**Proof.**
If $I$ is finite then the result follows from the Hilbert Nullstellensatz, Theorem 10.33.1. In the rest of the proof we assume $I$ is infinite. It suffices to prove the result for $\mathfrak m \subset k[\{ x_ i\} _{i \in I}]$ maximal in the polynomial ring on variables $x_ i$, since $S$ is a quotient of this. As $I$ is infinite the set of monomials $x_{i_1}^{e_1} \ldots x_{i_ r}^{e_ r}$, $i_1, \ldots , i_ r \in I$ and $e_1, \ldots , e_ r \geq 0$ has cardinality at most equal to the cardinality of $I$. Because the cardinality of $I \times \ldots \times I$ is the cardinality of $I$, and also the cardinality of $\bigcup _{n \geq 0} I^ n$ has the same cardinality. (If $I$ is finite, then this is not true and in that case this proof only works if $k$ is uncountable.)

To arrive at a contradiction pick $T \in \kappa (\mathfrak m)$ transcendental over $k$. Note that the $k$-linear map $T : \kappa (\mathfrak m) \to \kappa (\mathfrak m)$ given by multiplication by $T$ has the property that $P(T)$ is invertible for all monic polynomials $P(t) \in k[t]$. Also, $\kappa (\mathfrak m)$ has dimension at most the cardinality of $I$ over $k$ since it is a quotient of the vector space $k[\{ x_ i\} _{i \in I}]$ over $k$ (whose dimension is $\# I$ as we saw above). This is impossible by Lemma 10.34.10.

To show that $S$ is Jacobson we argue as follows. If not then there exists a prime $\mathfrak q \subset S$ and an element $f \in S$, $f \not\in \mathfrak q$ such that $\mathfrak q$ is not maximal and $(S/\mathfrak q)_ f$ is a field, see Lemma 10.34.5. But note that $(S/\mathfrak q)_ f$ is generated by at most $\# I + 1$ elements. Hence the field extension $k \subset (S/\mathfrak q)_ f$ is algebraic (by the first part of the proof). This implies that $\kappa (\mathfrak q)$ is an algebraic extension of $k$ hence $\mathfrak q$ is maximal by Lemma 10.34.9. This contradiction finishes the proof.
$\square$

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