Is every Artinian local ring a quotient of a nilpotent algebra over a field?

This question arose from a conversation with a friend where we tried to classify all one-point-schemes. Apologies if it's a totally stupid question.

If $R$ is a Noetherian commutative ring with identity and $\text{Spec } R$ has $1$ point, what is $R$? Since $R$ has exactly $1$ prime ideal, the ideal must be maximal: call it $\mathfrak{m}$ and let $k:= R/\mathfrak{m}$. Pick generators $u_1,u_2,\ldots,u_n$ of $\mathfrak{m}$ - they are obviously nilpotent and so we can choose $a_1, a_2, \ldots, a_n$ minimally so that $u_i^{a_i} = 0$.

I'm trying to construct a surjective map from $\displaystyle\frac{k[x_1,\ldots,x_n]}{(x_1^{a_1},x_2^{a_2},\ldots,x_n^{a_n})}$ to $R$ by sending $x_i$ to $u_i$, but I don't know if $R$ is a $k$-algebra. Obviously local rings don't have to be algebras over their residue fields (mixed characteristic being the natural example) but is this true in our more restricted setting? If not, can somebody come up with a counterexample?

Note: $R$ doesn't have to be Noetherian since we can consider $\text{ Spec } \displaystyle\frac{k[x_1,\ldots,x_n, \ldots]}{(x_1^{2},x_2^{2},\ldots,x_n^{2}, \ldots)}$. I'd be curious if people can address the non-Noetherian situation as well.

Edit 1: Obvious flaw pointed out. Assuming that $k$ has characteristic $0$, can the question be salvaged?

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What do you make of $\mathbb Z/4$? – Mariano Suárez-Alvarez Oct 11 '13 at 2:24
If I insist that $k$ has characteristic $0$, does that change things? – Ashwath Rabindranath Oct 11 '13 at 2:58
I suggest that you read about the Cohen Structure Theorem. – Jason Starr Oct 11 '13 at 3:21
Thanks for this answer, it was really helpful! – Ashwath Rabindranath Oct 11 '13 at 16:43