Ideals of $\mathbb{C}[x,y]$ contains $x^2 + y^2 - 5$ and $xy - 2$? This is a question from Artin's algebra Chapter 11, 9.8. I am trying to consider the quotient ring $\frac{\mathbb{C}[x,y]}{(x^2 + y^2 - 5, xy - 2)}$ which imposes the relation $x^2 + y^2 - 5 = 0$ and $xy - 2 = 0$ to the ring. I think there are 4 solutions to this system and the ideals containing $(x^2 + y^2 - 5, xy - 2)$ can be reverse constructed by picking one or more solutions to this system which gives 16 ideals in total (is this correct or there are repetitions?). Am I on the right track? Or am I completely in the wrong direction?
Note that the ideal does not have to be maximal.
 A: Comment: "I think there are 4 solutions to this system and the ideals containing (x2+y2−5,xy−2) can be reverse constructed by picking one or more solutions to this system which gives 16 ideals in total."
Answer: Let $A:=\mathbb{C}[x,y]/(x^2+y^2-5,xy-2)$ and let $k:=\mathbb{C}$.
Let $I$ be the system $x^2+y^2-5=0, xy-2=0$.
Given a solution $(a,b)\in \mathbb{C}^2:=k^2$ to the system $I$, you get a map of $k$-algebras
$$\phi_{a,b}:A \rightarrow k$$
defined by $\phi_{a,b}(x):=a, \phi_{a,b}(y):=b$. the map $\phi_{a,b}: A \rightarrow k$
is surjective hence its kernel $\mathfrak{m}_{a,b}:=(x-a,y-b) \subseteq A$ is a maximal ideal.
Conversely given a maximal ideal $\mathfrak{m} \subseteq A$ it follows the quotient
$A/\mathfrak{m}$ is a finite extension of $k$ - this is the "nullstellensatz". Hence since $k$ is algebraically closed it follows $p:A/\mathfrak{m}\cong k$ is an isomorphism. Let $a:=p(x), b:=p(y)$ it follows $(a,b)$ is a solution to the system $I$ and $\mathfrak{m}=(x-a,y-b)$. Hence there is a 1-1 correspondence between solutions to $I$ and maximal ideals in $A$. As you have observed: There is 4 solutions to the system $I$, hence there should be 4 maximal ideals in the ring $A$ - each corresponding to a solution $(a,b)$ of $I$. If $\mathfrak{m}_1,..,\mathfrak{m}_4$ are the maximal ideals in $A$ it follows any product $\mathfrak{m}_i\mathfrak{m}_j$ is a non-zero non-maximal ideal in $A$.  In order to determine all ideals in $A$ you must check explicitly what happens to higher powers $\mathfrak{m}_i^k\mathfrak{m}_j^l$ for $l+k > 2$ or triple products $\mathfrak{m}_i\mathfrak{m}_j\mathfrak{m}_k$. You may be aware of the Chinese remainder theorem (Atiyah-Macdonald Proposition 1.10): It says that if $I:=\cap \mathfrak{m}_i=(0)$ there is an isomorphism
$$ A \cong A/I \cong \oplus_i A/\mathfrak{m}_i \cong \mathbb{C}^4.$$
Hence if this is the case you can easily count the number of non-trivial ideals in $A$. You should check if $I=(0)$ holds in this case.
Example: Let us assume $I=(0)$ and let $A:=k \oplus k \oplus k\oplus k$
with $e_1:=(1,0,0,0),..,e_4:=(0,0,0,1)\in A$.
You get nontrivial ideals in $A$:  $(e_i), (e_i,e_j), (e_i,e_j,e_k)$ for all choices of $i,j,k$. Note that $e_i^2=e_i$. It seems to me yo get 14 non-trivial ideals but you should check this yourself.
