Determine all the ideals of $\mathbb{Q}[x,y]$ that contain the ideal $\langle x^2,y^2,xy\rangle$.

Question

Determine all the ideals of $\mathbb{Q}[x,y]$ that contain the ideal $I=\langle x^2,y^2,xy\rangle$.

Are these all: $$\langle x,y\rangle$$ $$I$$ $$\mathbb{Q}[x,y]$$

?

EDIT.... $$\langle x^2,y \rangle $$

$$\langle x,y^2\rangle$$

Are in the list too.

Answer 1

Hint: $I$ consists of polynomials with degree 2 terms and higher. So if $J$ is a strictly larger ideal, it contains a polynomial of the form $ax+by+c$. Consider the vector space $K = J \cap \left( \mathbb Q x + \mathbb Qy + \mathbb Q\right)$; it is of dimension at most $3$ over $\mathbb Q$.

If $\dim K = 0$, then $J=I$. If $\dim K = 3$, then $J = \mathbb Q[x,y]$.
Suppose $\dim K \geq 1$ and $ax+by+c \in K$. Multiply by $x$ and $y$ to get $cx, cy \in K$. If $c \neq 0$, then $1,x,y \in K$ and $\dim K = 3$.
So suppose $J$ contains only polynomials without constant term. Then $\dim K \leq 2$, and in fact every vector subspace $K \leq \mathbb Qx + \mathbb Qy$ gives an ideal: $$J = K + I$$ and this formula gives a one-to-one correspondence between ideals containing $J$ and subspaces $K \leq \mathbb Q x + \mathbb Qy + \mathbb Q$ with either $K = \mathbb Q x + \mathbb Qy + \mathbb Q$ or $K \leq \mathbb Qx + \mathbb Qy$. Concretely, we get: $$I,\mathbb Q[x,y], \langle x, y \rangle$$ and $$\langle ax+by, x^2,xy,y^2 \rangle$$ for all $a,b$; and $(a,b)$ is uniquely determined up to a scalar.

Answer 2

Because of that $\;xy\;$ term in $\;I\;$ , clearly any ideal $\;J\;$ containing it must have generators with maximal relative degree one in $\;x\;$ or in $\;y\;$ ( if all the generators of $\;J\;$ have $\;\deg_y\ge2\;$ say, then the element $\;xy\in I\;$ is not in $\;J\;$ , and likewise with $\;x\;$ ).

But if $\;p(x,y)\in J\;$ is a generator with $\;\deg_y(p)=n\ge 2\;$ , then we can write

$$p(x,y)=\sum_{k=0}^na_k(x)y^k=a_0(x)+a_1(x)y+a_2(x)y^2+\ldots\in\langle x,\,y\rangle$$.

Try to finish the argument now.