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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

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    $\begingroup$ How did you get $(0)=Nil(A)$? $\endgroup$ – Jyrki Lahtonen Jun 11 '14 at 6:59
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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.

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  • $\begingroup$ Why $A$ is a finite set of points $\implies$ $A=\mathbb C\times\ldots\times \mathbb C$? $\endgroup$ – user108205 Jun 11 '14 at 9:28

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