# Radicals and the Ring of Quotients

1. Dec 21, 2017

### Bashyboy

1. The problem statement, all variables and given/known data
Let $S$ be a multiplicative subset of a commutative ring $R$ with identity. If $I$ is an ideal in $R$, then $S^{-1}(\mbox{ Rad } I) = \mbox{Rad}(S^{-1}I)$.

2. Relevant equations

3. The attempt at a solution

If $x \in S^{-1}(\mbox{ Rad } I)$, then $x = \frac{r}{s}$ for some $r \in \mbox{ Rad } I$ and $s \in S$. Hence, there exists an $n \in \Bbb{N}$ such that $r^n \in I$; moreover, since $S$ is multiplicative, $s^n \in S$. Therefore $\left( \frac{r}{s} \right)^n = \frac{r^n}{s^n} \in S^{-1}I$ which means that $\frac{r}{s} \in \mbox{Rad}(S^{-1}I)$.

Now suppose that $x \in \mbox{Rad}(S^{-1}I)$. Then there exists a natural number $n$ such that $S^{-1}I \ni \left( \frac{r}{s} \right)^n = \frac{r^n}{s^n}$. Therefore $r^n \in I$ which implies $r \in \mbox{Rad } I$, and so $\frac{r}{s} \in S^{-1}(\mbox{ Rad } I)$.

Is it really that easy, or did I make a simple blunder?

2. Dec 21, 2017

### Staff: Mentor

I don't follow your conclusion $\operatorname{Rad}(S^{-1}I) \subseteq S^{-1}\operatorname{Rad}(I)$. With $x \in \operatorname{Rad}(S^{-1}I)$ we have $x^n=rs^{-1}\; , \;r \in I$. How can we assume, that $x$ is of the form $r^ns^{-n}\,$?

3. Dec 21, 2017

### Staff: Mentor

I don't follow your conclusion $\operatorname{Rad}(S^{-1}I) \subseteq S^{-1}\operatorname{Rad}(I)$. With $x \in \operatorname{Rad}(S^{-1}I)$ we have $x^n=rs^{-1}\; , \;r \in I\,.$ How can we assume, that $x$ is of the form $r^ns^{-n}\,$?

4. Dec 21, 2017

### Bashyboy

Let me try again. Suppose that $x \in \mbox{Rad}(S^{-1}I)$. Then $x = \frac{r}{s}$ with $s \in S$ and $r \in R$ such that $x^n = \frac{r^n}{s^n} \in S^{-1} I$. Then by definition $r^n \in I$ and $s^n \in S$ and therefore $r \in \mbox{Rad } I$. Hence $\frac{r}{s} \in S^{-1} \mbox{Rad}(I)$.

5. Dec 21, 2017

### Staff: Mentor

For $x \in \operatorname{Rad}(S^{-1}I)$ we have $x^n=s^{-1}r$ with $r \in I$ by definition. But how do we know that $x \in RS^{-1}\,?$ What have I missed?

6. Dec 21, 2017

### Bashyboy

Well, $I \subseteq R$, so $S^{-1} I \subseteq S^{-1} R$.

7. Dec 21, 2017

### Staff: Mentor

But you only have this for $x^n$ and not for $x$ itself. However, you keep making assumptions on $x$ which have to be proven.

Edit and hint: We want to have $x=s^{-1}r'$ with $r' \in \operatorname{Rad}(I)$ that is $r'^{\,n} =(sx)^n \in I$.

Last edited: Dec 21, 2017
8. Dec 23, 2017

### Bashyboy

Okay. Here is another try. If $x \in \mbox{ Rad }(S^{-1})$, then $x^n \in S^{-1}I$ for some $n \in \Bbb{N}$ and therefore $x^n = \frac{r}{s}$ with $r \in I$. But $x$ is also an element in $S^{-1}R$, so that $x = \frac{t}{k}$ for $t \in R$ and $k \in S$. This implies $\frac{t^n}{k^n} = \frac{r}{s}$ which happens if and only if there exists an $s_1 \in S$ such that $s_1(st^n - rk^n) = 0$ or $s_1s t^n = r s_1 k^n \in I$. Since $I$ is an ideal, $s_1^n s^n t^n = r s_1^n k^n s^{n-1} \in I$. Hence $x = \frac{t}{k} = \frac{s_1 s t}{s_1 sk}$ with $(s_1 s t)^n \in I$, which means $x \in S^{-1} \mbox{ Rad }(I)$.

How does this sound?

9. Dec 23, 2017

### Staff: Mentor

Certainly too complicated. And again, why is $x \in S^{-1}R\;$? Anyway, you don't need this detour.

$x \in \operatorname{Rad}(S^{-1}I)$ means $x^n=s^{-1}r\;(r \in I)$ and we get
$$(sx)^n=s^nx^n=s^{n-1}r \in I \Longrightarrow sx \in \operatorname{Rad}I \Longrightarrow x \in S^{-1}(\operatorname{Rad}I)$$

10. Dec 23, 2017