Given a field $K$, the two ideals $(x^2 ,xy,y^2)$ and $(x,y)^2$ in $K[x,y]$ are the same.

Question

I want to prove that given a field $K$, the two ideals $(x^2,xy,y^2)$ and $(x,y)^2$ in $K[x,y]$ are the same. For the inclusion $(\subseteq)$, it is obvious that $x^2,y^2 \in (x,y)^2$, but how do I have to show that $xy \in (x,y)^2$? If char$K \neq 2$, then I can use that $2xy=(x+y)^2-x^2-y^2$, but I cannot handle the case char$K=2$.

Answer 1

You even don't need $K$ to be a field. Any commutative ring would work. First of all, as Daniel recalled, $(x,y)^2$ is the ideal generated by $\{a\cdot b\mid a,b\in (x,y)\}$, which, in particular, means that $(x,y)^2\supsetneq \{a^2\mid a\in(x,y)\}$ and that elements in $(x,y)^2$ are of the form $$\sum_{i=1}^t (a_ix + b_iy)(c_ix+ d_iy)$$ for some $t\geq1$, $a_i,b_i,c_i,d_i \in K[x,y]$. Therefore, clearly, $x^2,y^2,x y\in(x,y)^2$ and hence $(x^2,x y,y^2)\subseteq (x,y)^2$. Conversely, $$\sum_{i=1}^t (a_ix + b_iy)(c_ix+ d_iy) = \sum_{i=1}^t a_ic_ix^2 + (b_ic_i + a_id_i)xy + b_id_i y^2\in (x^2,x y,y^2).$$