For an integral extension of domains, the upper ring is a field if and only if the lower ring is

Assume $A$ is a field, and let $0\mathrm{\ne }b\in B$. By [L1] there is a monic relation $b^n+a_{n-1}{b}^{n-1}+\cdots +a_0=0$ over $A$ of minimal degree. The constant term cannot vanish: if $a_0=0$, then $b(b^{n-1}+a_{n-1}{b}^{n-2}+\cdots +a_1)=0$, and [L2] with $b\mathrm{\ne }0$ would give a smaller monic relation, contradicting minimality. Since $a_0\mathrm{\ne }0$ and $A$ is a field, $a_0^{-1}\in A$, and rearranging gives $b^{-1}=-a_0^{-1}(b^{n-1}+a_{n-1}{b}^{n-2}+\cdots +a_1)\in B$. Thus every nonzero element of $B$ is invertible.

Assume $B$ is a field, and let $0\mathrm{\ne }a\in A$. Then $a^{-1}\in B$, and [L1] gives a monic equation $(a^{-1}{)}^n+c_{n-1}(a^{-1}{)}^{n-1}+\cdots +c_0=0$ with $c_i\in A$. Multiplying by $a^n$ yields $1+c_{n-1}a+\cdots +c_0{a}^n=0$, so $a(-c_{n-1}-c_{n-2}a-\cdots -c_0{a}^{n-1})=1$. Hence $a$ is invertible in $A$.

Step 1.1 proves that $A$ field implies $B$ field, and step 1.2 proves the converse. Therefore $A$ is a field if and only if $B$ is a field.