$\mathbb{C}[x,y]/(f,g)$ is an artinian ring, if $\gcd(f,g)=1$. [closed]

This problem extends the fact that $\mathbb{C}[x,y]/(x^n,y^m)$ is artinian ring.

Let $f,g \in \mathbb{C}[x,y]$ such that $\gcd(f,g)=1$. Show that $\mathbb{C}[x,y]/(f,g)$ is an artinian ring.

Solution 1:

I've decomposed Alex's argument into a classical version in case you haven't acquired the background for it yet.

Since $f$ and $g$ have no common factor in $\mathbb{C}[x,y],$ we see (by Gauss' lemma) that they have no common factor in $\left(\mathbb{C}(x)\right)[y].$ This is a PID, so there exist rational functions $a(x),b(x)$ such that $a(x) f(x,y) + b(x) g(x,y) =1.$ Clearing denominators, we have $$ \tilde{a}(x) f(x,y) + \tilde{b}(x) g(x,y) = d(x)$$ for some polynomials $\tilde{a}, \tilde{b}, d.$ From this we see that any common root $(x',y')$ of $f$ and $g$ forces $d(x)$ to have $x'$ as a root. Since $d$ has only finitely many roots, the set of common roots of $f$ and $g$ has finitely many $x$ values appearing in it.

Now suppose $x_1, \ldots, x_r$ is an enumeration of the $x$ values in the set of common roots. Consider the function $F = \prod_{i=1}^r (x-x_i).$ This vanishes on $V:=\mathbf{V}(f,g),$ so by Hilbert's Nullstellensatz there is some $N$ such that $F^N \in (f,g).$ Isolating the highest power of $x$ in the expansion of $F^N,$ we see that $x^{rN}+(f,g)$ is a linear combination of lower powers $x^i+(f,g).$ By induction it follows that all powers of $x+(f,g)$ are generated by finitely many elements of $\dfrac{\mathbb{C}[x,y]}{(f,g)}.$

The exact same logic applies when we switch the roles of $x$ and $y$, so $\dfrac{\mathbb{C}[x,y]}{(f,g)}$ is a finitely generated $\mathbb{C}$-module, which implies that it is Artinian.

Solution 2:

Here's another possible hint. Since $(f,g)=1$, you can see that $V(f,g)$ is a proper closed subset of $V(f)$. Since $V(f)$ is one-dimensional and Noetherian, you can decompose $V(f,g)$ into a finite product of points. So, then $\text{Spec}(\mathbb{C}[x,y]/(f,g))\to \text{Spec}(\mathbb{C})$ is quasifinite which, since $\mathbb{C}$ is a field forces it to be finite by (alternatively, just think of the Nullstellensatz). And, of course, finite dimensional $\mathbb{C}$-algebras are Artinian.

Solution 3:

Hint. The Krull dimension of a noetherian ring drops by at least one if specialize by a nonzero divisor, that is, $\dim R/xR\le\dim R-1$ provided $x$ is a nonzero divisor (and a non-unit, of course). (However, the equality holds for affine domains over any field.)

Added Later. A similar reasoning can be found in this answer.