= Solution
{c}
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 $n\geq1$ and $s_0,\ldots,s_{n-1}\in S$ such that
$$
\boxed{r^n+s_{n-1}r^{n-1}+\cdots+s_1r+s_0=0.}
$$
The 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 $1,r,\ldots,r^{n-1}$.
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