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For every polynomial with complex coefficients, the fundamental theorem of algebra guarantees the existence of complex numbers which happen to be roots of it. But is this everything? i.e. is the converse true? Suppose you give me a complex number: can I find a polynomial whose roots include it?

Trivially, if you allow for complex coefficients, the answer is yes: given a set of {w_n} in C, just build the polynomial up out of the factors (z - w_n).

But what if I restrict my coefficients to smaller sets? If I restrict to integer coefficients, the answer is already no, even on just the real line: real transcendental numbers exist.

What about other sets? What if I have rationals coefficients? Gaussian integers? How do I go about investigating these sorts of questions?

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Every complex number is the root of a polynomial with real coefficients: just consider $(x-w)(x-\bar{w})$ (exercise: why are all the coefficients here real)? As to the rest of your questions: there's no difference between rational coefficients and integer ones (just 'clear denominators' by multiplying each term of the polynomial by the LCM of all of the denominators) and, it turns out, no difference between gaussian integers and 'normal' integers (consider multiplying the entire polynomial by its conjugate; can you see how terms pair up to leave coefficients real?). – Steven Stadnicki Oct 24 '12 at 0:04
@StevenStadnicki, perhaps move your comment to an answer? – lhf Oct 24 '12 at 0:26
Indeed, if the coefficients are algebraic numbers, the roots will be algebraic numbers. – Gerry Myerson Oct 24 '12 at 1:20
@lhf Done, and expanded on in the process! – Steven Stadnicki Oct 24 '12 at 1:21
up vote 9 down vote accepted


As you noted, the answer is trivial for complex coefficients: if you want a polynomial with the root $w$, then $P(z) = z-w$ will do. What about a polynomial with real coefficients? There too, it's relatively straightforward; the polynomial $P(x) = (x-w)(x-\bar{w}) = x^2-2\mathcal{R}(w)x+\left|w\right|^2$ has real coefficients and has roots $w$ and $\bar{w}$.

What if I have rational coefficients?

The answer here is no, for the exact same reason as for integer coefficients; the two cases are equivalent. To see why, consider as an example the polynomial $P(x) = \frac{1}{5}x^3-\frac{7}{3}x^2+\frac{5}{4}x-\frac{17}{8}$. Then $P(x)$ has the same roots as the polynomial $Q(x) = 120P(x) = 24x^3-280x^2+150x-255$ obtained by multiplying $P$ by the LCM of the denominators of all its coefficients.

Gaussian integers?

Again no, by two separate results: one relatively straightforward, but one more subtle but much more far-reaching. The straightforward way of seeing this is again by using the complex conjugate: if $P(x)$ is a polynomial with Gaussian integers for coefficients, then $Q(x) = P(x)\bar{P}(x)$ has all of $P$'s roots among its roots, but has (real) integer coefficients; for instance, if $P(x) = x^2+(2+i)x+3i$, then $\bar{P}(x) = x^2+(2-i)x-3i$ and $P(x)\bar{P}(x) = x^4+(2+i)x^3+(2-i)x^3+3ix^2-3ix^2+(2+i)(2-i)x^2 + 3i(2+i)x -3i(2-i)x +(3i)^2 = x^4+4x^3+5x^2-6x-9$.

The more abstract reason, though, is that the algebraic numbers (defined as 'roots of polynomials with integer coefficients') are algebraically closed: any number that's the root of a polynomial with algebraic numbers for coefficients (which includes all the cases you mentioned above, as well as many more) is algebraic itself (and thus already the root of a polynomial with integer coefficients). This can be shown by, essentially, clearing coefficients: for an example of how this works, suppose we want to find an integer polynomial whose roots include the roots of the polynomial $P(x) = x^2-\sqrt{2}x+1$. Well, first rewrite the equation $P(x) = 0$ by moving the $\sqrt{2}$ term to the left, giving $\sqrt{2} x = x^2+1$; now by squaring both sides (which may introduce new roots, but won't cost the roots we have) we get $2x^2 = \left(x^2+1\right)^2 = x^4+2x^2+1$, or $x^4+1=0$; thus, any root of $P(x) = x^2-\sqrt{2}x+1$ is also a root of $Q(x) = x^4+1$. A (much!) more complicated but similar procedure will allow all algebraic coefficients to be 'cleared' from a polynomial, giving a polynomial with only integer coefficients and roots a superset of the original polynomial's roots. Taking the contrapositive of this indicates that no transcendental number can be the root of any polynomial with algebraic coefficients, whether real or complex.

In fact, it's possible to go one step farther still: we can say with confidence that, for instance, not every number is the root of a polynomial with coefficients any algebraic expression of whole numbers, $e$ and $\pi$! This time the argument is more abstract still: there are only countably many '$\pi e$-algebraic' expressions, so there are only countably many polynomials with $\pi e$-algebraic expressions for coefficients; and since each polynomial has finitely many roots, there are only countably many such roots. But because there are uncountably many real numbers, our set of roots must exclude some (in fact, almost all) real numbers. This argument shows that the reals have infinite (in fact, uncountably infinite) transcendence degree over the rationals; no finite (or countable) number of additional reals that are made available as coefficients will let you 'capture' all the real numbers (or complex numbers, of course) as roots of polynomials.

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This is fantastic! Thank you very much. I think I see the argument for the Gaussian integers reducing to (real) integers: split P(x) into polies in both real and imaginary parts, i.e. P(x) = q(x) + i r(x), then multiply out (q(x) + i r(x))(q(x) - i r(x)) = (q(x))^2 + (r(x))^2, which has only real integer coeffs. The more abstract material is exactly what I'm interested in, and this gives me plenty to look into. Thanks again! – AndrewG Oct 24 '12 at 2:11

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