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Let $f:X\to Y$ be a morphism of schemes and let $Z\subset Y$ be its scheme-theoretic image. If $T$ is any other scheme, consider the induced morphism $g=f\times 1_T:X\times T\to Y\times T$.

Question. Is the scheme theoretic image of $g$ equal to $Z\times T$?

Let $j:Z\times T\to Y\times T$ be the natural closed immersion. What I need to check is that the kernel of $$p:\mathcal O_{Y\times T}\to g_\ast\mathcal O_{X\times T},$$ which is the ideal of the scheme-theoretic image of $g$, agrees with the kernel of $$q:\mathcal O_{Y\times T}\to j_\ast\mathcal O_{Z\times T},$$ which is the ideal defining $Z\times T\subset Y\times T$.

To simplify matters, let us assume $X$ is reduced or noetherian, so that the scheme-theoretic image can be computed affine-locally. So, given an affine open subset $R=\textrm{Spec }B\subset Y\times T$, I am left with checking that the kernel of $$B\to \Gamma\bigl(R,g_\ast\mathcal O_{X\times T}\bigr)=\mathcal O_{X\times T}\bigl(g^{-1}(R)\bigr)$$ agrees with the kernel of $$B\to \Gamma\bigl(R\cap (Z\times T),j_\ast \mathcal O_{Z\times T}\bigr)=\mathcal O_{Z\times T}\bigl(R\cap (Z\times T)\bigr).$$ Now it is probably the time to use that $Z$ is the scheme-theoretic image of $f$, but I do not see exactly how this would help. Does anyone have any suggestion? Thanks in advance!

Armando j18eos
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Brenin
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  • Have you thought about the case where everything is affine? It seems to me that you are asking whether tensoring preserves kernels, or thereabouts. – Zhen Lin Feb 14 '16 at 14:13
  • I do not understand your comment. But here (http://math.stackexchange.com/questions/876229/scheme-theoretic-image-behaves-nicely-with-composition-base-change) probably my question is addressed. What do you think? In that case I will delete my question. – Brenin Feb 15 '16 at 08:24
  • @Brenin Zhen Lin basically said the same thing as that linked post. Namely, the scheme-theoretic image of $\text{Spec}(A)\to\text{Spec}(B)$ is $V(\ker(B\to A))$. So, to have this maintained under base change you'd need flat base change. – Alex Youcis Feb 19 '16 at 11:34
  • Yes, but isn't the projection $Y\times T\to Y$ flat? Then the ideal of $Z\subset Y$ should pullback to the ideal of $Z\times T\subset Y\times T$. – Brenin Feb 19 '16 at 12:29

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