Every element of a ring $A$ is integral over the ring $A$. How?

Question

I am learning Integral Dependence for the first time. Every book says that the answer to my question is trivial but I don't see it. Please help!

Answer 1

Recall the definition of integrality:

Definition: Let $B$ be a ring, and let $A\subseteq B$ be a subring of $B.$ An element $b\in B$ is integral over $A$ if there exists a monic polynomial $f\in A[x]$ such that $f(b) = 0.$

Certainly $A\subseteq A$ is a subring of itself, so $a\in A$ is integral over $A$ if there exists a monic polynomial $f\in A[x]$ such that $f(a) = 0.$ Setting $f(x) = x - a,$ we see that $f(a) = a - a = 0.$ As a monic polynomial is one whose leading coefficient is $1,$ this produces an explicit monic polynomial that has $a\in A$ as a root. As $a$ was arbitrary, we have shown that every element of $A$ is integral over $A.$