Corollary 4

Let $A$ be a ring, $A'$ a ring containing $A$ and integral over $A$ and $f$ a homomorphism from $A$ to an algebraically closed field $L$. Then $f$ can be extended to a homomorphism from $A'$ to $L$.

Let $\mathfrak{p}$ be the kernel of $f$, which is a prime ideal since $f(A) \subset L$ is an integral domain; let $\mathfrak{p}'$ be a prime ideal of $A'$ lying above $\mathfrak{p}$ (Theorem 1). Then $A/\mathfrak{p}$ is canonically identified with a subring of $A'/\mathfrak{p}'$ and $A'/\mathfrak{p}'$ is integral over $A/\mathfrak{p}$ ($§ 1$, no. 1, Proposition 2). The homomorphism $f$ defines, by taking the quotient, an isomorphism of $A/\mathfrak{p}$ onto the subring $f(A)$ of $L$, which can be extended to an isomorphism $g$ of the field of fractions $K$ of $A/\mathfrak{p}$ onto a subfield of $L$. As the field of fractions $K'$ of $A'/\mathfrak{p}'$ is algebraic over $K$, $g$ can be extended to an isomorphism $g'$ of $K'$ onto a subfield of $L$ (Algebra, Chapter V, $§ 4$, no. 2, Corollary to Theorem 1); if $\pi': A' \to A'/\mathfrak{p}'$ is the canonical homomorphism, $g' \circ \pi'$ is a homomorphism from $A'$ to $L$ extending $f$.