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Consider two morphisms $T\to Z$ and $Y\to Z$ of varieties over an algebraically closed field $k$ where $Z$ is an affine space. If $Y\to Z$ is flat, is it always true that the fiber product of $T\times_Z Y$ is a complete intersection in $T\times_k Y$?

The motivation comes from an argument of Knop in his paper On the Set of Orbits for a Borel Subgroup. Let $G$ be a reductive group with Lie algebra $\mathfrak{g}$, and $X$ be a spherical variety on which $G$ acts. In the proof of Lemma 6.5, where $Y=\mathfrak{t}$ is a Cartan subalgebra of $\mathfrak{g}$, $Z=\mathfrak{t}/W$ is the quotient by the Weyl group, and $T=T^\ast X$ is the cotangent bundle over $X$, the above statement is claimed for $$T^\ast X\times_{\mathfrak{t}/W}\mathfrak{t}\subset T^\ast X\times_k \mathfrak{t}.$$ EDIT: I have edited to include a flatness assumption to avoid simple counterexamples, as pointed out by @Alexander Braverman.

I would like to understand this point a bit better. Is this a standard argument, and if so is there a good reference?

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    $\begingroup$ The answer is obviously "no" in general. For example, consider the case when $Y$ is one point in $Z$. Then $T\times _Z Y$ is the preimage of this point in $T$ which is not necessarily a complete intersection in $T$ in general. However, I think that the statemet is true if one of the maps is flat (and the map ${\mathfrak t}\to {\mathfrak t}/W$ is flat. $\endgroup$ – Alexander Braverman Oct 4 '18 at 21:23
  • $\begingroup$ Yes of course, thanks for the simple counterexample. I have edited the question to include a flatness assumption. $\endgroup$ – WSL Oct 4 '18 at 21:33
  • $\begingroup$ If $Y=Z$ then $T\times_ZY=T$ so you need $T$ to be a CI in T\times_kY$. That is, you need a point to be CI in $Y$, so it is necessary that $Y$ be smooth. $\endgroup$ – inkspot Oct 4 '18 at 21:53
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Let $Z=\mathbb{A}^n$. The map $T\times _ZY\rightarrow T$ is the pull back over $T$ of $Y\rightarrow Z$, hence it is flat, of relative dimension $\dim(Y)-n$. Therefore $$\dim(T\times _ZY)=\dim(T)+(\dim(Y)-n)=\dim(T\times Y)-n\, .$$ On the other hand, $T\times _ZY$ is defined in $T\times Y$ by the $n$ equations $f_i-g_i=0$, where $(f_1,\ldots ,f_n)$ and $(g_1,\ldots ,g_n)$ are the two maps from $T$ and $Y$ to $Z=\mathbb{A}^n$. Thus $T\times _ZY$ is a (global) complete intersection in $T\times Y$.

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