Radical
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Radical
Let \(\displaystyle{\left(A,+,\cdot\right)}\) be a commutative ring with unity and \(\displaystyle{I\,,J}\) be ideals of this ring.
1. The ideal \(\displaystyle{I}\) is radical (\(\displaystyle{I=\sqrt{I}}\)) if, and only if, the ring
\(\displaystyle{A/I}\) has not nonzero nilpotents elements.
So, the prime ideals, are radical.
2. If \(\displaystyle{I\,,J}\) are radical, then so is \(\displaystyle{I\cap J}\) but \(\displaystyle{I+J}\) need not be.
1. The ideal \(\displaystyle{I}\) is radical (\(\displaystyle{I=\sqrt{I}}\)) if, and only if, the ring
\(\displaystyle{A/I}\) has not nonzero nilpotents elements.
So, the prime ideals, are radical.
2. If \(\displaystyle{I\,,J}\) are radical, then so is \(\displaystyle{I\cap J}\) but \(\displaystyle{I+J}\) need not be.
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- Community Team
- Posts: 314
- Joined: Tue Nov 10, 2015 8:25 pm
Re: Radical
Let us prove the following:
$ " \implies " $ : Consider $ x \in A / I $ such that $ x^{n} = 0 $ for some positive integer $n$. We will show that $ x = 0 $. Since $ x = a + I $ for some $a \in A$, we have that \[ x^{n} = 0 \implies a^{n} + I = 0 + I \implies a^{n} \in I \]By definition of the radical of $I$, we conclude that $ a \in \sqrt{I} = I $ (by hypothesis). This means that $ x = 0 $.
$ " \impliedby " $ : By definition of the radical, it is clear that $ I \subseteq \sqrt{I} $. We will show that $ \sqrt{I} \subseteq I $. Consider $ x \in \sqrt{I} $. Then there exists a positive integer $n$ such that $ x^{n} \in I $, so $ x^{n} + I = 0 + I $ in the quotient $ A / I $. Set $ \xi = x + I $. Then $ \xi $ is a nilpotent element of $ A / I $, since \[ \xi^{n} = (x + I)^{n} = x^{n} + I = 0 + I = 0_{A/I} \]By hypothesis, $ \xi = 0 $, which means that $ x \in I $.
Now, let $ \mathfrak{p} $ be a prime ideal of $A$. Then the quotient $ A / \mathfrak{p} $ is an integral domain, so it contains no non-zero nilpotent elements. By the above, $ \mathfrak{p} $ is a radical ideal.
Recall that the radical $ \sqrt{\mathfrak{a}} $ of an ideal $ \mathfrak{a} $ of a ring $A$ is defined as follows \[ \sqrt{\mathfrak{a}} = \left\{ x \in A \ \big| \ x^{n} \in \mathfrak{a} \text{ for some positive integer } n \right\} \]Papapetros Vaggelis wrote: 1. The ideal \(\displaystyle{I}\) is radical (\(\displaystyle{I=\sqrt{I}}\)) if, and only if, the ring
\(\displaystyle{A/I}\) has not nonzero nilpotents elements.
So, the prime ideals, are radical.
$ " \implies " $ : Consider $ x \in A / I $ such that $ x^{n} = 0 $ for some positive integer $n$. We will show that $ x = 0 $. Since $ x = a + I $ for some $a \in A$, we have that \[ x^{n} = 0 \implies a^{n} + I = 0 + I \implies a^{n} \in I \]By definition of the radical of $I$, we conclude that $ a \in \sqrt{I} = I $ (by hypothesis). This means that $ x = 0 $.
$ " \impliedby " $ : By definition of the radical, it is clear that $ I \subseteq \sqrt{I} $. We will show that $ \sqrt{I} \subseteq I $. Consider $ x \in \sqrt{I} $. Then there exists a positive integer $n$ such that $ x^{n} \in I $, so $ x^{n} + I = 0 + I $ in the quotient $ A / I $. Set $ \xi = x + I $. Then $ \xi $ is a nilpotent element of $ A / I $, since \[ \xi^{n} = (x + I)^{n} = x^{n} + I = 0 + I = 0_{A/I} \]By hypothesis, $ \xi = 0 $, which means that $ x \in I $.
Now, let $ \mathfrak{p} $ be a prime ideal of $A$. Then the quotient $ A / \mathfrak{p} $ is an integral domain, so it contains no non-zero nilpotent elements. By the above, $ \mathfrak{p} $ is a radical ideal.
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