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) $?


Thanks in advance.



Codimension in zero and positive characteristic
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 $\langle x\_1,x\_2 \rangle$.


**Edit**. I should make clear that, in the above, by "codimension" of an ideal, I mean what is usually called "height", i.e., the minimal height of a prime ideal that contains the ideal (assuming the ideal is not the entire ring).


