# The maximal ideal in a local ring is finitely generated

Assume $m<R$ is the maximal ideal of a commutative local ring with identity, such that $m=m^2$. Is $m$ finitely generated? Is the condition $m=m^2$ redundant?

I am trying to apply Nakayama's lemma to the maximal ideal $m$, but I can't choose a finite generating system for it.

• The condition $m = m^2$ is certainly not redundant. Just take any of the standard examples to see this. – Tobias Kildetoft Feb 4 '14 at 14:13
• Perhaps I should elaborate that by "not redundant" I mean that $m = m^2$ is not automatic in local rings, and by "standard example" I mean the localization of the integers at a prime. – Tobias Kildetoft Feb 5 '14 at 8:31

Let $k$ be a field and $A=k\times k\times ...$ the product of denumerably many many copies of $k$.
Let $I\subset A$ be the ideal of eventually zero sequences and $\mathfrak m\supset I$ a maximal ideal containing it.
Since in $A$ every element $a$ is multiple of $a^2$, we certainly have $\mathfrak m=\mathfrak m^2$ but $\mathfrak m$ is not finitely generated: else it would be generated by an idempotent ( by Nakayama ).

Edit
Since the OP has edited his question, requesting an example with a local ring, here is such an example.

Consider the domain $A=\mathbb Q[X^{1/n}|\; n=1,2,\cdots]$ consisting of "polynomials" over a field $k$ with positive rational exponents, and its maximal ideal $M=\langle X^{1/n}|n=1,2,\cdots\rangle\subset A$.
Obviously $M=M^2$.
If we now localize at $M$ we get the required local ring $R=A_M$, with maximal ideal $\mathfrak m=MA_M$.

Indeed, $\mathfrak m=\mathfrak m^2$ is clear and that ideal is not finitely generated: the simplest argument is again that if it were, it would be generated by a single idempotent element (Nakayama).
But this is impossible, because $R$ is a domain and thus has only $1$ and $0$ as idempotents.

• I changed the brakets to $M=\langle\cdots\rangle$ to denote the ideal generated by the elements. – superAnnoyingUser Feb 5 '14 at 9:39
• @Student Why? You should not edit to impose your own stylistic preferences on others. Also, using round brackets for ideals is quite standard in ring theory. – Tobias Kildetoft Feb 5 '14 at 9:59
• I did it because I thought it is the conventional notation. I had never seen anyone use round brackets before, Mr. @Kildetoft. If you wish, you will roll back to the original. – superAnnoyingUser Feb 5 '14 at 10:42

Georges has fully answered the question, but I want to point out another example familiar from number theory. Let $\mathbf C_p$ denotes the $p$-adic complex numbers (i.e. the completion of the algebraic closure of $\mathbf Q_p$). Its ring of integers $\mathcal O_{\mathbf C_p}$ is a local ring with maximal ideal $\mathfrak m$ which satisfies $\mathfrak m = \mathfrak m^2$ (because every element of $\mathfrak m$ has a square root in the algebraically closed field $\mathbf C_p$, which necessarily also lies in $\mathfrak m$). On the other hand, $\mathfrak m$ is not finitely generated, because it contains elements of arbitrarily small absolute value.