For $\mathfrak{m}$ maximal and principal, there's no ideal between $\mathfrak{m}^2$ and $\mathfrak{m}$ 
Let $R$ be a commutative ring with unity. If a maximal ideal $\mathfrak{m}$ of $R$ is principal, prove that there is no ideal $I$ with $\mathfrak{m}^2\subsetneq I\subsetneq \mathfrak{m}$.

I have no idea how to start this one - I can't see why we can say anything specific about $I$.
 A: Solution $1$.
Suppose $\mathfrak m=(t)$, and $t\notin I$. Let's show that $I\subseteq\mathfrak m^2$. Pick $a\in I$. Then $a=tb$, $b\in R$. If $b\in\mathfrak m$, then $b=tc$ and thus $a=t^2c\in\mathfrak m^2$. Otherwise, $\mathfrak m+(b)=R$, so $1=tx+by$. Then $t=t^2x+tby$, so $t=t^2x+ay\in I$, a contradiction.
A little more general:

Let $(A,\mathfrak n)$ be an artinian local ring and $\mathfrak n=(t)$. If $\mathfrak n^r=0$, and $\mathfrak n^{r-1}\ne0$, then the ideals of $A$ are: $(0)$, $\mathfrak n^{r-1}$, $\dots$, $\mathfrak n$, $A$.

In your case $A=R/\mathfrak m^2$.
Solution $2$.
$\mathfrak m/\mathfrak m^2$ is an $R/\mathfrak m$-vector space, and the subspaces of $\mathfrak m/\mathfrak m^2$ correspond to the ideals $I\subseteq R$ such that $\mathfrak m^2\subseteq I\subseteq\mathfrak m$. Since $\mathfrak m$ is principal we have $\dim_{R/\mathfrak m}\mathfrak m/\mathfrak m^2=1$, so $\mathfrak m/\mathfrak m^2$ has only the trivial subspaces.
A: Hint: you may already know this, but if not show that $\mathfrak m$ is principal if and only if there exists a surjection of $R$-modules $\pi: R \twoheadrightarrow \mathfrak m$. What is $\pi(\mathfrak m)$?
