# $A$ is normal $\Rightarrow \left (A[x]/(f)\right )_{f'}$ is normal.

Let $$A$$ be a ring and $$f\in A[x]$$ be a monic polynomial. Let's consider the algebra $$B=A[x]/(f).$$ I want to prove that if $$A$$ is normal (i.e for any prime $$p\subset A,\ A_p$$ is integrally closed) then so is $$B_{f'}.$$

Let $$q\subset B_{f'}$$ be a prime ideal and $$p$$ its image in $$A.$$ We want to prove that $$B_q$$ is integrally closed. We localise at $$p$$, $$(B_{f'})_p\cong \left (A_p[x]/(f)\right )_{f'}$$ and so, by replacing $$A$$ with $$A_p$$ and $$B$$ by $$B_p$$, we can suppose that $$A$$ is a local normal ring.
In the proof that I have, the author then argues as follows:
He considers the quotient field $$k$$ of $$A$$ and the algebra $$C=B\otimes_Ak.$$ He says the $$k$$-algebra $$C$$ is finite and contains $$B$$ (since the canonical morphism $$B\rightarrow C$$ is injective). He then proceeds to prove that $$B_{f'}$$ is integrally closed in $$C_{f'}.$$ For that he considers the integral closure $$B'$$ of $$B$$ in $$C$$ and proves that $$f'(x)B'\subset B.$$ He then directly concludes that $$B_{f'}$$ is normal.
My question is how does he conclude that $$B_{f'}$$ is normal from that ? How is proving $$B_{f'}$$ is integrally closed in $$C_{f'}$$ sufficient to prove that $$B_{f'}$$ is normal ?
This is proposition 2, chapter 7 in Raynaud's "anneaux locaux hénseliens". Any help would be appreciated !

• Where $f'$ is?? Jan 22, 2022 at 20:13
• $f'$ is the derivative of $f.$ Jan 23, 2022 at 9:04

Let me fix the notation to that which is used by Raynaud: $$A$$ is a normal local domain with field of fractions $$K$$. $$C = A[X]/(f)$$. We are interested in proving that $$C_{f'}$$ is normal. Raynaud's proof shows that the integral closure $$C'$$ of $$C$$ in the finite $$K$$-algebra $$L = C \otimes_A K$$ satisfies $$f'(x)C' \subset C$$. We are suppposed to conclude from this that $$C_{f'}$$ is normal.
Since $$C_{f'}$$ is reduced (proved right before this one in Raynaud's book) and quasifinite over $$A$$, it suffices to show that $$C_{f'}$$ is integrally closed in its total ring of fractions. But in fact, the total ring of fractions is exactly the same as $$C_{f'} \otimes_A K$$. We know that it is supposed to include this, but we already have $$C_{f'} \otimes_A K = (K[X]/(f))_{f'},$$ which is already a product of fields (localizing at $$f'$$ gets rid of all the double factors of $$f$$), hence there are no non-zerodivisors left to localize with. So in fact the only thing we need to prove is that $$C_{f'}$$ is integrally closed in $$C_{f'} \otimes_A K$$. I will let you deduce this from what Raynaud proved (do the same type of thing as e.g. the standard texts on algebraic number theory or commutative algebra, where you prove that localization preserves the property of being integrally closed essentially by clearing denominators).