I have the ring $R = \frac{k[x,y]}{\langle x^2\rangle}$ where $k$ is a field. But I'm having some trouble understanding exactly what this means. So, as I understand it, in general the ring $R = \frac{R}{I}$ would be the set of all equivalence classes $[f]$ such that $f \in R$.
So in this specific case, would $[f]$ be in the form of all $f(x^2)$? Further, would this mean that $f(x^2) = 0?$
Any suggestions would be appreciated, thank you.
Well, $k[x,y]$ is the ring of all polynomials in the (commuting) variables $x$ and $y$. Modding out by $\langle x^2\rangle$ means that in all of those elements, we're setting $x^2 = 0$. So, given any $f(x,y) \in k[x,y]$, we kill every term with $x^k$, $k \geq 2$, and regroup the remaining terms according to whether they have an $x$ factor or not. We conclude that the elements in $k[x,y]/\langle x^2\rangle$ are of the form $a(y) + b(y)x$, where $a(y),b(y) \in k[y]$. That is to say, any element in $k[x,y]/\langle x^2\rangle$ can be represented by a unique element in $k[x,y]$ of the previously given form.