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

 A: 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. 
A: 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.
A: 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.
A: Here's another hint, in case you cannot use that Artinian is equivalent to Noetherian and of dimension $0$.
The ideal $(f,g)$ has a finite set of common zeros, which we number $p_1, \ldots, p_m,$ and so its radical is equal to $\bigcap_{i=1}^m {\mathfrak m}_ i= {\mathfrak m}_1 \cdots {\mathfrak m}_m$ (Chinese Remainder Theorem, see e.g. Atiyah-Macdonald).  We now consider the following formula: for a Noetherian ring $A$, the fact that $I$ and $J$ have the same radical implies that $I^a\subset J$ and $J^b \subset I$.  Applying this to the above we obtain:
$$\bigcap_{i=1}^m {\mathfrak m}_ i \subset (f,g),$$ which yields an epimorphism
$$\frac{k[x,y]}{\bigcap_{i=1}^m {\mathfrak m}_ i}\twoheadrightarrow A=\frac{k[x,y]}{(f,g)},$$
thus showing the target ring to be a finite-dimensional vector space over $k$, hence Artinian.
Regarding one of the above responses, the exact Bézout-type identities can be produced by means of the resultant, see e.g. Walker's Algebraic Curves or Fulton's book of the same title, or Queysanne's book on Basic Algebra.
By using resultants one achieves precise computations on the dimension over $k$ of $A$ as a vector space (which is the same as its length as an Artinian module over itself), after learning some geometric data (see e.g. Fulton's book).
