# Understanding a specific quotient ring

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.

• The typical element of $R$ is a coset $f(x,y)+\left<x^2\right>$ of $\left<x^2\right>$ in $k[x,y]$. – Lord Shark the Unknown May 3 at 5:25

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.
• So would that mean that $f(x^2)$ has to be equal to zero ? – Masha May 3 at 5:28
• You are confusing a polynomial with a polynomial function, maybe. No, this is not the case. For example, take $f(x,y) = 1 \in k[x,y]$. Then $f(x^2,y) = 1 \neq 0$. – Ivo Terek May 3 at 5:29
• The equivalence class of an arbitrary polynomial $f(x,y)$ is not zero, but indeed we have $[x^2] = $ in $k[x,y]/\langle x^2\rangle$, by the very definition of the quotient. I'll rephrase my answer above in the following way: given $f(x,y) \in k[x,y]$, there are unique $a(y),b(y) \in k[y]$ such that $[f(x,y)] = [a(y)+b(y)x]$ in $k[x,y]/\langle x^2\rangle$. Better? – Ivo Terek May 3 at 5:40
• I would avoid using the word "function" here, but deep down this is what quotients are about: when considering $R/I$, you take every element in $I$ and set it equal to zero. In our case, we take polynomials in $k[x,y]$ and set $x^2 = 0$. As a concrete example, $[1 + xy^2 + x^3y^2] = [1+xy^2]$, because $[x^2]=$ kills the last term in the left side. – Ivo Terek May 3 at 5:43