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Can you check my Work or mention a link for a proof.
Let $R$ be Noetherian ring. Then if $M$ is a maximal ideal in $R$. Prove that $M^2 \ne M$.
Proof:
Since $R$ is Noetherian ring then $M$ is finitely generated. Thus $M = (a_1, a_2 , \cdots, a_k)$ we can choose the $a_i's $ which are minimal in the sense that $M$ can't be generated if we ignore one of the $a_i's$. Also that means one $a_i$ can be generated by other $a's $.
Suppose on the contrary that $M^2 = M$. Hence $a_1 \in M^2 $. But elements of $M^2$ has the form $\displaystyle \sum_{i=1}^{n} k_i a_j a_h$ for some $n$
Therefore
$\displaystyle a_1 = \sum_{i=1}^{n} k_i a_j a_h $ the right hand side is divisible by $a_1$ since the left hand side is. and since $a_1$ can't be generated by other $a_i's$ that means $a_1$ should appear on terms of the right hand side moving these terms to the left and factoring $a_1$ we will get someone like this
$\displaystyle a_1 ( 1 - \sum_{i=1}^{n} c_i a_i) = \sum_{i=1}^{m} k_i a_j a_h ,~~~ i, j \ne 1$
I think the right hand side is equal to zero if it is then we will get $1 = \sum_{i=1}^{n} c_i a_i $, it follows that $1 \in M$ a contradiction. But is it zero ?. if it is not then how i can prove it.
Thanks
Let $R$ be Noetherian ring. Then if $M$ is a maximal ideal in $R$. Prove that $M^2 \ne M$.
Proof:
Since $R$ is Noetherian ring then $M$ is finitely generated. Thus $M = (a_1, a_2 , \cdots, a_k)$ we can choose the $a_i's $ which are minimal in the sense that $M$ can't be generated if we ignore one of the $a_i's$. Also that means one $a_i$ can be generated by other $a's $.
Suppose on the contrary that $M^2 = M$. Hence $a_1 \in M^2 $. But elements of $M^2$ has the form $\displaystyle \sum_{i=1}^{n} k_i a_j a_h$ for some $n$
Therefore
$\displaystyle a_1 = \sum_{i=1}^{n} k_i a_j a_h $ the right hand side is divisible by $a_1$ since the left hand side is. and since $a_1$ can't be generated by other $a_i's$ that means $a_1$ should appear on terms of the right hand side moving these terms to the left and factoring $a_1$ we will get someone like this
$\displaystyle a_1 ( 1 - \sum_{i=1}^{n} c_i a_i) = \sum_{i=1}^{m} k_i a_j a_h ,~~~ i, j \ne 1$
I think the right hand side is equal to zero if it is then we will get $1 = \sum_{i=1}^{n} c_i a_i $, it follows that $1 \in M$ a contradiction. But is it zero ?. if it is not then how i can prove it.
Thanks