# Codimension in zero and positive characteristic

Let $F_0,\ldots,F_m\in\mathbb{Z}[x_0,\ldots,x_n]$ be polynomials with integer coefficients and let $p$ be a prime integer. Consider the two ideals: $$I_0:=(F_0,\ldots,F_m)\subset \mathbb{Q}[x_0,\ldots,x_n]$$ and $$I_p:=(F_0\,\mathrm{mod}\,p,\ldots,F_m\,\mathrm{mod}\,p)\subset \mathbb{Z}/(p)[x_0,\ldots,x_n].$$

Is it true that $\mathrm{codim}(I_p) \leq \mathrm{codim}(I_0)$?

That depends on what you mean by "codimension", but I think most interpretations give an answer of "no". For instance, let $m=n$ be $2$, let $F_0$ be $p(px_0-1)$, let $F_1$ be $(px_0 - 1)x_1$, and let $F_2$ be $(px_0-1)x_2$. Then the codimension of $\langle F_0,F_1,F_2 \rangle$ in $\mathbb{Q}[x_0,x_1,x_2]$ equals $1$; indeed, the ideal is just the principal prime ideal $\langle px_0 - 1 \rangle$. However, the codimension of $\langle F_0,F_1,F_2 \rangle$ in $(\mathbb{Z}/p\mathbb{Z})[x_0,x_1,x_2]$ equals $2$; indeed, the ideal is just the prime ideal $\langle x_1,x_2 \rangle$.
• Is this a properness over $Spec\ \mathbb Z$ issue? Dec 13 '14 at 14:04
• @AllenKnutson: "Is this a properness over $\text{Spec} \mathbb{Z}$ issue?" Yes, precisely. The example above only works because the induced morphism from the zero scheme of $F_0$, $F_1$, $F_2$ to $\text{Spec} \mathbb{Z}$ is not proper. Dec 13 '14 at 14:08
• With "codimension" I just mean "height". Thank you. Is it possible to find a counterexample when the $F_i$'s are homogeneous polynomials?
• @gio: If the polynomials $F_i$ are homogeneous, then the semicontinuity theorem holds. This was also Allen Knutson's point. I suggest you look up in a textbook "semicontinuity" (for dimension of fibers) and "properness". Since semicontinuity is a result on the domain of the morphism, and you want a result on the target, you need to use properness. Dec 14 '14 at 12:17