MHB Proof Noetherian Ring: $M^2 \ne M$

[Amer]

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

 

[Euge]

Hi Amer, is it assumed that $R$ is a commutative?

 

[Amer]

Hi Amer, is it assumed that $R$ is a commutative?

yea commutative with 1

 

[Euge]

Ok. The proof you've provided is incorrect. Here's something to keep in mind. If $M^2 = M$, then since $M$ is finitely generated, Nakayama's lemma gives an $e \in M$ for which $m = em$ for all $m \in M$. In particular, $e^2 = e$, i.e., $e$ is idempotent. Furthermore, $M = (e)$.