Lemma 33.37.11. Let $A$ be a Noetherian local ring of dimension $1$. Let $L = \prod A_\mathfrak p$ where the product is over the minimal primes of $A$. Let $K \to L$ be an integral ring map. Then there exist $a \in \mathfrak m_ A$ and $x \in K$ which map to the same element of $L$ such that $\mathfrak m_ A = \sqrt{(a)}$.

Proof. By Lemma 33.37.10 we may replace $A$ by $A/(\bigcap \mathfrak p)$ and assume that $A$ is a reduced ring (some details omitted). We may also replace $K$ by the image of $K \to L$. Then $K$ is a reduced ring. The map $\mathop{\mathrm{Spec}}(L) \to \mathop{\mathrm{Spec}}(K)$ is surjective and closed (details omitted). Hence $\mathop{\mathrm{Spec}}(K)$ is a finite discrete space. It follows that $K$ is a finite product of fields.

Let $\mathfrak p_ j$, $j = 1, \ldots , m$ be the minimal primes of $A$. Set $L_ j$ be the fraction field of $A_ j$ so that $L = \prod _{j = 1, \ldots , m} L_ j$. Let $A_ j$ be the normalization of $A/\mathfrak p_ j$. Then $A_ j$ is a semi-local Dedekind domain with at least one maximal ideal, see Algebra, Lemma 10.120.18. Let $n$ be the sum of the numbers of maximal ideals in $A_1, \ldots , A_ m$. For such a maximal ideal $\mathfrak m \subset A_ j$ we consider the function

$v_{\mathfrak m} : L \to L_ j \to \mathbf{Z} \cup \{ \infty \}$

where the second arrow is the discrete valuation corresponding to the discrete valuation ring $(A_ j)_{\mathfrak m}$ extended by mapping $0$ to $\infty$. In this way we obtain $n$ functions $v_1, \ldots , v_ n : L \to \mathbf{Z} \cup \{ \infty \}$. We will find an element $x \in K$ such that $v_ i(x) < 0$ for all $i = 1, \ldots , n$.

First we claim that for each $i$ there exists an element $x \in K$ with $v_ i(x) < 0$. Namely, suppose that $v_ i$ corresponds to $\mathfrak m \subset A_ j$. If $v_ i(x) \geq 0$ for all $x \in K$, then $K$ maps into $(A_ j)_{\mathfrak m}$ inside the fraction field $L_ j$ of $A_ j$. The image of $K$ in $L_ j$ is a field over $L_ j$ is algebraic by Algebra, Lemma 10.36.18. Combined we get a contradiction with Algebra, Lemma 10.50.7.

Suppose we have found an element $x \in K$ such that $v_1(x) < 0, \ldots , v_ r(x) < 0$ for some $r < n$. If $v_{r + 1}(x) < 0$, then $x$ works for $r + 1$. If not, then choose some $y \in K$ with $v_{r + 1}(y) < 0$ as is possible by the result of the previous paragraph. After replacing $x$ by $x^ n$ for some $n > 0$, we may assume $v_ i(x) < v_ i(y)$ for $i = 1, \ldots , r$. Then $v_ j(x + y) = v_ j(x) < 0$ for $j = 1, \ldots , r$ by properties of valuations and similarly $v_{r + 1}(x + y) = v_{r + 1}(y) < 0$. Arguing by induction, we find $x \in K$ with $v_ i(x) < 0$ for $i = 1, \ldots , n$.

In particular, the element $x \in K$ has nonzero projection in each factor of $K$ (recall that $K$ is a finite product of fields and if some component of $x$ was zero, then one of the values $v_ i(x)$ would be $\infty$). Hence $x$ is invertible and $x^{-1} \in K$ is an element with $\infty > v_ i(x^{-1}) > 0$ for all $i$. It follows from Lemma 33.37.5 that for some $e < 0$ the element $x^ e \in K$ maps to an element of $\mathfrak m_ A/\mathfrak p_ j \subset A/\mathfrak p_ j$ for all $j = 1, \ldots , m$. Observe that the cokernel of the map $\mathfrak m_ A \to \prod \mathfrak m_ A/\mathfrak p_ j$ is annihilated by a power of $\mathfrak m_ A$. Hence after replacing $e$ by a more negative $e$, we find an element $a \in \mathfrak m_ A$ whose image in $\mathfrak m_ A/\mathfrak p_ j$ is equal to the image of $x^ e$. The pair $(a, x^ e)$ satisfies the conclusions of the lemma. $\square$

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