$\mathbb C[x_1,\ldots,x_n]/I=\mathbb C\times\cdots\times\mathbb C$. 
Let $A=\mathbb C[x_1,\ldots,x_n]/I$ and for every $y\neq 0$, we have $y^2\neq 0$ and $\dim A=0$. I would like to prove that $A=\mathbb C\times\cdots\times\mathbb C$.

Attempt of a solution
$\dim A=0$ and $A$ Noetherian $\implies$ $A$ Artinian $\implies$ there are maximal ideals $\mathfrak m_1\ldots \mathfrak m_n$. Then
$$A=A/(0)=A/Nil(A)=A/\cap\mathfrak m_i=A/\Pi\mathfrak m_i=(A/\mathfrak m_1)\times\ldots\times(A/\mathfrak m_n)$$
the last isomorphism is due by chinese remainder theorem.
I need help to finish this proof, I found strange this solution because I didn't use the fact $y\neq 0\implies y^2\neq 0$.
Is there a simpler solution?
I really need help
Thanks in advance
 A: The $y\neq 0 \implies y^2\neq 0$ says $A$ has no non-zero nilpotents, which is true if and only if $I$ is a radical ideal. Hence $A$ is the coordinate ring of the algebraic set $V(I),$ which is a finite set of points since the dimension of $A$ is $0.$ Therefore $A \cong \mathbb{C}\times \mathbb{C} \times \ldots \times \mathbb{C}$ where the number of copies is the number of points in $V(I).$

Added: Some words about why $A \cong \mathbb{C}\times \mathbb{C} \times \ldots \times \mathbb{C}$. Intuitively, we should expect this: $A$ is the ring of polynomial functions restricted to $V(I),$ and such a function is basically determined by its values at those points. 
Suppose $V(I) = \{ x_1, \ldots, x_n\}.$ Define $\phi: A \to \mathbb{C}^n$ by $p + I \mapsto (p(x_1) \ , \ p(x_2)\ , \ \ldots \ , \ p(x_n)  ).$
Check that this map is well defined (does not depend on the coset representative $p$ from $p+I$), and a ring homomorphism. 
It's not hard to check it's injective: If a polynomial $p$ is such that $(p(x_1) \ , \ p(x_2)\ , \ \ldots \ , \ p(x_n)  ) = (0,0, \ldots, 0)$ then $p\in I(V(I)) = I$ so $p+I=0+I.$
The only thing left is to show that it is surjective: We must show that for any $n$-tuple of complex numbers $(a_1,\ldots, a_n),$ we can find a polynomial function such that $p(x_i)=a_i.$ 
Lemma: If $W_1,W_2$ are algebraic sets, then $W_1=W_2$ if and only if $I(W_1)=I(W_2).$ Proof: One direction is obvious, the other follows from the fact $V(I(V(J))) = V(J).$
Let $W_i = V(I)\setminus \{ x_i \}.$
From the lemma, we see that there exists a polynomial $f_i$ that vanishes on $W_i$ and is non-zero at $x_i.$ Multiplying by a scalar, we can assume $f_i(x_i)=1.$ Then the polynomial $p(x)=\sum_{i=1}^n a_i f_i(x)$ is such that $p(x_i)=a_i,$ as required. 
