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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.

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

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  • $\begingroup$ So would that mean that $f(x^2)$ has to be equal to zero ? $\endgroup$ – Masha May 3 at 5:28
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    $\begingroup$ 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 $f(x^2,y) = 1 \neq 0$. $\endgroup$ – Ivo Terek May 3 at 5:29
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    $\begingroup$ The equivalence class of an arbitrary polynomial $f(x,y)$ is not zero, but indeed we have $[x^2] = [0]$ 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? $\endgroup$ – Ivo Terek May 3 at 5:40
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    $\begingroup$ 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 $[x^2]=[0]$ kills the last term in the left side. $\endgroup$ – Ivo Terek May 3 at 5:43
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    $\begingroup$ Got it, thank you! $\endgroup$ – Masha May 3 at 5:44

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