Let $R$ be a regular local ring, $X$ and $Y$ smooth $R$-schemes, $T\to Y$ a regular closed immersion over $R$ with $T$ smooth over $R$, and $f: X\to Y$ an $R$-morphism.



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> Is the fiber product scheme $X\_T:=X\times\_YT$ Cohen-Macaulay?
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What conditions on $f$ would ensure that $X\_T$ is Cohen-Macaulay?



Cohen-Macaulay fiber products
Ok, $T \hookrightarrow Y$ is locally a complete intersection, right? (Closed regular subscheme of a regular scheme). So you'd want that regular sequence (locally) to become a regular sequence on $X$. That's not always true, but if $f : X \to Y$ is *flat* it is ok (since a weakly regular sequence will stay weakly regular...).


Without the flatness though I don't see how to guarantee it. Explicitly, let $R = k$ be a field, let $Y = \mathbb{A}^3\_k = \mathrm{Spec}\, k[x,y,z]$ and let $f : X \to Y$ be the blowup of the origin with exceptional divisor $E \cong \mathbb{P}^2$. Let $T = V(x,y) \subseteq Y$, note $T \cong \mathbb{A}^1\_k$. Then $X\_T$ is not equidimensional (it's a union of $E$ with the strict tranform of $T$), and hence it can't be Cohen-Macaulay.


Similar constructions work over other regular $R$s too (ie, $R$ need not be a field).


