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Let $k$ be a field and consider the ideal $I=(x,y) \subset k[x,y]$. Am I correct in saying that $(x,y)/(x,y)^{2}$ is generated as a $k$-vector space by the class of $x$ and $y$?

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up vote 2 down vote accepted

Yes. We have that $$(x,y)=\{f\in k[x,y]\mid f \text{ has no terms of degree}\leq0\}$$

and that $$(x,y)^2=(x^2,xy,y^2)=\{f\in k[x,y]\mid f \text{ has no terms of degree}\leq1\}$$

so that any $f\in(x,y)$ is equivalent modulo $(x,y)^2$ to one having terms only in degree 1, i.e. an element of $(x,y)$ of the form $ax+by$.

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