2
$\begingroup$

I need some help to solve the second part of this problem. Also I will appreciate corrections about my solution to the first part. The problem is the following.

Let $\sigma$ be an automorphism of the integral domain $R$. Show that $\sigma$ extends in an unique way to the integral closure of $R$ in it's field of fractions.

My solution: denote $\overline{R}$ for the normalization of $R$ and $\overline{\sigma}$ for the extension of $\sigma$. Given some $r/s\in \overline{R}$, we can define $\overline{\sigma}(r/s) = \sigma(r)/\sigma(s)$. From this I can show that $\overline{\sigma}(r/s)\in\overline{R}$. So this looks like a decent extension. I guess the extension is supposed to be a homomorphism (or even an automorphism), althought this is not said explicity.

The problem comes when I try to show it is unique. I don't know what to do, all my approaches failed.

$\endgroup$
0

2 Answers 2

3
$\begingroup$

Let $\sigma$ be an automorphism of an integral domain $R$. Then this extends uniquely to an automorphism $\overline{\sigma}$ of $K$, where $K$ is the field of fractions of $R$.

This follows easily from the Universal Property of the Ring of Fractions noticing that $\sigma(a)=0\iff a=0$. Moreover, the construction of $\overline{\sigma}$ tells us that $\overline{\sigma}(a/b)=\sigma(a)/\sigma(b)$.

Now let's observe that $x\in\overline{R}\Longrightarrow\overline{\sigma}(x)\in\overline R$: if $x^n+a_1x^{n-1}+\cdots+a_n=0$ with $a_i\in R$ then $\overline{\sigma}(x)^n+\sigma(a_1)\overline{\sigma}(x)^{n-1}+\cdots+\sigma(a_n)=0$ with $\sigma(a_i)\in R$. Thus $\overline{\sigma}_{|\overline R}$, the restriction of $\overline{\sigma}$ to $\overline R$, is an automorphism of $\overline R$.

Let $\tau$ be an automorphism of $\overline R$ which extends $\sigma$. We want to show $\tau=\overline{\sigma}_{|\overline R}$. But $K$ is also the field of fractions of $\overline R$, so it follows that $\tau$ extends (uniquely) to an automorphism $\overline{\tau}$ of $K$. Since $\overline{\tau}$ extends $\sigma$ to $K$ we must have $\overline{\tau}=\overline{\sigma}$, and therefore $\tau=\overline{\sigma}_{|\overline R}$.

$\endgroup$
3
$\begingroup$

The uniqueness is fairly trivial. Since suppose $\bar{\sigma}$ is an extension then it must be defined as you said. Indeed $\bar{\sigma}(r/s)=\bar{\sigma}(r\cdot s^{-1})=\bar{\sigma}(r)\cdot\bar{\sigma}(s)^{-1}=\sigma(r)\cdot\sigma(s)^{-1}=\sigma(r)/\sigma(s)$. And indeed $\bar{\sigma}$ is an automorphism if $\sigma$ is since the inverse is given by the unique extension of $\sigma^{–1}$.

$\endgroup$
3
  • $\begingroup$ "Since suppose $\bar{\sigma}$ is an extension then it must be defined as you said." Why? $\endgroup$
    – user26857
    May 12, 2015 at 14:41
  • $\begingroup$ I suppose this definition is the only one giving an automorphism, that's why. But I'm not sure the argument above proves that. $\endgroup$
    – Integral
    May 13, 2015 at 0:58
  • $\begingroup$ I did'nt answered because I have written it after the "indeed". $\sigma$ and $\bar{\sigma}$ coincide on $r$ and $s$. $\endgroup$ May 15, 2015 at 10:51

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .