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Let B be an A-algebra where A is a noetherian ring

Prove that E the set of integral elements over A in B form a subring of B.

What I did:

I tried to prove that E is a subalgebra of B. I proved by induction that E is a finitely-generated sub-module of B. Does it make it a sub-algebra of B ? on which conditions?

Conjecture
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2 Answers2

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Not necessarily. Consider for example $B=A[X]$. The submodule generated by $1$ and $X$ is not closed under multiplication and so is not a subalgebra.

hmakholm left over Monica
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I'm not sure how you proved that $E$ is a finitely generated $B$-submodule, which it isn't generally; it is generally not a finitely generated $A$-module either.

The condition for an element $b\in B$ to be integral over $A$ is that $A[b]$ is a finitely generated $A$-module.

If $b,c\in E$, then $c$ is integral over $A[b]$, so $A[b][c]$ is a finitely generated $A[b]$-module; hence also a finitely generated $A$-module (prove it). Since $b+c$ and $bc$ belong to $A[b][c]$, we see that both $A[b+c]$ and $A[bc]$ are finitely generated $A$-modules. Try and see where $A$ being Noetherian is used.

egreg
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  • If $A[b+c]$ and $A[bc]$ are finitely generated A-modules, then ab and a+b are integral over A since A is noetherian. Thank you both for your help. – Conjecture Dec 22 '18 at 16:16