Integral extensions of rings, when one of the rings is a field

Integral extensions of rings, when one of the rings is a field 5.2. going up Proposition. Suppose that $A, B$ are integral domains and $B$ is an integral extension of $A$. Then $B$ is a field iff $A$ is a field. Proof. Suppose $B$ is a field and $a \neq 0 \in A$. Then $B$ contains the inverse $a^{-1}$ of $a$, and since $B$ is an integral extension, $a^{-1}$ is a root of monic polynomial $f(x) \in A[x]$: \[a^{-n}+a_1 a^{1-n}+a_2 a^{2-n}+\cdots+a_n=0\] Multiplying by $a^{n−1}$, we get: \[a^{-1} + a_1 + a_2a+ \cdots+ a_na^{n−1} = 0.\] Solving for $a^{-1}$, we see that $a^{-1} \in A$. So, $A$ is a field. Conversely suppose that $A$ is field. Let $b \neq 0 \in B$. Then $b \in A[b]$ which is a finitely generated module. In other words, $A[b] \cong A^n$ is finite dimensional vector space over field. Multiplication by $b$ gives linear mapping: \[\mu_b :A[b] \cong A^n \to A^n\cong A[b].\] Since $B$ is an integral domain, the kernel of $\mu_b$ is $0$. This means that $\mu_b$ has rank $n$, and therefore an isomorphism of vector spaces. So, it is onto. So there is some $c \in B$ so that $\mu_b(c)=bc= 1$. So, $c=b^{-1} \in A[b] \subseteq B$ showing that $B$ is field.