Given a prime ideal $I(V)\subset k[x_1,\ldots, x_n]$, where $V$ is some affine curve (living in $\mathbb{A}_k^n$) and a maximal ideal $m=(x_1 - a_1,\ldots, x_n - a_n)$, where $(a_1,\ldots, a_n)$ lies on the curve is, it true that $m$ strictly contains $I(V) + m^2$?
I believe this must be the case because the dimension of $m/(I(V) + m^2)$ as a $k$-vector space turns out to be the dimension of the cotangent space of the curve at the point $(a_1,\ldots, a_n)$, which I suspect must always be at least 1?
And if this is true, for what rings does this generalize to (i.e. perhaps the statement that if $\mathfrak{p}$ is a prime strictly contained in a maximal ideal $\mathfrak{m}$ that $\mathfrak{m}$ strictly contains $\mathfrak{p} + \mathfrak{m}^2$)?
For the sake of completeness, here's a solution that works, I think.
If our assumption is that our ring $A$ is Noetherian and that $m=p+m^2$, we know that $mA_m=pA_m+(mA_m)(mA_m)$ and $A_m$ is also Noetherian and local, so it follows that $mA_m$ is the Jacobson radical of $A_m.$.
Then, Nakayama tells us that $mA_m$ and $pA_m$ are the same. Moreover, using the bijection of prime ideals of $A_m$ and those that don't intersect the associated multiplicative set of $A_m$, it follows that $m$ and $p$ are actually the same.