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We consider $R=\mathbb{R}[x^2-1]\subset \mathbb{R}[x]=R'$ a ring extension, which is easy to see that is integral. Now we consider the ideal $P'=(x-1)$ of $R'$, and $P=P'\cap R$. We claim that $R_P\subset R'_{P'}$ is not an integral extension.

My attempt: I consider the element $\frac{1}{x+1}\in R'_{P'}$. Suppose that this element is integral, then there exists a relation $$ \frac{1}{(x+1)^n}+\sum_{i=0}^{n-1}\frac{1}{(x+1)^i}\frac{a_i}{b_i}=0$$ for $n\in \mathbb{N}$, $\frac{a_i}{b_i}\in R'_{P'}$. Making common denominator and calling $c_n=b_1\cdots b_{n-1}$, $c_i=\prod_{j\ne i}b_j\notin P$, we get $$c_n+\sum_{i=0}^{n-1}(x+1)^{n-i}a_ic_i=0.$$ I think that from here I can deduce an absurd, but I am stucked on this point...

Lios
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1 Answers1

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The prime ideals of $R’$ lying over $P = (x^2-1)$ are $P’ = (x-1)$ and $(x+1)$.

Since $R \subset R’$ is integral, $R_P \subset R’_P$ is integral. Here, $R’_P = (R\setminus P)^{-1} R’$. By the above observation $R’_P$ has two maximal ideals, and $R’_{P’}$ is an integrally closed ring having one maximal ideal.

If $R_P \subset R’_{P’}$ were integral, then $R_P \subset R’_P \subset R’_{P’}$, and in particular, the second inclusion is integral. However, there is no prime in $R’_{P’}$ lying over the prime ideal $(x+1)R’_P$. This is a contradiction.

Youngsu
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