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I am trying to prove the following statement:

Let $k$ be an algebraically closed field, and let $A = k[ X_1, \ldots , X_n]$. If $I$ is an ideal of $A$ or $A$, define $V(I) = \{ \mathbf{x}\in k^n | f( \mathbf{x} ) = 0 \hspace{5pt} \forall \hspace{5pt} f \in I\}$.

Then $V(I) = \emptyset \implies I = A$

Attempt at proof:

Let $k$ be an algebraically closed field, and let $A_n = k[ X_1, \ldots , X_n]$. We proceed by induction on $n$. The base case of $n=1$ follows (with some effort) from $k$ being algebraically closed.

So assume $n>1$. Suppose for contradiction that there exists a proper ideal $I$ such that $V(I)=\emptyset$. Then we may assume $I$ maximal because all ideals are contained in maximal ideals and inclusion on ideals corresponds to reverse-inclusion on $V$.

Consider then $\varphi: A_n \rightarrow A_{n-1}$, such that $X_i\mapsto X_i \hspace{5pt} \forall 1\leq i<n$ and $X_n \mapsto 0$, and $\varphi$ fixes $k$. This is a surjective ring homomorphism, so the image of $I$, call it $I'$, is a maximal ideal of $A_{n-1}$ or $A_{n-1}$ itself. At the same time, $V(I') \hookrightarrow V(I)$ by $(x_1,x_2, ... , x_{n-1}) \mapsto (x_1,x_2, ... , x_{n-1},0)$, so $V(I)= \emptyset \implies V(I')=\emptyset$. Now by our induction hypothesis we have that $I'=A_{n-1}$. Therefore $1 \in I'$, which in turn implies $\exists f \in I$ such that $\varphi(f)=1$, so $f$ must be a polynomial in $X_n$ only. If $f$ is constant then we are done. If not then by the maximality of $I$ and algebraic closedness of $k$, we have that $X_n - c \in I$ for some $c \in k$.

Note also that $c \neq 0$ because then $\varphi(f)=0 \neq 1$.

Now repeat the same argument in the paragraph above except letting $\varphi(X_n)=c$ instead. This yields that $X_n-c' \in I$ for some $c' \neq c$. $I$ is closed under addition so $X_n-c - (X_n - c') = c'-c \in I$. Hence $I=(1)$. Done.

So is this a valid proof? I suspect that it is too simple to be correct, but I cannot find any problem with it.

Thanks very much in advance.

(Edit: resolved edge-case.)

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The error is your claimed inference

. . . $\varphi(f)=1$, so $f$ must be a polynomial in $X_n$ only.

From $\varphi(f)=1$, all you get is $f=gX_n+1$, for some $g\in A_n$.

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