What are the ideals in ${\Bbb C}[x,y]$ that contain $f_1,f_2\in{\Bbb C}[x,y]$? This question is based on an exercise in Artin's Algebra:

Which ideals in the polynomial ring $R:={\Bbb C}[x,y]$ contain $f_1=x^2+y^2-5$ and $f_2=xy-2$?

Using Hilbert's (weak) nullstellensatz, one can identify all the maximal ideals of $R$ that contain $f_1$ and $f_2$. For the general ideals contain $(f_1, f_2)$, it suffices to identify ideals in $R/(f_1,f_2)$ by the correspondence theorem. But I don't see how to go on. Is there a systematic way to do it?
 A: I don't know if this counts as a "systematic" way of doing it or not, but you can do the following.
First, in the ring $R$ you have $y = 2/x$, so you have the isomorphism $$R = \frac{\mathbb{C}[x,y]}{(f_1,f_2)}\cong \frac{\mathbb{C}[x,x^{-1}]}{(x^2 + 4x^{-2} - 5)}.$$ Observe that you have the factorization of ideals $$x^2 + 4x^{-2} - 5 = (x-2)(x+2)(x-1)(x+1)$$ in the ring $\frac{\mathbb{C}[x,x^{-1}]}{(x^2 + 4x^{-2} - 5)}$. These are pairwise coprime maximal ideals, so the Chinese remainder theorem says that $$\frac{\mathbb{C}[x,x^{-1}]}{(x^2 + 4x^{-2} - 5)}\cong \frac{\mathbb{C}[x,x^{-1}]}{(x-2)}\times \frac{\mathbb{C}[x,x^{-1}]}{(x+2)}\times \frac{\mathbb{C}[x,x^{-1}]}{(x-1)}\times \frac{\mathbb{C}[x,x^{-1}]}{(x+1)}$$$$\cong \mathbb{C}\times \mathbb{C}\times \mathbb{C}\times \mathbb{C}.$$ Unraveling definitions, we see that $R\cong \mathbb{C}^4$, the isomorphism being $$f(x,y)\mapsto (f(2,1), f(-2, -1), f(1,2), f(-1,-2)).$$
Now, to understand what the ideals of $R$ are, you need to know what the ideals of $\mathbb{C}^4$ are. There are $2^4$ ideals of $\mathbb{C}^4$, corresponding to subsets $S\subseteq\{1,2,3,4\}$; namely, the ideals $$I_S = \{(a_1,a_2,a_3,a_4) : a_i = 0\,\,\,\mathrm{ if }\,\,\,i\in S\}.$$ There are therefore $2^4$ ideals of $R$, corresponding to subsets $S\subseteq\{1,2,3,4\}$; namely the ideals $$J_S = \{f : (f(2,1), f(-2,-1), f(1,2), f(-1,-2))\in I_S\}.$$ I hope this analysis works!
A: Here's a procedure that works more generally: the ideal $I := (x^2 + y^2 - 5, xy - 2)$ has height $2$ in $\mathbb{C}[x,y]$. To see this, note that $x^2 + y^2 - 5$ (or $xy - 2$) is irreducible over $\mathbb{C}$ (e.g. by Eisenstein), hence generates a height $1$ prime ideal in $\mathbb{C}[x,y]$, which does not contain the other generator. More generally yet, the generators of $I$ form a regular sequence, and any ideal generated by a regular sequence has height equal to the length of the sequence.
Thus, the quotient $R := \mathbb{C}[x,y]/I$ has Krull dimension $\le \dim \mathbb{C}[x,y] - \text{ht}(I) = 0$, so $R$ is an Artinian ring. Every Artinian ring is a finite product of Artinian local rings, and ideals in a finite product are products of ideals, so writing $R = \prod_{i=1}^n R_i$, the number of $R$-ideals is the product of the numbers of $R_i$-ideals.
In this particular case, $R = R_1 \times \cdots \times R_4$, each $R_i \cong \mathbb{C}$ by the Nullstellensatz, and $\mathbb{C}$ has precisely $2$ ideals, so $R$ has a total of $2^4 = 16$ ideals.
