 Solution

For a subring $S\subseteq R$ with the same identity, $R$ is an integral extension of $S$ if every $r\in R$ is an integral element over $S$: there exist $n\ge 1$ and $s_0,\dots ,s_{n-1}\in S$ such thatThe requirement that the polynomial be monic is essential. Merely satisfying some polynomial equation over $S$ is insufficient: for instance $1/2\in \mathbb{Q}$ satisfies $2T-1=0$ over $\mathbb{Z}$, but is not integral over $\mathbb{Z}$. To see this directly, a hypothetical monic equation for $1/2$, multiplied by $2^n$, would say that an odd integer $1$ plus a sum of even integers is zero.

The degree and coefficients may depend on $r$. Integrality does not by itself assert that the entire ring $R$ is a finite $S$-module; it does imply that the subalgebra $S[r]$ is generated as an $S$-module by $1,r,\dots ,r^{n-1}$.