideals of $R$ (when they exist).

Indeed, $\PPind(R) = 0$ if and only if there are no s (which are non-prime by definition), if and only if $R$ satisfies $\PP$. Now assume $R$ does not satisfy . Consider the set ${{\{\PPind(R/I)\}}}$ ranging over the s $I$. If this set is unbounded, then clearly $\PPind(R) = \infty$. Otherwise, take a $I$ such that $n = \PPind(R/I)$ is maximal among the -indices of the quotients. If $J_1/I \subset \cdots \subset J_n/I$ are s in a chain of primes in $R/I$, then $I \subset J_1
\subset \cdots \subset J_n$ are in a chain in $R$. On the other hand if $J_0 \subset J_1 \subset \cdots \subset J_n$ are in a chain in $R$, then $\PPind(R/J_0) \geq n$.

For example, $\PPind(R) = 1$ if and only if the union of an ascending chain of primes starting from a  ideal is necessarily prime.

The property $\PP$ in PI-rings {#sec:PI}
==============================

In this section we show that for PI-rings, the $\PP$-index is bounded by the PI-class.

\[Azumaya\_CP\] Any Azumaya algebra satisfies (and ).

Let $A$ be an Azumaya algebra over a commutative ring $C$. There is a 1:1 correspondence between ideals of $A$ and the ideals of $C$, preserving inclusion, primality and semiprimality. The claim follows since the center satisfies  (and ).

Recall that by Posner’s theorem ([@RowenPI]), a prime PI-ring $R$ is [*representable*]{}, namely embeddable in a matrix algebra ${{\operatorname{M}_{n}}}(C)$ over a commutative ring $C$. The minimal such $n$ is the PI-class of $R$, denoted $\PIdeg(R)$.

Although PI-rings do not necessarily satisfy the property , we show that the PI-class bounds the extent in which  may fail. We are now ready for our main positive result about PI-rings.

\[mainPI\] Let $R$ be a (prime) PI-ring. Then $\PPind(R) < \PIdeg(R)$.

Let $R$ be a prime PI-ring of PI-class $n$. If the PI-class is $1$ then $R$ is commutative, and has $\PPind(R) = 0$. We continue by induction on $n$. Let $$0 = P_0 \subset P_1 \subset \cdots$$ be an ascending chain of primes, and assume that $\bigcup P_n$ is not a prime ideal. Let $Q \supset \bigcup P_n$ be a prime ideal. We want to prove that the PI-class of $R/Q$ is smaller than that of $R$.

Assume otherwise. Let $g_n$ be a central polynomial for $n \times n$ matrices (see [@RowenPI p. 26]). Since $\PIdeg(R/Q) = n$, there is a value $\gamma \neq 0$ of $g_n$ in the center of $R$, which is not in $Q$. Since the center is a domain we can consider the localization $A[\gamma^{-1}]$ (see [@R-RT Section 2.12]), which is Azumaya by Rowen’s version of the Artin-Procesi Theorem [@RowenPI Theorem 1.8.48], since $1$ is a value of $g_n$ on this algebra. But then the union of $$0
\subset P_1[\gamma^{-1}] \subset P_2[\gamma^{-1}] \subset \cdots$$ is prime by [[Proposition \[Azumaya\_CP\]]{}]{}, so $\bigcup P_n$ is prime as well, contrary to assumption.

We now show the bound is tight. Notice that the ring constructed in [[Example \[Ex4\]]{}]{} has PI-class $2$ and is not (and thus has $\PPind(R) = 1$). Let us generalize this.

\[PI\_non\_CP\] Let $A_{(n)}=F\left[\lambda_i^{(j)} : 1 \leq j < n,\, i =
1,2,\dots\right]$. Let $M_n = 0$ and, for $j = n-1,n-2,\dots,1$, take $M_j=M_{j+1}+\left<\lambda_1^{(j)},\lambda_2^{(j)},\cdots\right>$, so that $0 = M_n \subset M_{n-1} \subset \cdots \subset M_1 \normali A_{(n)}$. Let $e_{ij}$ denote the matrix units of the matrix algebra over $A_{(n)}$. Let $J_{(n)} =
\sum_{i,j} e_{ij}M_{\max(i,j)-1}$, $S_{(n)} = \sum e_{ii} A_{(n)}$. Let $$R_{(n)} = J_{(n)}+S_{(n)} =\begin{pmatrix}
A_{(n)} & M_1 & M_2 & \cdots & M_{n-1} \\
M_1 & A_{(n)} & M_2 & \cdots & M_{n-1} \\
M_2 & M_2 & A_{(n)} & \cdots & M_{n-1} \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
M_{n-1} & M_{n-1} & M_{n-1} & \cdots & A_{(n)} \\
\end{pmatrix}.$$

Clearly $S_{(n)} J_{(n)}, \, J_{(n)} S_{(n)} \sub J_{(n)}$, and $S_{(n)}S_{(n)}= S_{(n)}$. Moreover $M_{k}M_{\ell} \sub
M_{\max{{\{k,\ell\}}}}$ for every $k,\ell$, so that $J_{(n)}J_{(n)}
\sub J_{(n)}$. It follows that $R_{(n)} = J_{(n)}+S_{(n)}$ is a ring. The ring of central fractions of $R_{(n)}$ is the simple ring ${{\operatorname{M}_{n}}}(\operatorname{q}(A_{(n)}))$, so $R_{(n)}$ is prime, of PI-class $n$.

When $n = 2$ we obtain the ring of [[Example \[Ex4\]]{}]{}, so $\PPind(R_{(2)}) =
1$. For arbitrary $n$, consider the chain $I_1 \subset I_2 \subset
\cdots$ of ideals of $A$ defined by $I_i =
{{\left<\lam_1^{(n-1)},\dots,\lam_i^{(n-1)}\right>}}$; thus $\bigcup I_i =
M_{n-1}$. Let $\tilde{I}_i = {{\operatorname{M}_{n}}}(I_i)$. Each ideal $\tilde{I}_i$ is prime (again by central fractions), and their union is the set of matrices over $M_{n-1}$. The quotient ring is therefore $R_{(n)}/\bigcup_i \tilde{I}_i \,{{\,\cong\,}}\, R_{(n-1)} \times
A_{(n-1)}$, which is not prime, and $\PPind(R_{(n-1)}) = n-2$ by induction. We conclude that $\PPind(R_{(n)}) = n-1$.

\[end\]

1.  \[AffinePIhasACCsp\] Although PI-rings do not necessarily satisfy (see [[Example \[Ex4\]]{}]{}), affine PI-rings over commutative Noetherian rings do satisfy ACC on semiprime ideals, and in particular are  and (by Schelter’s theorem, [@RowenPI Thm 4.4.16]).

2.  PI-rings of finite Gel’fand-Kirillov dimension satisfy ACC on primes, since a prime PI-ring is Goldie, and then every prime ideal contains a regular element which reduces the dimension.

3.  On the other hand, we have examples of a (non-affine) locally nilpotent monomial algebra, which does not satisfy , and of affine algebras of $\GK=2$ which do not satisfy (details will appear elsewhere.) In both cases the -index is uncountable.

4.  Graded affine algebras of quadratic growth and graded affine domains of cubic growth have finite Krull dimension and so satisfy , [@BS; @GLSZ].

Further ideas
=============

(This section will be omitted from the submitted paper).

Normalizing extensions
----------------------

The most general version:

If $S$ is a subring of $R$ and there is a finite $S$-module $M$ such that $aR \subseteq Ma$ for all $a\in R$, then the P-index of $R$ does not the P-index of exceed $S$.

This should cover all the cases that we know. If what I wrote is true, I suggest that Beeri write down a short exposition of this, together with the examples about PI-rings and submit it on the ArXiv in parallel to our article, so that nobody from the outside jumps in with the result (which is not difficult to discover given the thrust of the article).

\[first.1\] Every ring $R$ which is a finite normalizing extension satisfies .

Write $R = \sum_{i=1}^t Cr_i$ where $R$ normalizes $C$ (namely $aC = Ca$ for every $a \in R$). Suppose $P_1 \subset P_2 \subset \cdots$ is a chain of prime ideals, with $P = \cup P_i$. If $a,b \in R$ with $$\sum C ar_i b = \sum aCr_i b = aRb \subseteq P$$ then there is $n$ such that $ar_i b \in P_n$ for $1 \le i \le t,$ implying $aRb = \sum aCr_i b \subseteq P_n$, and thus $a \in P_n$ or $b \in P_n$.

\[first+\] Let $R$ be a normalizing extension of $C$, and $Q \normali R$ prime. Then $Q \cap C$ is prime.

Indeed, assume $aCb \sub Q \cap C$. Then $aRb = aCRb = aCbR \sub Q$ so $a \in Q$ or $b \in Q$.

Assume $R$ is a finite normalizing extension of a ring $C$. Assume $R$ is PI. Then $C$ (and $R$) satisfy .

The claim for $R$ is [[Proposition \[first.1\]]{}]{}.

Let $P_n$ be an ascending chain of ideals in $C$. By the LO for PI rings, there is a chain $Q_n$ in $R$ such that each $P_n$ is minimal prime over $Q_n \cap C$, and since $Q_n \cap C$ is prime by the remark, $P_n = Q_n \cap C$. But $\bigcup Q_n$ is prime since $R$ is a finite normalizing extension ([[Proposition \[first.1\]]{}]{}), so $\bigcup P_n = \bigcup (Q_n \cap C) = (\bigcup Q_n) \cap C$ is prime by the [[Remark \[first+\]]{}]{}.

Comments on [[Section \[sec:example2.2\]]{}]{}
----------------------------------------------

((This is only needed if more details are required above))

-   In \[Hungerford, 1967\], he defines the free product for augmented algebras. So the definition depends on the augmentation, and one needs to be extra careful about this, because $F(y)$ has no natural augmentation. It seems best to take $F[x] = F \oplus xF[x]$ and $F(y) = F \oplus {\operatorname{span}}_F {{\{\frac{f}{g} {{\,:\ }}(f,g) = 1, fg \not \in {{{F}^{\times}}}\}}}$. Finally $F[x] *_F F(y)$ is well defined.

-   We could extend $\nu$ to $F[x] *_F F(y)$ by taking $\nu()$ of a monomial to be the maximal $\nu(a)$ for $xax$ within the monomial; and $\nu()$ of a sum of monomials to be the minimal $\nu()$ of a monomial. Since $\nu(xa_1xa_2\cdots a_kx) = \max(\nu(a_1),\dots,\nu(a_k))$, we have that $\nu(xa_1xa_2\cdots a_kx) \geq -n$ iff for some $a_i$ we have $\nu(a_i) \geq -n$, iff some factor $xa_i x \in P_n$. Which is fine. In this definition of $\nu()$ on monomials we ignore initial or terminal coefficients from $F(y)$. So for example $\nu(axbx) = \nu(b)$, regardless of $a$. Likewise $\nu(\sum w_i) = \min(\nu(w_i))$ where $w_i$ are monomials; so $\nu(\sum w_i) \geq -n$ iff $\nu(w_i) \geq -n$ for each $i$, iff $w_i \in P_n$ for each $i$; which is fine again, because we want $f \in P_n$ iff $\nu(f) \geq -n$ for every $f \in R$. Of course there is still a problem of defining everything; what are monomials, etc.

-   Write $F[x] = F \oplus A$ and $F(y) = F\oplus B$. As in \[Hungerford, 1967\], define $T_n$ and $T_n'$ to be the tensor products $A {{\otimes_{}}}B {{\otimes_{}}}\cdots$ and $B {{\otimes_{}}}A {{\otimes_{}}}\cdots$ of length $n$, respectively. The free product is, by definition, $F[x]*_FF(y) = F\oplus \bigoplus(T_n \oplus T_n')$. We suppress the tensor notating for elements. Since $A = \bigoplus Fx^i$, every simple tensor can be written as a sum of [**[monomials]{}**]{}, which have the form $x^{i_0}a_1x^{i_1}a_2x^{i_3}a_3\cdots x^{i_{t-1}}a_{t}x^{i_t}$, where $t \geq 0$, $i_0, i_t \geq 0$, $i_1,\dots,i_{t-1} > 0$ and $a_1,\dots,a_t \in B$. The components $a_i$ of a monomial are uniquely determined up to (balanced) multiplication by a nonzero scalar. The presentation of an element as a sum of monomials is not unique, but the sum of monomials of any given [**[signature]{}**]{} $(t;i_0,\dots,i_t)$ – is. Every monomial can be uniquely written in the form $w = x^{j_0}a_1 x a_2 x a_3\cdots x a_{\ell}x^{j_\ell}$, where $\ell \geq 0$, $j_0, j_t \in {{\{0,1\}}}$, and $a_1,\dots,a_\ell \in B \cup F$ (the number of appearances of $x$ in $w$ is $\ell+j_0+j_{\ell}-1$ if $\ell > 0$, or $j_0$ for $\ell = 0$).

Let us now compute $P_n$. By definition, $$P_n = \begin{cases} \sum_{a \in B, \nu(a) \geq -n} R xax R & n < 0 \\ P_n = \sum_{a \in B, \nu(a) \geq -n} R xax R + RxxR & n \geq 0\end{cases}.$$

$P_n$ is monomial (in the sense that if it contains an element, it contains all of its monomials).

A monomial is in $P_n$ iff it has $xax$ as a subword where $\nu(a) \geq -n$.

We extend the definition of $\nu \co F(y) \ra \Z$ to $F[x]*_F F(y)$ as follows. For a monomial, we set $$\nu(x^{j_0}a_1 x a_2 x a_3\cdots x a_{\ell}x^{j_\ell}) = \begin{cases}\max{{\{\nu(a_2),\dots,\nu(a_{\ell-1})\}}} & j_0 = 0, j_{\ell} = 0
\\ \max{{\{\nu(a_2),\dots,\nu(a_{\ell})\}}} & j_0 = 0, j_{\ell} = 1
\\ \max{{\{\nu(a_1),\dots,\nu(a_{\ell-1})\}}} & j_0 = 1, j_{\ell} = 0
\\ \max{{\{\nu(a_1),\dots,\nu(a_{\ell})\}}} & j_0 = 1, j_{\ell} = 1
\end{cases}$$

Proofs for Section \[sec:mat\]
------------------------------

Recall that for ideals $I,M$ in a commutative ring, $(I:M) = {{\{x \in A{{\,:\ }}xM \sub I\}}}$.

\[compactness\] Let $A$ be a commutative ring, with an ideal $M \normali A$. For any semiprime $I \normali A$, $$M \cap (I:M) = M \cap I.$$

The inclusion $I \sub (I:M)$ is trivial. In the other direction let $x \in M \cap (I:M)$, then $x^2 \in xM \sub I$, so $x \in I$ by assumption.

\[prep\] Let $A$ be a commutative ring with an ideal $M \normali A$. Let $I,J \sub M$ be ideals of $A$ such that $I$ is semiprime, $M I\sub J$ and $M J\sub I$.

1.  \[x1\] For $b \in M$, $b^2 M \sub J$ iff $b \in I$.

2.  \[x4\] The condition “For every $b \in M$, if $b^2M \sub J$, then $b \in J$" is equivalent to $M \cap I \sub J$.

3.  \[x3\] Let $I'$ be another semiprime ideal such that $MI' \sub J$ and $MJ \sub I'$. Then $M \cap I = M \cap I'$.

<!-- -->

1.  If $b^2 M \sub J$ then $(bM)^2 \sub JM \sub I$ so $bM \sub I$ and $b \in M \cap (I:M) \sub I$ by [[Proposition \[compactness\]]{}]{}. On the other hand if $b \in M \cap I$ then $b^2 \in MI \sub J$ and $b^2M \sub J$.

2.  This is [[(\[x1\])]{}]{}.

3.  Indeed, by [[Proposition \[compactness\]]{}]{} and [[(\[x1\])]{}]{}, $b \in M \cap I$ iff $b \in M \cap (I:M)$ iff ($b \in M$ and $b^2M \sub J$) iff $b \in M \cap (I':M)$ iff $b \in M \cap I'$.

\[main0FI\]

1.  \[Y1\] The ideals of $\hat{A}$ are the subsets $\hat{I} = {{\left(\begin{array}{cc} {I_{11}} & {I_{12}} \\ {I_{21}} & {I_{22}} \end{array}\right)}}$, where for every $i,j$, $I_{ij} \normal A$ (not necessarily proper), $I_{ii'} \sub M$, and $MI_{ij} \sub I_{i'j} \cap I_{ij'}$ (where $1' = 2$ and $2' = 1$).

2.  \[Y3\] $\hat{I}$ is semiprime iff

    -   $I_{11}$ and $I_{22}$ are semiprime, and

    -   $M \cap I_{11} \sub I_{12} \cap I_{21}$;

    iff

    -   $I_{11}$ and $I_{22}$ are semiprime, and

    -   $I_{12} = I_{21} = M \cap I_{11} = M \cap I_{22}$.

3.  \[Y33\] The semiprime ideals of $\hat{A}$ are of the form ${{\left(\begin{array}{cc} {I} & {M\cap I} \\ {M \cap I} & {I'} \end{array}\right)}}$ where $I,I'$ are semiprime, and $M \cap I' = M \cap I$.

4.  \[Y7\] The ring $\hat{A}$ satisfies .

<!-- -->

1.  Follows by computing the principal ideals generated by monomial matrices.

2.  Write $A_{ij} = A$ if $i = j$ and $A_{ij} = M$ otherwise. Clearly $\hat{I}$ is semiprime if for every $a_{11} \in A_{11}$,..., $a_{22} \in A_{22}$, if $(\sum a_{ij}e_{ij})(\sum A_{rs}e_{rs})(\sum a_{k\ell}e_{k\ell}) \sub \sum I_{i\ell}e_{i\ell}$ then $a_{ij} \in I_{ij}$ for each $i,j$. In other words, if

    (for every $i,\ell$, $\sum_{j,k} A_{jk} a_{ij}a_{k\ell} \sub I_{i\ell}$) implies (for every $i,\ell$, $a_{i\ell} \in I_{i\ell}$).

    Assuming this is the case, fix $i,j$ and choose $a_{k\ell} = 0$ for every $(k,\ell) \neq (i,j)$; then

    $\COND \quad A_{ji} a_{ij}^2 \sub I_{ij}$ implies $a_{ij} \in I_{ij}$.

    On the other hand if Condition $\COND$ holds and for every $i,\ell$, $\sum_{j,k} A_{jk} a_{ij}a_{k\ell} \sub I_{i\ell}$, then in particular $A_{ji} a_{ij}^2 \sub I_{ij}$ so each $a_{i\ell} \in I_{i\ell}$. Therefore, $\hat{I}$ is semiprime iff $\COND$ holds for every $i,j$.

    Let us interpret Condition $\COND$. For $i = j$ it requires that $I_{ii}$ are semiprime. Assuming this is the case, for $i \neq j$ the condition is “$Ma_{ij}^2 \in I_{ij}$ implies $a_{ij} \in I_{ij}$”, which in light of the standing assumption that $a_{ij} \in A_{ij}$, is equivalent by [[Lemma \[prep\]]{}]{}.[[(\[x4\])]{}]{} to $M \cap I_{11} \sub I_{ij}$.

    Now assume that $I_{ii}$ are semiprime, and that $M \cap I_{11} \sub I_{12} \cap I_{21}$. Since $I_{12} M \sub I_{11}$, we have that $I_{12}\sub M \cap (I_{11}:M) = M \cap I_{11} \sub I_{12}$ so $I_{12} = M \cap I_{11}$ and likewise $I_{21} = M\cap I_{11}$. Finally the equality $M \cap I_{11} = M \cap I_{22}$ is [[Lemma \[prep\]]{}]{}.[[(\[x3\])]{}]{}.

3.  Clear from [[(\[Y3\])]{}]{}, noting that $(M \cap I)M \sub I,I'$ and $MI,MI' \sub M\cap I$ by [[(\[Y1\])]{}]{}.

4.  By [[(\[Y3\])]{}]{} every chain of semiprimes $T_1 \sub T_2 \sub \cdots$ in $\hat{A}$ has the form $T_n = {{\left(\begin{array}{cc} {I_n} & {J_n} \\ {J_n} & {I'_n} \end{array}\right)}}$, $I_n$ and $I'_n$ are ascending chains of semiprimes, and $J_n = M \cap I_n = M \cap I_n'$. The union of this chain is ${{\left(\begin{array}{cc} {\bigcup I_n} & {L} \\ {L} & {\bigcup I_n'} \end{array}\right)}}$ where $L = M \cap \bigcup I_n = M \cap \bigcup I_n'$. Again by [[(\[Y3\])]{}]{} the union is semiprime.

\[main0.1\]

1.  \[Y4\] An ideal $\hat{I}$ (as in [[Proposition \[main0\]]{}]{}.[[(\[Y1\])]{}]{}) is prime iff

    -   $I_{11}$ and $I_{22}$ are prime,

    -   $I_{12} = I_{21} = M \cap I_{11} = M \cap I_{22}$.

    -   If $M \sub I_{ii}$ for some $i$, then $I_{jj} = A$ for some $j$.

    -   $I_{11} \sub I_{22}$ or vise versa.

2.  \[Y5FI\] The prime ideals of $\hat{A}$ are ${{\left(\begin{array}{cc} {J} & {M} \\ {M} & {A} \end{array}\right)}}$ and ${{\left(\begin{array}{cc} {A} & {M} \\ {M} & {J} \end{array}\right)}}$ for prime ideals $J \normali A$ containing $M$, and $I^{0} = {{\left(\begin{array}{cc} {I} & {M \cap I} \\ {M \cap I} & {I} \end{array}\right)}}$ for prime ideals $I \normali A$ not containing $M$.

3.  $\hat{A}$ is prime.

4.  The almost prime ideals of $\hat{A}$ are of the form ${{\left(\begin{array}{cc} {M'} & {M} \\ {M} & {M'} \end{array}\right)}}$ where $M' \normali A$ is a prime containing $M$, which can be presented as a union over an ascending chain of primes not containing $M$.

<!-- -->

1.  As in [[Proposition \[main0\]]{}]{}.[[(\[Y3\])]{}]{}, $\hat{I}$ is prime if for every $a_{11},a_{11}' \in A_{11}$,..., $a_{22},a_{22}' \in A_{22}$, the following condition holds:

    ($\forall i,j,k,\ell$, $A_{jk} a_{ij}a'_{k\ell} \sub I_{i\ell}$) implies ($\forall i,\ell$, $a_{i\ell} \in I_{i\ell}$ or $\forall i,\ell$, $a_{i\ell}' \in I_{i\ell}$).

    Assuming this is the case, fix $i,j$ and choose $a_{k\ell} = a'_{k\ell} = 0$ for every $(k,\ell) \neq (i,j)$; then

    $\COND' \quad A_{ji} a_{ij}a_{ij}' \sub I_{ij}$ implies $a_{ij} \in I_{ij}$ or $a'_{ij} \in I_{ij}$,

    and in particular $I_{ii}$ is prime (for each $i$). Since $\hat{I}$ is semiprime, we also have that $I_{12} = I_{21} = M \cap I_{ii}$, as claimed. If, moreover, $M \sub I_{ii}$ for some $i$, then $I_{12} = I_{21} = M$ and $\hat{A}/\hat{I} = (A/I_{11}) \times (A/I_{22})$ so one of the components is zero. Finally suppose there are $a \in I_{22}$ and $a' \in I_{11}$ such that $a \not \in I_{11}$ and $a' \not \in I_{22}$. Then $(ae_{11})\hat{A}(a'e_{22}) = aa'Me_{12} \sub aI_{11}Me_{12}\sub aI_{12}e_{12} \sub \hat{I}$, whereas $ae_{11}, a'e_{22} \not \in \hat{I}$, a contradiction. This proves the fourth condition.

    On the other hand, assume the four conditions hold. If $M \sub I_{ii}$ then we may assume $I_{ii} = A$ and $\hat{A}/\hat{I} = A/I_{i'i'}$ is prime. So we assume $M \nsub I_{ii}$ for $i = 1,2$.

    Suppose $\forall i,j,k,\ell$, $A_{jk} a_{ij}a'_{k\ell} \sub I_{i\ell}$, but for some (fixed) $i,j,k,\ell$, $a_{ij} \not \in I_{ij}$ and $a'_{k\ell} \not \in I_{k\ell}$. We have that $A_{jk} a_{ij}a'_{k\ell} \sub I_{i\ell}$, contained in the prime ideals $I_{ii}, I_{\ell\ell}$ (whether or not $i = \ell$). But since $M \nsub I_{ii}, I_{\ell\ell}$, we must have $a_{ij}a'_{k\ell} \in I_{ii}, I_{\ell\ell}$.

    We claim that $a_{ij} \not \in I_{ii}$. Indeed if $j = i$ this is the assumption, while if $j \neq i$ the claim follows since $a_{ij} \not \in I_{ij} = M \cap I_{ii}$. Likewise $a'_{k\ell} \not \in I_{\ell\ell}$. Therefore $a_{ij} \in I_{\ell\ell}$ and $a_{k\ell}' \in I_{ii}$.

    Now, if $j \neq i$, then $a_{ij} \in M$, so $a_{ij} \in M \cap I_{\ell\ell} = I_{ij}$, a contradiction, so we may assume $i = j$, and likewise $k = \ell$. Since $a_{ij} \in I_{k\ell}$ but $a_{ij} \not \in I_{ij}$, we must have $i \neq k$, but $a_{ij} \in I_{k \ell}\setminus I_{ii}$ and $a_{k\ell}' \in I_{ij} \setminus I_{k\ell}$, contradicting the fourth condition.

2.  Following the conditions given in [[(\[Y4\])]{}]{}, assume $I_{11} \sub I_{22}$. Then $I_{22}M \sub I_{22} \cap M = I_{11} \cap M \sub I_{11}$, so since $I_{11}$ is prime there are two options: either $I_{22} \sub I_{11}$, in which case $I_{22} = I_{11}$; or $M \sub I_{11}$, in which case $I_{22} = A$ by the third condition.

3.  Zero is a prime ideal of $\hat{A}$ by [[(\[Y4\])]{}]{}.

4.  If the chain of primes includes an ideal with $A$ in one of the corners, then every higher term has the same form, and the union is determined by the union of entries in the other corner, which is prime since $A$ is commutative. We thus assume the chain has the form $I_1^0 \sub I_2^0 \sub \cdots$ where $I_1 \sub I_2 \sub \cdots$ are primes in $A$, not containing $M$. The union is clearly $M'^0$ where $M' = \bigcup I_n$ is prime, and by [[(\[Y5FI\])]{}]{}, $M'^0$ is not a prime iff $M \sub M'$.

Chain of primitive ideals
-------------------------

It would be nice to find an example of a chain of primitive ideals with prime non primitive union.

This one does not work (the ideals $P_{ij}$ are not prime): In $F{{\left<e,y\right>}}$ with $e$ idempotent define $I_{i,j} = {{\left<y^iey^j-y^jey^i\right>}}$. Then, taking lexicographic order we can make a chain $P_{ij} = \sum_{(i',j')\leq (i,j)} I_{i'j'}$. I didn’t check details but the idea is that the corner is primitive (is it?) modulo every ideal in the chain but isomorphic to a polynomial ring (hence non primitive) modulo the union.

P-index of a module
-------------------

(BTW, what about P-index of a module?)

up and down
-----------

We might also want to consider how the property lifts and descends.

GU
--

If $R \subset T$ satisfies GU and every prime ideal of $T$ intersects $R$ nontrivially, then descends from $T$ to $R$. I think this holds for finite normalizing extensions, maybe finite extensions in general by work of Letzter.

Compactness
-----------

(One could define ‘compactness’ with respect to semiprime ideals: $M$ is compact if whenever it is contained in a union of a chain of semiprime ideals, it is contained in one of the ideals. Perhaps our ring $\hat{A}$ has non-semiprime chains iff $M$ is not compact?)

From the section on matrices
----------------------------

\[main0.2FI\] Let $A$ be an integral domain, and $M$ an ideal. Let $\hat{A} = {{\left(\begin{array}{cc} {A} & {M} \\ {M} & {A} \end{array}\right)}}$.

1.  \[z1\] The left ideals of $\hat{A}$ are the subsets $\hat{I} = {{\left(\begin{array}{cc} {I_{11}} & {I_{12}} \\ {I_{21}} & {I_{22}} \end{array}\right)}}$, where for every $i,j$, $I_{ij} \normal A$ (not necessarily proper), $I_{ii'} \sub M$, and $MI_{ij} \sub I_{i'j}$.

2.  The maximal left ideals of $\hat{A}$ are of the form ${{\left(\begin{array}{cc} {I} & {M} \\ {M \cap I} & {A} \end{array}\right)}}$ and ${{\left(\begin{array}{cc} {A} & {M \cap I} \\ {M} & {I} \end{array}\right)}}$ for maximal ideals $I \normali A$.

3.  In light of [[(\[z1\])]{}]{}, a maximal left ideal has the form ${{\left(\begin{array}{cc} {I} & {M} \\ {J} & {A} \end{array}\right)}}$ for $I \normali A$ and $J \sub M$; or have the symmetric form. If $I$ is not maximal and $I \subset I'$, this is contained in ${{\left(\begin{array}{cc} {I'} & {M} \\ {J+MI'} & {A} \end{array}\right)}}$, so we assume $I$ is maximal. But then $J^2 \sub MJ \sub I$ implies $J \sub I$ and since $J \sub M$, ${{\left(\begin{array}{cc} {I} & {M} \\ {J} & {A} \end{array}\right)}} \sub {{\left(\begin{array}{cc} {I} & {M} \\ {I \cap M} & {A} \end{array}\right)}}$, so we have an equality. Since ideals of the form $  {{\left(\begin{array}{cc} {I} & {M} \\ {I \cap M} & {A} \end{array}\right)}}$ do not contain each other, and every maximal has this form, they are all maximal.

4.  The Jacobson radical of $\hat{A}$ is ${{\left(\begin{array}{cc} {J(A)} & {J(A) \cap M} \\ {J(A) \cap M} & {J(A)} \end{array}\right)}}$.

5.  $\hat{A}$ is semiprimitive iff $A$ is semiprimitive.

6.  \[z5\] $\hat{A}$ is not primitive.

7.  Every maximal left ideal contains a (nonzero) two-sided ideal: for ${{\left(\begin{array}{cc} {I} & {M \cap I} \\ {M\cap I} & {I} \end{array}\right)}} \sub {{\left(\begin{array}{cc} {I} & {M} \\ {M\cap I} & {A} \end{array}\right)}}$ is an ideal.

8.  The only primitive ideals of $\hat{A}$ are the maximal ones, which are of the forms: ${{\left(\begin{array}{cc} {J} & {M} \\ {M} & {A} \end{array}\right)}}$ and ${{\left(\begin{array}{cc} {A} & {M} \\ {M} & {J} \end{array}\right)}}$ for $J \normali A$ maximal; and ${{\left(\begin{array}{cc} {I} & {M \cap I} \\ {M\cap I} & {I} \end{array}\right)}}$ where $I+M = A$.

9.  Primitive PI-rings are simple. But here is a direct proof: For prime ideals of the form ${{\left(\begin{array}{cc} {J} & {M} \\ {M} & {A} \end{array}\right)}}$ or ${{\left(\begin{array}{cc} {A} & {M} \\ {M} & {J} \end{array}\right)}}$, the quotient is commutative, so the ideal is primitive if and only if it is maximal. For the prime ideals of the other form, ${{\left(\begin{array}{cc} {I} & {M \cap I} \\ {M \cap I} & {I} \end{array}\right)}}$ where $I$ is a prime not containing $M$, the quotient is isomorphic to ${{\left(\begin{array}{cc} {A/I} & {(M+I)/I} \\ {(M+I)/I} & {A/I} \end{array}\right)}}$. If $M+I = A$ this is a simple ring so the ideal is maximal; otherwise, it is of the same structure as $\hat{A}$ (since $0 \neq (M+I)/I \normali A/I$), so it is not primitive by [[(\[z5\])]{}]{}.

Questions:

-   How do semiprime quotients look like?

-   Which semiprime ideals are semiprimitive?

-   Is the union over a semiprimitive chain necessarily semiprimitive?

Stability
---------

Question: assuming $A$ is , does $A[\lam]$ have ?

\*Noncomm. Hilbert basis thm? A is CP (or ACC(primes)). Is the same true for A\[X\]?

Proof that commutative has 
---------------------------

Given an ascending chain of prime ideals in a commutatuve ring, the union is prime too (indeed, an ideal in a commutative ring is prime if and only if $ab\in I$ implies either $a\in I$ or $b\in I$, so if $ab\in \bigcup{Q_i}$ then for some $i$ we have $ab\in Q_i$, which is prime, hence we may assume $a\in Q_i\subseteq \bigcup{Q_i}$).

Finite GK-dim
-------------

We pose the following question.

Is there an affine ring of finite  which does not have ?

Note that all quotient rings $R/P_n$ considered in our examples \[Ex1\],\[Ex3\],\[PI\_non\_CP\] are large - in the sense that they have infinite . On the other hand, many division rings (obviously ) contain a free subalgebra. It is reasonable therefore to ask whether there are natural restrictions on a ring in order to consist of a chain of prime ideals with non-prime union.

\*An example of an affine algebra with GKdim=2 which has infinite P-index (based on an earlier construction in locally finite nil algebras. This is not \*that\* short, so might not suite; on the other hand, this could not hold an independent paper but it would be a shame to burry it.)

locally primeness
-----------------

Here are some properties of ideals $I$ ($\exists a_i$ marks finitely many, $P$ is always a prime). The rightmost one is called ‘locally prime’ in [@TK]. (I didn’t give the reverse implications much thought).

$$\xymatrix{
{} & \mbox{prime} \ar@{=>}[d] & {} & {} & {} \\
{} & \bigcup_{\mbox{chain}}\mbox{primes} \ar@{=>}[dr] & {} & {} & {} \\
{} & {}  & {{\mbox{$\begin{array}{c} \forall a_i \in I\\ \exists P \sub I: a_i \in P \end{array}$}}} \ar@{=>}[dl] \ar@{=>}[dr] & {} & {} \\
{} & \bigcup \mbox{primes} = {{\mbox{$\begin{array}{c} \forall a \in I\\ \exists P \sub I: a \in P \end{array}$}}} \ar@{=>}[dr] & {} & {{\mbox{$\begin{array}{c} \forall a_i \in I, b_j \not \in I\\ \exists P: a_i \in P, b_j \not \in P \end{array}$}}} \ar@{=>}[dl] \ar@/_25pt/@{..>}[lluuu]|{quasi-commutative} & {} \\
{} & {}  & {{\mbox{$\begin{array}{c} \forall a \in I, b \not \in I\\ \exists P: a \in P, b \not \in P \end{array}$}}} & {} & {} \\
}$$ In [@TK] they give examples of rings which are not quasi-commutative. We should check their examples for our ‘union of primes is prime’.

A ring is ‘spectral’ if $\operatorname{spec}R$ is a spectral topological space (a property characterized by being the spectrum of some commutative ring, but which also has topological characterization). The space of locally prime ideals is spectral. So quasi-commutative implies spectral. Again, how does  fit in?

By [@TK Rem. 15], $R$ is quasy-commutative iff:

QC: For every $x,y$ there are $r_1,\dots,r_n$ such that if $x,y \not \in P$ then some $xr_iy \not \in P$.

Equivalently,

QC: $\forall x,y$ there are $r_i$ such that if ${{\left<xr_1y,\dots,xr_ny\right>}} \sub P$ then $x\in P$ or $y \in P$.

Weaker version of quasi-commutative
-----------------------------------

What about the following \*weaker\* condition:

QCF: $\forall x,y$ $\exists$ f.g. ideal $I$ such that for every $P$, $xIy \sub P$ implies $x\in P$ or $y \in P$

What does it imply?

Is our chain condition the only way to violate quasi-commutativity?
-------------------------------------------------------------------

Let us then phrase ‘not quasi-commutative’:

$\sim{}QC$: For some $x,y$, for every finite set $r_1,\dots,r_n$, there is a prime $P$ such that all $xr_iy \in P$ but $x,y \not \in P$.

Reformulation:

$\sim{}QC$: For some $x,y$, in every quotient of the form $\bar{R} = R/{{\left<xr_1y,\dots,xr_ny\right>}}$, there is a prime $\bar{P}$ such that $x,y \not \in \bar{P}$.

The challenge: prove that if $\sim{}QC$, then there is a chain of primes whose union is not prime.

Notice that if $x,y \not \in P$ then there is some $r$ such that $xry \not \in P$.

Let’s see what the condition gives, for some $x,y$.

-   There is $P_1$ such that $x,y \not \in P_1$.

    So there is some $r_1$ such that $xr_1y \not \in P_1$.

-   There is $P_2$ such that $xr_1y \in P_2$ but $x,y \not \in P_2$. \[So $P_2 \not \sub P_1$\]

    So there is some $r_2$ such that $xr_2y \not \in P_2$.

-   There is $P_3$ such that $xr_2y \in P_3$ but $x,y \not \in P_3$. \[So $P_3 \not \sub P_1 \cup P_2$\]

    So there is some $r_3$ such that $xr_3y \not \in P_3$.

-   There is $P_4$ such that $xr_3y \in P_4$ but $x,y \not \in P_4$. \[So $P_4 \not \sub P_1 \cup P_2 \cup P_3$\]

    So there is some $r_4$ such that $xr_4y \not \in P_4$.

-   ...

The key issue is the following property:

(\*\*) For every prime $P$ and $a\not \in P$, $x,y \not \in {{\left<P,a\right>}}$, there is a prime $Q \supset P$ such that $a \in Q$ and $x,y \not \in Q$.

If this is granted, we can construct our chain. Note that passing to the quotient, what we ask is this (in a prime ring):

(\*\*) For every $a \neq 0$, $x,y \not \in {{\left<a\right>}}$, there is a prime $Q$ such that $a \in Q$ and $x,y \not \in Q$.

Passing to the quotient modulo ${{\left<a\right>}}$, what we ask is this (in an arbitrary ring):

(\*\*) For every $x,y \neq 0$, there is a prime $Q$ such that $x,y \not \in Q$.

(Take $Q$ maximal with respect to not containing $x,y$? the monoid generated by $x,y$?)

Compare and consider the condition:

(\*) For every $x \neq 0$, there is a prime $Q$ such that $x \not \in Q$.

(obviously, even in a commutative ring this would require $x$ to be non-nil. So are we asking too much?)

Primes are f.g.
---------------

\* Consider the class of rings where primes are f.g.

Neocommutativity
----------------

In [@FS] it is mentioned that ACC(ideals) implies ‘neocommutativity’ (the product of f.g. ideals is f.g.; the notion is due to Kaplansky, unpublished).

LO, GU etc.
-----------

\* Consider comparison properties such as LO, GU, GD, INC with respect to .

-index
------

\* Concerning the “CP dimension”: we want $\PPind(A) \leq \PPind(A/I) + \PPind(I)$. What about $\PPind(A\otimes B)$?

\* We can ask questions like catenarity on the -index.

regular rings
-------------

(Remark: it seems that ${{\operatorname{End}}}(V)$ (vN-regular) have all ideals prime and principal, so in particular  holds. )

Finite module over center
-------------------------

\* Question: suppose $A$ is a finite module over its center. Does it satisfy ?

Strongly primes
---------------

\* The union of a chain of strongly prime is always strongly prime.

An example to check
-------------------

(This example does not hold).

\*An example (of something) in finite GK. Take the ring spanned - as a module - by all subwords of $xy^2xy^4xy^8\cdots$ and obtain a chain of ideals by ’shaving’ the word from the left to the right. These ideals are \*not\* prime, but modulo each one of them the ideal generated by $x$ is not nilpotent, whereas modulo the union, $xRx=0$.

\*An attempt in finite GKdim: Define ’Agata numbers’ - a sequence $\{a_1,a_2,...\}$ with incredible growth. Let $R$ be the quotient of the free algebra obtained by setting $x^2=0$ and also $xy^mx=0$ for all $m$ not an Agata number. What about GKdimR? Let $I_n$ be the ideal of $R$ generated by all $xy^{a_i}x$ for $1 \leq i \leq n$. They are prime but their union is not even semiprime.

ACC(primes)
-----------

Does an affine  ring necessarily satisfy ACC(primes)?

Affine algebras
---------------

\*Is it true that any PI ring finitely generated as a ring over some commutative ring is ? If not, there exists a nice, quite surprising example of a ring which is a finite module both over a commutative ring and over a non-CP ring. This is obtained by a combination of a theorem of Schelter and Noether normalization lemma (which does not exist yet). This also means that if we prove that it is possible to raise and lower CP between finite modules then we prove that PI f.g. over comm. ring is CP.

Crossed products
----------------

Use properties of normalizing extensions.

Generic flatness
----------------

?

Further ideas for another article
=================================

Chains of rings
---------------

It is also possible to discuss chains of rings. It is well known that the union of a chain of prime (semiprime, simple) rings is prime (semiprime, simple respectively). The case of primitive rings is slightly different, however.

Consider the Leavitt path algebras $L(E_{\alpha})$ of the graphs $E_{\alpha}$, where $E_{\alpha}$, for $\alpha$ an ordinal, is the graph whose vertices are the elements of $\alpha$, with an edge from $\beta$ to $\gamma$ iff $\beta<\gamma$. As in [@Leavitt], for all countable ordinals $\alpha$, the algebra $L(E_\alpha)$ is primitive but the algebra $E_{\aleph_1}$ is not primitive (as it does not satisfy the countable separation property), resulting in a chain of primitive rings with non-primitive union. However, the chain is uncountable and the rings are not finitely generated.

Is there a countable chain of affine primitive rings with non-primitive union?

A general theory for monomials
------------------------------

Let $F$ be a field, $S$ an algebra over $F$, and $R = F[x] * S$ the amalgamated product (which is the co-product in the category of algebras).

The notion of a monomial is clearly defined. It is easy to see that if $I$ is generated by monomials then if $f \in I$, every monomial of $f$ is in $I$.

There is also the notion of a leading monomial, but we might need some assumption on $S$ for this to satisfy the conditions below. We denote by $\overline{g}$ the leading monomial. Assume for any monomial $w$ and elements $f,g \in R$, we have that $\overline{fwg} = \overline{f}w \overline{g}$. We claim that if $P$ is ‘monomially prime’ (if $uRv \sub P$ then $u\in P$ or $v \in P$, for any two monomials $u,v$), then $P$ is prime. Indeed, assume $f R g \sub P$. Then for every monomial $w$ we have that $fwg \in P$ so $\overline{f}w\overline{g} = \overline{fwg} \in P$, which proves $\overline{f}R\overline{g} \sub P$, so by assumption $\overline{f} \in P$ or $\overline{g} \in P$, and the claim on $f,g$ follows by induction on the number of monomials.

((In order to produce the required technique, we need to define a leading monomial in a way that $\overline{fwg} = \overline{f}w \overline{g}$.))

[99]{} J. Bell and P. Colak, [*[Primitivity of finitely presented monomial algebras]{}*]{}, Journal of Pure and Applied Algebra [**[213]{}**]{}, (2009), 1299–1305.

J. Bell, A. Smoktunowicz, [*[The prime spectrum of algebras of quadratic growth]{}*]{}, J. Algebra [**[319]{}**]{}(1) (2008), 414–431.

A. Braun and N. Vonessen, [*[Integrality for PI-rings]{}*]{}, [**[151]{}**]{}(1), (1992), 39–79.

J.W. Fisher and R.L. Snider, [*[On the von Neumann regularity of rings with regular prime factor rings]{}*]{}, Pacific J. Math. [**[54]{}**]{}(1), (1974), 135–144.

K.R. Goodearl, “Von Neumann regular rings”, Monographs and Studies in Math. [**[4]{}**]{}, Pitman, London, 1979.

Abrams, Bell and Rangswami on Leavitt path algebras.

The Prime Spectrum and the Extended Prime Spectrum of Noncommutative Rings, Igor Klep and Marcus Tressl, (2007). \[http://personalpages.manchester.ac.uk/staff/Marcus.Tressl/papers/Klep,Tressl-ThePrimeSpectrumandtheExtendedPrimeSpectrumofNoncommutativeRings.pdf\],

T.Y. Lam, “A First Course in Noncommutative Rings”, Springer, 2001.

C. Lanski, R. Resco and L. Small, [*[On the primitivity of prime rings]{}*]{}, J. Algebra [**[59]{}**]{}(2), (1979), 395-398.

J. Okniński, [*[Structure of prime finitely presented monomial algebras]{}*]{}, J. Algebra [**[320]{}**]{}, (2008), 3199–3205.

D. Passman, [*[Prime Ideals in Normalizing Extensions]{}*]{}, J. Algebra [**[73]{}**]{} (1981), 556–572.

L.H. Rowen, “Polynomial Identities in Ring Theory”, 1980.

L.H. Rowen, “Ring Theory”, Acedemic Press, 1988.

S. Sarussi, [*[Maximal covers of chains of prime ideals]{}*]{}, preprint, [[`arXiv:1301.4340`]{}]{}.

B. Greenfeld, A. Leroy, A. Smoktunowicz and M. Ziembowski, [*[Chains of Prime Ideals and Primitivity of $\mathbb Z$-Graded Algebras]{}*]{}, Alg. Rep. Theory [**[18]{}**]{}(3), 777–800, (2015).

[^1]: For simplicity we deal only with countable chains throughout the paper, but the arguments are general.
