Let $k[x_1,\ldots,x_n]$ be a polynomial ring, $k$ be an algebraically closed field. Suppose $k[T_1,\ldots,T_m]$ is a finitely generated $k$-subalgebra such that for any proper ideal $I$ of $k[T_1,\ldots,T_m]$, $Ik[x_1,\ldots,x_n]\neq(1)$ .

I want a counter example such that the above is not true. Namely, find a finitely generated $k$-subalgebra $k[T_1,\ldots,T_m]$ of $k[x_1,\ldots,x_n]$ such that there exists a proper ideal $I$ of $k[T_1,\ldots,T_m]$, but the extended ideal $Ik[x_1,\ldots,x_n]=k[x_1,\ldots,x_n]$.

I believe there exists many examples, but I have no one in hand now.

Motivation: Consider $\varphi:\mathbb{R}\to \mathbb{R}$, $x\mapsto x^2$. Then it is clear that $\mathrm{\varphi}$ is not an algebraic set. So I assume $k$ be an algebraically field, and want to know if an image of a polynomial map is an algebraic set.

Let $k$ be an algebraically closed field. Suppose $\varphi:k^n\to k^m$ is a polynomial map defined by $(T_1,\ldots,T_m)$, I want to know if the $\mathrm{Im}(\varphi)$ is always an algebraic set. The polynomial map $\varphi$ induces a ring homomorphism $\widetilde{\varphi}$ from $k[x_1,\ldots,x_m]$ to $k[x_1,\ldots,x_n]$. Let $\mathfrak{p}=\mathrm{Ker}\widetilde{\varphi}$, then $Z(\mathfrak{p})=\mathrm{Im}(\varphi)^-$(the closed closure of $\mathrm{Im}\varphi$). We know that {points in $Z(\mathfrak{p})$} 1:1 correspond {maximal ideals of $k[x_1,\ldots,x_m]/\mathfrak{p}$} 1:1 correspond {maximal ideals $(T_1-b_1,\ldots, T_m-b_m)$ of $k[T_1,\ldots,T_m]$}, then we know $Z(\mathfrak{p})=\mathrm{Im}\varphi$ iff for every maximal ideal $\mathfrak{m}$ of $k[T_1,\ldots,T_m]$, $\mathfrak{m}^e\neq(1)$ in $k[x_1,\ldots,x_n]$.

So I ask the above question.

When I considered one indeterminate polynomial ring, I found that there exist no counter examples. As a consequence, I got an interesting question for linear algebra as following.

Let $f,g\in \mathbb{Q}[x]$, suppose $(f,g)=1$, then there will exist $h(u,v)\in \mathbb{Q}[u,v]$ and $l(u,v)\in \mathbb{Q}[u,v]$ such that $f(x)h(f,g)+g(x)l(f,g)=1$. I will ask how to find $h,l$ ?



Let your larger ring be $k[x,y]$, and your subring be $k[x, xy]$. We will also write the smaller ring as $k[u,v]$, with $u=x$ and $v=xy$.

Consider the ideal $I=\langle u, v-1 \rangle$. Clearly, $k[u,v]/I \cong k$, so $I$ is a proper ideal. However, in $k[x,y]$, we have $$y \cdot u - v = xy-(xy-1)=1.$$

The geometric intuition here is that we are mapping $k^2$ to $k^2$ by $(x,y) \mapsto (x, xy)$. The image is all points where $x \neq 0$, and the point $(0,0)$. So $(0,1)$, which corresponds to our ideal $I$, is in the closure of the image but not in the image.

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    $\begingroup$ thanks. this is really a good example! $\endgroup$ – wxu Jun 23 '11 at 22:37

I'll only consider the very last question, and even that, indirectly.

If $a$ and $b$ are relatively prime integers then there exist integers $r$ and $s$ such that $ar+bs=1$. Clearly $r$ and $s$ depend on $a$ and $b$; we may say $r=r(a,b)$, $s=s(a,b)$, and I'd be surprised if we didn't have $r(u,v)$ and $s(u,v)$ in ${\bf Q}[u,v]$. But the way one actually goes about finding $r$ and $s$ is by applying the Euclidean algorithm to $a$ and $b$, and then back-solving, which is to say, by an algorithm rather than by a formula. I think the point is that $r$ and $s$ depend on $a$ and $b$ and should really be written $r_{a,b}(u,v)$ and $s_{a,b}(u,v)$.

Presumably, what goes for my $a$ and $b$ also goes for your $f$ and $g$.

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