Basic Ring Theory - 16 (Radical Of An Ideal)

[Tsakanikas Nickos]

Let $A$ be a ring and let $\mathfrak{a}, \mathfrak{b}$ be ideals of $A$. Show that

1. $\sqrt{ \sqrt{ \mathfrak{a} } } = \sqrt{ \mathfrak{a} }$
2. $\sqrt{ \mathfrak{a} \mathfrak{b} } = \sqrt{ \mathfrak{a} \cap \mathfrak{b} } = \sqrt{ \mathfrak{a} } \cap \sqrt{ \mathfrak{b} }$
3. $\sqrt{ \mathfrak{a} } = A \Leftrightarrow \mathfrak{a} = A$

Moreover, if $\mathfrak{p}$ is a prime ideal of $A$, then $\sqrt{ \mathfrak{p}^{n} } = \mathfrak{p} \, , \, \forall n>0 $.

[Papapetros Vaggelis]

i. Obviously, \(\displaystyle{\sqrt{a}\subseteq \sqrt{\sqrt{a}}}\) .

On the other hand, if \(\displaystyle{f\in\sqrt{\sqrt{a}}}\), then, \(\displaystyle{f^m\in\sqrt{a}}\)

for some \(\displaystyle{m\in\mathbb{N}}\). But then, \(\displaystyle{(f^m)^n\in a}\) for some \(\displaystyle{n\in\mathbb{N}}\), so

\(\displaystyle{f^{m\,n}\in a\implies f\in\sqrt{a}}\).

Therefore, \(\displaystyle{\sqrt{\sqrt{a}}\subseteq \sqrt{a}}\) and we are done.

ii. \(\displaystyle{a\cap b\subseteq b\implies \sqrt{a\cap b}\subseteq \sqrt{b}}\)

and similarly \(\displaystyle{\sqrt{a\cap b}\subseteq \sqrt{a}}\), so \(\displaystyle{\sqrt{a\cap b}\subseteq \sqrt{a}\cap \sqrt{b}\,\,(I)}\).

Now, if \(\displaystyle{f\in\sqrt{a}\cap \sqrt{b}}\), then \(\displaystyle{f^m\in a}\) and \(\displaystyle{f^n\in b}\)

for some \(\displaystyle{m\,,n\in\mathbb{N}}\) .

Since \(\displaystyle{a\,,b}\) are ideals of \(\displaystyle{A}\), we have that

\(\displaystyle{f^{m+n}=f^n\cdot f^m\in a}\) and \(\displaystyle{f^{n+m}=f^{m+n}=f^m\cdot f^n\in b}\)

that is \(\displaystyle{f^{m+n}\in a\cap b\implies f\in \sqrt{a\cap b}}\).

So, \(\displaystyle{\sqrt{a}\cap \sqrt{b}\subseteq \sqrt{a\cap b}\,\,(II)}\).

Also, \(\displaystyle{f^{n+m}=f^n\cdot f^m=f^m\cdot f^n\in a\,b}\).

By this way, \(\displaystyle{\sqrt{a}\cap \sqrt{b}\subseteq \sqrt{a\,b}\,\,(III)}\).

Conversely, if \(\displaystyle{f\in \sqrt{a\,b}}\), then \(\displaystyle{f^k\in a\,b}\). So,

\(\displaystyle{f^k=x_1\,y_1+...+x_n\,y_n\,\,,x_i\in a\,,y_i\in b\,,1\leq i\leq n}\).

But, \(\displaystyle{x_{i}\,y_{i}\in a\cap b\,,\forall\,i\in\left\{1,...,n\right\}}\) because \(\displaystyle{a\,,b}\)

are ideals of \(\displaystyle{A}\) and \(\displaystyle{x_{i}\in a\,,y_{i}\in b}\), which means that

\(\displaystyle{\sqrt{a\,b}\subseteq \sqrt{a}\cap\sqrt{b}\,\,(IV)}\).

The relations \(\displaystyle{(I)\,,(II)\,,(III)\,,(IV)}\) give us the result.

iii. If \(\displaystyle{a=A}\), then \(\displaystyle{\sqrt{a}=\sqrt{A}=A}\).

Suppose now that \(\displaystyle{\sqrt{a}=A}\). Then,

\(\displaystyle{1\in A\implies 1\in \sqrt{a}\implies 1^k\in a\implies 1\in a\implies a=A}\).

For the last question

Let \(\displaystyle{p}\) be a prime ideal of \(\displaystyle{A}\). Let \(\displaystyle{n>0}\).

\(\displaystyle{p\subseteq p^n\implies \sqrt{p}\subseteq \sqrt{p^n}}\).

On the other hand, if \(\displaystyle{f\in\sqrt{p^n}}\), then \(\displaystyle{f^k\in p^n}\)


for some \(\displaystyle{k\in\mathbb{N}}\). Since \(\displaystyle{p}\) is prime, we get

\(\displaystyle{f^k\in p\implies f\in\sqrt{p}}\), so \(\displaystyle{\sqrt{p^n}\subseteq \sqrt{p}}\)

and finally \(\displaystyle{\sqrt{p^n}=\sqrt{p}}\).

[Tsakanikas Nickos]

Thank you very much for your answer, Vaggelis.

Allow me to conclude the proof of the last question and prove an easy fact that you used throughout your proof:

• If $\mathfrak{p}$ is a prime ideal, then $\mathfrak{p} = \sqrt{ \mathfrak{p} } $ :
  
  Indeed, it is obvious that $\mathfrak{p} \subseteq \sqrt{ \mathfrak{p} } $. On the other hand, let $ f \in \sqrt{ \mathfrak{p} } $. Then there exists an integer $r>0$ such that $f^{r} \in \mathfrak{p}$. But since $\mathfrak{p}$ is a prime ideal, it follows that $f \in \mathfrak{p}$. Thus $\mathfrak{p} \supseteq \sqrt{ \mathfrak{p} } $.
• If $\mathfrak{a},\mathfrak{b}$ are ideals of $A$ such that $\mathfrak{a} \subseteq \mathfrak{b}$, then $\sqrt{\mathfrak{a}} \subseteq \sqrt{\mathfrak{b}}$ :
  
  Indeed, let $a \in \sqrt{\mathfrak{a}}$. Then there exists an integer $r>0$ such that $a^{r} \in \mathfrak{a}$. By hypothesis, $a^{r} \in \mathfrak{b}$, so $a \in \sqrt{\mathfrak{b}}$.