Lemma 10.36.19. Let $k$ be a field. Let $S$ be a $k$-algebra over $k$.
If $S$ is a domain and finite dimensional over $k$, then $S$ is a field.
If $S$ is integral over $k$ and a domain, then $S$ is a field.
If $S$ is integral over $k$ then every prime of $S$ is a maximal ideal (see Lemma 10.26.5 for more consequences).
Proof. The statement on primes follows from the statement “integral $+$ domain $\Rightarrow $ field”. Let $S$ integral over $k$ and assume $S$ is a domain, Take $s \in S$. By Lemma 10.36.4 we may find a finite dimensional $k$-subalgebra $k \subset S' \subset S$ containing $s$. Hence $S$ is a field if we can prove the first statement. Assume $S$ finite dimensional over $k$ and a domain. Pick $s\in S$. Since $S$ is a domain the multiplication map $s : S \to S$ is surjective by dimension reasons. Hence there exists an element $s_1 \in S$ such that $ss_1 = 1$. So $S$ is a field. $\square$
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