My first thought was $(2,x,y)$. But as in this post, Show that any ideal in $\mathbb{C}[x,y]$ containing $y$ can be generated by $2$ elements, $(2,x,y)$ would be generated by 2 elements. Does anyone know how to prove the statement in general? I am practicing for my qualifying exam next month. I think the problem is designed to be solved without using knowledge from algebraic geometry.
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2https://math.stackexchange.com/questions/1622081/ideal-of-mathbbcx-y-not-generated-by-two-elements – hunter Aug 12 '20 at 02:47
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1The ideal $(2,x,y)$ is generated by one element because $(2,x,y)=(1)$. – pancini Aug 12 '20 at 02:51
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Take $I=(X^2,XY,Y^2)\subset \mathbb Q[X,Y]$. Since $I$ is homogeneous, if $f(X,Y)\in I$, the homogeneous components of $f$ also belong to $I$. If $I$ could be generated by $2$ elements, $f_1,f_2\in I$, then $f_i$ has no homogeneous component of degree $0$ or $1$. So the homogeneous elements of degree $2$ in $I$ can be generated by $2$ elements as a $\mathbb Q$-vector space. But $\dim_\mathbb Q I\cap \{aX^2+bXY+cY^2 \ : \ a,b,c \in \mathbb Q \}=3$. So we get a contradiction.
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1Not sure this adds anything, but it helps me to picture $\Bbb Q[x,y]$ as a graded ring. Then the short version of the story is that, if $f_1,f_2$ generate $I$ over $\Bbb Q[x,y]$, then $f_1$ and $f_2$ generate the grade-$2$ component of $I$ over $\Bbb Q$. – pancini Aug 12 '20 at 03:02