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I'm trying to solve exercise 8.5 from the book "A course in Commutative algebra" by Gregor Kemper. It says

Let $R$ be a ring and $a\in R$ such that $a$ is not a zero divisor, the ideal $(a)$ is a radical ideal and the localization $R_a$ (localization with respect $U= \left\{ 1,a,a^2,\ldots\right\}$) is a normal domain. Then $R$ is normal.

What i have done so far is the following.

Since $\text{Quot}(R_a)=\text{Quot}(R)$ and $R_a$ is normal, if we take an element $x \in \text{Quot}(R_a)$ which is integral over $R_a$ we must have $x \in R_a$ so $x = \frac{r}{a^m}$ for some element $r\in R$ and an integer $m \in \mathbb{N}.$ Now suppose that $x$ is integral over $R,$ so we have a polynomial with coefficients in $R$ $$t^n +c_1 t^{n-1}+\cdots +c_1t+c_n$$ such that $$x^n +c_1 x^{n-1}+\cdots +c_1x+c_n=0.$$ From the above we know that $x \in R_a$ so in order to show that $x\in R$ we must show that $\frac{r}{a^m}= \frac{r'}{1}$ which means that there is a $a^{\nu}$ such that $a^{\nu} r =a^{\nu} a^m r'.$

I cant really proceed from here, any hints?

user26857
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Nikitas Nikandros
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1 Answers1

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All you need to show is that $R$ is integrally closed in $R_a = R[\frac{1}{a}]$. Once you have done that it is easy: If $x \in Quot(R)$ is integral over $R$, it is in particular integral over $R_a$ and thus we have $x \in R_a$ by assumption. If you have managed to show the above, you deduce $x \in R$ and you are done.

So let us take $x=\frac{r}{a^m} \in R_a$ integral over $R$. We can choose $m$ minimal, hence there is no representation of $x$ with the exponent of $a$ in the denominator being smaller than $m$.

We have an equation:

$$0 = x^n + c_{n-1}x^{n-1} + \dotsb + c_0= \frac{r^n}{a^{mn}} + \dotsb + c_0$$

After multiplication with $a^{mn}$ we deduce $r^n \in (a)$ (provided that $m>0$), hence $r \in (a)$ since $(a)$ is radical. Write $r = r'a$ and get the contradiction $x = \frac{r'}{a^{m-1}}$.

MooS
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