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Math Help - Radical and Ideal

  1. #1
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    Radical and Ideal

    Let I be an ideal of a commutative ring R with identity. Prove that the radical of I is an ideal of R and each prime ideal that contains I, also contains  \sqrt {I} . Also prove that  \frac { \sqrt {I} }{I} is the nilradical of  \frac {R}{I}

    Proof.

    Now the radical of I is  \sqrt {I} = \{ r \in R : r^n \in I \ , \ n \geq 1 \}

    So pick  i \in \sqrt {I} and  j \in R , so i^n = k \in I , (ij)^n=i^nj^n=kj^n \in I since I is an ideal. Therefore  \sqrt {I} is also an ideal.

    Now, suppose that Q is a prime ideal of R such that  I \subseteq Q . Pick  i \in \sqrt {I} , so then  \exists n such that  i^n \in I So I need to show that  i \in Q ... How should I start?
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  2. #2
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    Quote Originally Posted by tttcomrader View Post
    Proof.

    Now the radical of I is  \sqrt {I} = \{ r \in R : r^n \in I \ , \ n \geq 1 \}

    So pick  i \in \sqrt {I} and  j \in R , so i^n = k \in I , (ij)^n=i^nj^n=kj^n \in I since I is an ideal. Therefore  \sqrt {I} is also an ideal.
    Yes, this is good.

    Now, suppose that Q is a prime ideal of R such that  I \subseteq Q . Pick  i \in \sqrt {I} , so then  \exists n such that  i^n \in I So I need to show that  i \in Q ... How should I start?
    Let R be a commutative (unitary) ring. An ideal P is called a prime ideal iff ab\in P \implies a\in P \text{ or }b\in P.

    Lemma: If P is a prime ideal and a_1\cdot ... \cdot a_k \in P then a_1\in P or a_2\in P or ... or a_k \in P.
    Proof: The proof is by induction. If n=2 that is true because that is the definition of what it means being a prime ideal. So say it is true for n=k. Now say a_1\cdot ... \cdot a_k \cdot a_{k+1} \in P then it means \left( a_1\cdot ... \cdot a_k \right) \cdot a_{k+1} \in P then by property of being a prime ideal means a_1\cdot ... \cdot a_k \in P or a_{k+1} \in P. By the inductive step we get a_1\in P or ... or a_k \in P or a_{k+1}\in  P. And that completes the proof.

    Therefore given Q and ideal containing I let a\in \sqrt{I} then it means a^n \in I \subseteq Q for some n\geq 1. However, a^n = a\cdot ... \cdot a \in Q \implies a\in Q by above.
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