Show that an integral domain with elements integral over $\mathbb{Q}$ is a field.

Question

Let $A$ be an integral domain which contains the field of rational numbers $\mathbb{Q}$. Suppose that every element $a\in A$ is algebraic over $\mathbb{Q}$. Prove that $A$ is a field.

Answer 1

If $a\ne0$, then $a^n+b_{n-1}a^{n-1}+\dots+b_1a+b_0=0$ for some $b_0,\dots,b_{n-1}\in\mathbb{Q}$, with $n\ge1$ and $b_0\ne0$ (prove it). Then $$ a(a^{n-1}+b_{n-1}a^{n-2}+\dots+b_1)=-b_0 $$ Can you finish?

Answer 2

Let $0\neq x\in A$. We will be done if we show that there exists $y\in A$ such that $xy=1$. By hypothesis, there exists $p\in \Bbb Q [X]$ such that $p(x)=0$. Without loss of generality, assume $p$ is the minimal polynomial of $x$. It has a nonzero constant term, hence we can subtract it over and multiply by the negative of the inverse. On the left-hand side, we now have an $x$ in common with all of the terms, so we may factor it out. What remains is the inverse of $x$.