Lemma 37.22.6. Let $f : X \to Y$ be a morphism of schemes. Assume that all the fibres $X_ y$ are locally Noetherian schemes. Let $Y' \to Y$ be locally of finite type. Let $f' : X' \to Y'$ be the base change of $f$. Let $x' \in X'$ be a point with image $x \in X$.

1. If $f$ is Cohen-Macaulay at $x$, then $f' : X' \to Y'$ is Cohen-Macaulay at $x'$.

2. If $f$ is flat at $x$ and $f'$ is Cohen-Macaulay at $x'$, then $f$ is Cohen-Macaulay at $x$.

3. If $Y' \to Y$ is flat at $f'(x')$ and $f'$ is Cohen-Macaulay at $x'$, then $f$ is Cohen-Macaulay at $x$.

Proof. Note that the assumption on $Y' \to Y$ implies that for $y' \in Y'$ mapping to $y \in Y$ the field extension $\kappa (y')/\kappa (y)$ is finitely generated. Hence also all the fibres $X'_{y'} = (X_ y)_{\kappa (y')}$ are locally Noetherian, see Varieties, Lemma 33.11.1. Thus the lemma makes sense. Set $y' = f'(x')$ and $y = f(x)$. Hence we get the following commutative diagram of local rings

$\xymatrix{ \mathcal{O}_{X', x'} & \mathcal{O}_{X, x} \ar[l] \\ \mathcal{O}_{Y', y'} \ar[u] & \mathcal{O}_{Y, y} \ar[l] \ar[u] }$

where the upper left corner is a localization of the tensor product of the upper right and lower left corners over the lower right corner.

Assume $f$ is Cohen-Macaulay at $x$. The flatness of $\mathcal{O}_{Y, y} \to \mathcal{O}_{X, x}$ implies the flatness of $\mathcal{O}_{Y', y'} \to \mathcal{O}_{X', x'}$, see Algebra, Lemma 10.100.1. The fact that $\mathcal{O}_{X, x}/\mathfrak m_ y\mathcal{O}_{X, x}$ is Cohen-Macaulay implies that $\mathcal{O}_{X', x'}/\mathfrak m_{y'}\mathcal{O}_{X', x'}$ is Cohen-Macaulay, see Varieties, Lemma 33.13.1. Hence we see that $f'$ is Cohen-Macaulay at $x'$.

Assume $f$ is flat at $x$ and $f'$ is Cohen-Macaulay at $x'$. The fact that $\mathcal{O}_{X', x'}/\mathfrak m_{y'}\mathcal{O}_{X', x'}$ is Cohen-Macaulay implies that $\mathcal{O}_{X, x}/\mathfrak m_ y\mathcal{O}_{X, x}$ is Cohen-Macaulay, see Varieties, Lemma 33.13.1. Hence we see that $f$ is Cohen-Macaulay at $x$.

Assume $Y' \to Y$ is flat at $y'$ and $f'$ is Cohen-Macaulay at $x'$. The flatness of $\mathcal{O}_{Y', y'} \to \mathcal{O}_{X', x'}$ and $\mathcal{O}_{Y, y} \to \mathcal{O}_{Y', y'}$ implies the flatness of $\mathcal{O}_{Y, y} \to \mathcal{O}_{X, x}$, see Algebra, Lemma 10.100.1. The fact that $\mathcal{O}_{X', x'}/\mathfrak m_{y'}\mathcal{O}_{X', x'}$ is Cohen-Macaulay implies that $\mathcal{O}_{X, x}/\mathfrak m_ y\mathcal{O}_{X, x}$ is Cohen-Macaulay, see Varieties, Lemma 33.13.1. Hence we see that $f$ is Cohen-Macaulay at $x$. $\square$

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