Lemma 48.25.8. 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 Gorenstein at $x$, then $f' : X' \to Y'$ is Gorenstein at $x'$.

2. If $f$ is flat at $x$ and $f'$ is Gorenstein at $x'$, then $f$ is Gorenstein at $x$.

3. If $Y' \to Y$ is flat at $f'(x')$ and $f'$ is Gorenstein at $x'$, then $f$ is Gorenstein 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 Gorenstein 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 Gorenstein implies that $\mathcal{O}_{X', x'}/\mathfrak m_{y'}\mathcal{O}_{X', x'}$ is Gorenstein, see Lemma 48.25.1. Hence we see that $f'$ is Gorenstein at $x'$.

Assume $f$ is flat at $x$ and $f'$ is Gorenstein at $x'$. The fact that $\mathcal{O}_{X', x'}/\mathfrak m_{y'}\mathcal{O}_{X', x'}$ is Gorenstein implies that $\mathcal{O}_{X, x}/\mathfrak m_ y\mathcal{O}_{X, x}$ is Gorenstein, see Lemma 48.25.1. Hence we see that $f$ is Gorenstein at $x$.

Assume $Y' \to Y$ is flat at $y'$ and $f'$ is Gorenstein 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 Gorenstein implies that $\mathcal{O}_{X, x}/\mathfrak m_ y\mathcal{O}_{X, x}$ is Gorenstein, see Lemma 48.25.1. Hence we see that $f$ is Gorenstein at $x$. $\square$

Comment #7719 by Andrew on

I am confused by (2). Let $f \colon \mathrm{Spec}\, A \to \mathrm{Spec}\, \mathbb Z$ where $A$ is an order in a number field. The fiber of $f$ over $Y' = \mathrm{Spec}\, \mathbb F_p$ is Gorenstein, for its localisation at any maximal ideal $x$ is, after a finite basechange along $k/\mathbb F_p$, of the form $\mathrm{Spec}\, k[x]/(x)^e$ which has the simple socle $kx^{e-1}$, and is therefore Gorenstein by, e.g., Prop. 21.5 of Eisenbud. Assertion (2) implies that $f$ is Gorenstein at $x$ but $x$ was arbitrary, and number fields of degree $>2$ contain non-Gorenstein orders. For the same reason, I am also confused by Lemma 0C05.

Comment #7720 by on

If you compute through an example, you will find that your assertion about the base change does not hold --- that is the point of this lemma.

Comment #7721 by Laurent Moret-Bailly on

Typo in (2): "$f$ is flat and $x$".

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