Sum of ideals in the polynomial ring 
Could someone explain to me how to find a sum of ideals where $I=(x+y)$ and $J=(x)$? The answer to this is $I+J=(x,y)$ and we work in the polynomial ring $k[x,y]/(xy)$.

I know that the definition of the sum of ideals is $I+J=\{i+j\mid i\in I,j\in J\}$.
I know that for the ring $\mathbb{Z}$ the sum of ideals is the greatest common divisor of them but how does it work for polynomial rings?
 A: HINT:
Show the two inclusions $I+J\subseteq(x,y)$ and $(x,y)\subseteq I+J$ separately.
The first one should be obvious.
For the second one, all boils down to show that $y$ is in $I+J$, i.e. that $y$ can be recovered from $x$ and $x+y$ via sums and products.
A: I'd just like to add to Andrea's excellent answer:
If $R$ is a PID (that is, principal ideal domain), and if $x,y\in R$, then the sum of the ideals they generate, $(x)+(y)$, is an ideal in $R$. Of course, this ideal is principal so there exists $z\in R$ such that $(z)=(x)+(y)$. You might think of $z$ as the "Greatest Common Divisor" (GCD) of $x$ and $y$ in $R$. (If $R=\mathbb{Z}$, then this agrees with the usual definition of "GCD" up to sign.)
Unfortunately, if $R$ is not a PID, then this construction fails. For example, in the polynomial ring $k[x,y]$ ($k$ is a field), the sum of the ideals $(x)$ and $(y)$ is the ideal $(x,y)$ which isn't principal! You might say that $x$ and $y$ don't have a GCD in $k[x,y]$. So that's why it's not as easy to add two ideals in $k[x,y]$ as it is to add ideals in $\mathbb{Z}$!
The good news is that a polynomial ring $k[x]$ over a field $k$ is a PID so you can speak of the notion of GCD (again, up to multiplication by a unit, i.e., an element of $k$). 
I'll sign off by noting that there are general notions such as that of "Bezout domain" (my apologies to Bezout; I don't know how to add an accent on the "e") and "GCD domain" which generalize these ideas.
I hope this helps!
