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Let $X$ be an integral Noetherian scheme. If $Z \subset \mathbb{P}^n_{\mathbb{Z}}$ is a closed subscheme, under what conditions can we say that the codimension of $X \times Z \subset \mathbb{P}^n_X$ is equal to the codimension of $Z \subset \mathbb{P}^n_{\mathbb{Z}}$? I am really interested in the case that $Z$ has codimension one, i.e. $Z$ is a Weil divisor on $\mathbb{P}^n_{\mathbb{Z}}$.

Edit: this is not always true. If the subscheme is of an "arithmetic" nature, e.g. $\mathbb{P}^n_{\mathbb{F}_p} \subset \mathbb{P}^n_{\mathbb{Z}}$, then the result depends very much on the characteristic in which $X$ lives.

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Exercise III.12.6 in Hartshorne says that $\mathrm{Pic}(X\times \mathbb{P}^n)\cong\mathrm{Pic}(X)\times\mathrm{Pic}(\mathbb{P}^n)$ if we assume $h^1(X,\mathcal{O}_X)=0$. Thus, if $\mathcal{J}$ is the ideal sheaf of $Z$, the ideal sheaf $\mathcal{O}_X\times\mathcal{J}$ corresponds to $X\times Z$, which means that its codimension is the same as the codimension of $Z$.

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Thanks, but this seems a bit backwards. It is my fault for not giving enough background to my question. The question occurred to me while doing Exercise II.6.1, and I was hoping there is a more straightforward argument using commutative algebra. Also, I am talking about Weil divisors, which only correspond to invertible sheaves under some strong assumptions (locally factorial I think). – Justin Campbell Dec 12 '11 at 20:49

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