Show that $A\simeq k[x_1,...,x_d]/I$ Let $k$ be a field and $A$ be an Artin local $k$-algebra such that $k\simeq A/M$. Then one fact is that $M/M^2$ is a finite dimensional $k$-vector space. I've  saw that if $A =k[x]/(x^2)$ then $\dim_{k}(M/M^2) =1$ and if $A = k[x,y]/(x^2,y^2)$ then $\dim_k(M/M^2) = 2$, if $A = k[x,y,z]/(x^2,y^2,z^2)$ then $\dim_k(M/M^2) = 3$ ... etc. I wonder if the converse is true i.e.

If $A$ is an Artin local $k$-algebra such that $k\simeq A/M$ and $d:=\dim_k(M/M^2)$ then $A\simeq k[x_1,...,x_d]/I$ for some ideal $I$?

Maybe to prove this, we need proper surjection $k[x_1,...,x_d]\to A$. How can I show this? Is this statement true?

Let $m_1,\dots,m_d\in M$ such that their images in $M/M^2$ generate $M/M^2$ as a $k=A/M$-vector space. Then by Nakayama's Lemma $m_1,\dots,m_d$ generate the ideal $M$. Define the map $f:k[x_1,\dots,x_d]\to A$ by $x_i\mapsto m_i$. We need to show that this is surjective. We show by 'backwards' induction on $n$ that $M^n\subseteq \operatorname{im} f$ for $n\geq0$. As $A$ is Artinian some power $M^m$ of $M$ is zero. This will be the base case. Now for $n+1\to n$ let $x\in M^n$ be an element of the form $am_{i_1}\cdots m_{i_n}$ with $a\in A$ for some (not necessarily distinct) indices $i_1,\dots,i_n\in\{1,\dots,d\}$. By assumption we may write $a = u+m$ with $u\in k$ and $m\in M$. It follows that $$am_{i_1}\cdots m_{i_n}=\underbrace{um_{i_1}\cdots m_{i_n}}_{=f(ux_{i_1}\cdots x_{i_n})\in \operatorname {im} f}+\underbrace{mm_{i_1}\cdots m_{i_n}}_{\in M^{n+1}\subseteq \operatorname{im} f}\in \operatorname{im} f$$
As every element in $M^n$ is the sum of elements in the above form we conclude $M^n\subseteq \operatorname{im} f$ which we wanted to prove.
