Prove: Integrality of $S[y]$ over $S$

[Fermat1]

Let $S$ be a sub ring of a commutative ring $R$ let $y$ be in $R$. Prove that if $y$ is integral over $S$, then $S[y]$ is integral over $S$. I'm puzzled as how to start this one.

 

[jvicens]

Thank you for your question. This is a very interesting problem in commutative algebra. Let me guide you through the proof step by step.

First, let's recall the definition of integral elements. An element $y$ in a ring $R$ is said to be integral over a subring $S$ if there exists a monic polynomial $f(x) \in S[x]$ such that $f(y) = 0$. In other words, $y$ satisfies a monic polynomial with coefficients in $S$.

Now, let's consider the subring $S[y]$ generated by $y$. We want to show that every element in $S[y]$ is integral over $S$. To do this, we will show that $S[y]$ is contained in a larger ring $T$ such that every element in $T$ is integral over $S$. This will then imply that $S[y]$ is integral over $S$.

So, let $T$ be the ring generated by $S$ and $y$. In other words, $T$ is the smallest ring containing $S$ and $y$. We claim that every element in $T$ is integral over $S$. To see this, consider an element $z \in T$. Then, by definition, $z$ can be expressed as a polynomial in $y$ with coefficients in $S$. In other words, $z = a_0 + a_1y + a_2y^2 + \dots + a_ny^n$ for some $a_0, a_1, \dots, a_n \in S$. Now, since $y$ is integral over $S$, it satisfies a monic polynomial $g(x) \in S[x]$. Therefore, we can rewrite $z$ as $z = a_0 + a_1y + a_2y^2 + \dots + a_ny^n = g(y)$. This shows that every element in $T$ is integral over $S$.

Now, since $S[y] \subseteq T$ and every element in $T$ is integral over $S$, it follows that every element in $S[y]$ is also integral over $S$. Therefore, $S[y]$ is integral over $S$, as desired.

I hope this explanation helps you understand the proof. Please let me know if you have any further questions.