Lemma 37.31.1. Let $R$ be a ring. Let $\mathfrak p_1, \ldots , \mathfrak p_ r$ be prime ideals of $R$ with $\mathfrak p_ i \not\subset \mathfrak p_ j$ if $i \not= j$. Let $k_ i \subset \kappa (\mathfrak p_ i)$ be subfields such that the extensions $\kappa (\mathfrak p_ i)/k_ i$ are not algebraic. Let $J \subset R$ be an ideal not contained in any of the $\mathfrak p_ i$. Then there exists an element $x \in J$ such that the image of $x$ in $\kappa (\mathfrak p_ i)$ is transcendental over $k_ i$ for $i = 1, \ldots , r$.
Proof. The ideal $J_ i = J \mathfrak p_1 \ldots \hat{\mathfrak p}_ i \ldots \mathfrak p_ r$ is not contained in $\mathfrak p_ i$, see Algebra, Lemma 10.15.1. It follows that every element $\xi $ of $\kappa (\mathfrak p_ i) = \text{Frac}(B/\mathfrak p_ i)$ is of the form $\xi = a/b$ with $a, b \in J_ i$ and $b \not\in \mathfrak p_ i$. Choosing $\xi $ transcendental over $k_ i$ we see that either $a$ or $b$ maps to an element of $\kappa (\mathfrak p_ i)$ transcendental over $k_ i$. We conclude that for every $i = 1, \ldots , r$ we can find an element $x_ i \in J_ i = J \mathfrak p_1 \ldots \hat{\mathfrak p}_ i \ldots \mathfrak p_ r$ which maps to an element of $\kappa (\mathfrak p_ i)$ transcendental over $k_ i$. Then $x = x_1 + \ldots + x_ r$ works. $\square$
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