Lemma 29.56.2. Let $f : X \to Y$ be a morphism of schemes. Let $n \geq 0$. The following are equivalent:

the integer $n$ bounds the degrees of the fibres of $f$, and

for every morphism $\mathop{\mathrm{Spec}}(k) \to Y$, where $k$ is a field, the fibre product $X_ k = \mathop{\mathrm{Spec}}(k) \times _ Y X$ is finite over $k$ of degree $\leq n$.

In this case the fibres of $f$ are universally bounded and the schemes $X_ k$ have at most $n$ points. More precisely, if $X_ k = \{ x_1, \ldots , x_ t\} $, then we have

\[ n \geq \sum \nolimits _{i = 1, \ldots , t} [\kappa (x_ i) : k] \]

**Proof.**
The implication (2) $\Rightarrow $ (1) is trivial. The other implication holds because if the image of $\mathop{\mathrm{Spec}}(k) \to Y$ is $y$, then $X_ k = \mathop{\mathrm{Spec}}(k) \times _{\mathop{\mathrm{Spec}}(\kappa (y))} X_ y$. By definition the fibres of $f$ being universally bounded means that some $n$ exists. Finally, suppose that $X_ k = \mathop{\mathrm{Spec}}(A)$. Then $\dim _ k A = n$. Hence $A$ is Artinian, all prime ideals are maximal ideals $\mathfrak m_ i$, and $A$ is the product of the localizations at these maximal ideals. See Algebra, Lemmas 10.53.2 and 10.53.6. Then $\mathfrak m_ i$ corresponds to $x_ i$, we have $A_{\mathfrak m_ i} = \mathcal{O}_{X_ k, x_ i}$ and hence there is a surjection $A \to \bigoplus \kappa (\mathfrak m_ i) = \bigoplus \kappa (x_ i)$ which implies the inequality in the statement of the lemma by linear algebra.
$\square$

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