Lemma 29.42.7. Let $f : X \to S$ be a morphism of schemes. The following are equivalent

$f$ is integral, and

$f$ is affine and universally closed.

** integral $=$ affine $+$ universally closed **

Lemma 29.42.7. Let $f : X \to S$ be a morphism of schemes. The following are equivalent

$f$ is integral, and

$f$ is affine and universally closed.

**Proof.**
Assume (1). An integral morphism is affine by definition. A base change of an integral morphism is integral so in order to prove (2) it suffices to show that an integral morphism is closed. This follows from Algebra, Lemmas 10.35.22 and 10.40.6.

Assume (2). We may assume $f$ is the morphism $f : \mathop{\mathrm{Spec}}(A) \to \mathop{\mathrm{Spec}}(R)$ coming from a ring map $R \to A$. Let $a$ be an element of $A$. We have to show that $a$ is integral over $R$, i.e. that in the kernel $I$ of the map $R[x] \to A$ sending $x$ to $a$ there is a monic polynomial. Consider the ring $B = A[x]/(ax -1)$ and let $J$ be the kernel of the composition $R[x]\to A[x] \to B$. If $f\in J$ there exists $q\in A[x]$ such that $f = (ax-1)q$ in $A[x]$ so if $f = \sum _ i f_ ix^ i$ and $q = \sum _ iq_ ix^ i$, for all $i \geq 0$ we have $f_ i = aq_{i-1} - q_ i$. For $n \geq \deg q + 1$ the polynomial

\[ \sum \nolimits _{i \geq 0} f_ i x^{n - i} = \sum \nolimits _{i \geq 0} (a q_{i - 1} - q_ i) x^{n - i} = (a - x) \sum \nolimits _{i \geq 0} q_ i x^{n - i - 1} \]

is clearly in $I$; if $f_0 = 1$ this polynomial is also monic, so we are reduced to prove that $J$ contains a polynomial with constant term $1$. We do it by proving $\mathop{\mathrm{Spec}}(R[x]/(J + (x))$ is empty.

Since $f$ is universally closed the base change $\mathop{\mathrm{Spec}}(A[x]) \to \mathop{\mathrm{Spec}}(R[x])$ is closed. Hence the image of the closed subset $\mathop{\mathrm{Spec}}(B) \subset \mathop{\mathrm{Spec}}(A[x])$ is the closed subset $\mathop{\mathrm{Spec}}(R[x]/J) \subset \mathop{\mathrm{Spec}}(R[x])$, see Example 29.6.4 and Lemma 29.6.3. In particular $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(R[x]/J)$ is surjective. Consider the following diagram where every square is a pullback:

\[ \xymatrix{ \mathop{\mathrm{Spec}}(B) \ar@{->>}[r]^ g & \mathop{\mathrm{Spec}}(R[x]/J) \ar[r] & \mathop{\mathrm{Spec}}(R[x])\\ \emptyset \ar[u] \ar[r] & \mathop{\mathrm{Spec}}(R[x]/(J + (x)))\ar[u] \ar[r] & \mathop{\mathrm{Spec}}(R) \ar[u]^0 } \]

The bottom left corner is empty because it is the spectrum of $R\otimes _{R[x]} B$ where the map $R[x]\to B$ sends $x$ to an invertible element and $R[x]\to R$ sends $x$ to $0$. Since $g$ is surjective this implies $\mathop{\mathrm{Spec}}(R[x]/(J + (x)))$ is empty, as we wanted to show. $\square$

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