- $\sqrt{ \sqrt{ \mathfrak{a} } } = \sqrt{ \mathfrak{a} }$
- $\sqrt{ \mathfrak{a} \mathfrak{b} } = \sqrt{ \mathfrak{a} \cap \mathfrak{b} } = \sqrt{ \mathfrak{a} } \cap \sqrt{ \mathfrak{b} }$
- $\sqrt{ \mathfrak{a} } = A \Leftrightarrow \mathfrak{a} = A$
Basic Ring Theory - 16 (Radical Of An Ideal)
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Basic Ring Theory - 16 (Radical Of An Ideal)
Let $A$ be a ring and let $\mathfrak{a}, \mathfrak{b}$ be ideals of $A$. Show that
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Re: Basic Ring Theory - 16 (Radical Of An Ideal)
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 \(\displaystyle{\sqrt{p^n}=\sqrt{p}}\).
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 \(\displaystyle{\sqrt{p^n}=\sqrt{p}}\).
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Re: Basic Ring Theory - 16 (Radical Of An Ideal)
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:
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 $a \in \sqrt{\mathfrak{b}}$.
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