# Find an ideal in $K[x,y]$ that is maximal but not principal.

Let $K$ be a field. Find an ideal of $K[x,y]$ that is maximal but not principal. Prove your claims.(Here we are working in a commutative ring with $1\neq 0.$)

My idea:

Choose $K=\mathbb{Q}.$ Then we claim that an ideal $I\subset K[x,y]$ which is maximal but not principal is $I=(x,y)$.

First I will prove that $(x,y)$ is a prime ideal of $\mathbb{Q}[x,y]$. Let $a,b\in (x,y)$ be such that $p=ab\in (x,y)$. Any element in $(x,y)$ is going tobe of the form: $p=Ax+By$ where $A,B\in \mathbb{Q}[x,y].$ If $a'b'=0$ where $a',b'$ are constant terms of $a$ and $b$ respectively then since $\mathbb{Q}[x,y]$ is an integral domain either $a'=0$ or $b'=0.$ Hence either $a$ or $b$ is of the form $Ax+By$ where $A,B\in \mathbb{Q}[x,y]$ which in turn means that either $a\in (x,y)$ or $b\in (x,y)$.

Now since $(x,y)$ is a prime ideal , $\mathbb{Q}[x,y]/(x,y)$ is an integral domain. Now if we can show that $\mathbb{Q}[x,y]/(x,y)$ is a field then we are done. Not sure how to show this. If $1\in \mathbb{Q}[x,y]/(x,y)$ then we can conclude that it is a field right? So, observe that $1-f(x,y)\in \mathbb{Q}[x,y]$ and $f(x,y)\in (x,y)$ then $1-f(x,y)+f(x,y)=1\in \mathbb{Q}[x,y]/(x,y)$.

$(x,y)$ is not principal in $\mathbb{Q}[x,y]$. Observe that $(x,y)=\{xp(x,y)+yq(x,y)| p(x,y),q(x,y)\in \mathbb{Q}[x,y]\}$. Assume by way of contradiction that $(x,y)=(a(x,y))$ for some $a(x,y)\in \mathbb{Q}[x,y].$ Since $x\in (a(x,y))$ there must be $p(x,y)$ such that $x=p(x,y) a(x,y).$ Since degree $x$= degree $p(x,y)$+degree $a(x,y)$ we conclude that $p(x,y)$ must be a constant polynomial. I don't know how to continue from here.

Also, if there is a simple proof for this problem please share it with me. Thank you.

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First, your solution should work for every field, not a particular one, so you should not "choose $K=\mathbb{Q}$.

Second: it's easier to prove that $(x,y)$ is a maximal ideal by showing that $K[x,y]/(x,y)$ is a field (which is very easy).

Third: to show that $(x,y)$ is not principal, note that if $(x,y)=(a)$, then $a$ divides $x$ and $a$ divides $y$. What are the elements that divides $x$ and what are the elements that divide $y$? The only elements that divide $x$ are nonzero constants and nonzero constant multiples of $x$; the only elements that divide $y$ are nonzero constants and nonzero constant multiples of $y$. So then $a$ would have to be a nonzero constant, but then $(a)=(1)$, which is impossible since $(x,y)\neq (1)$.

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First, you are probably being asked to show this for any field $K$, rather than some particular field $K$. But the proof should be the same for general $K$ as when $K=\mathbb Q$.

Second, you can show $(x,y)$ is maximal (hence prime) by noting that $K[x,y]/(x,y)\cong K$, as each $k+xp(x,y)+yq(x,y)\in K[x,y]$ is mapped to the equivalence class of $k$ when you pass to the quotient, and for $k\neq k'$ the equivalence classes of $k$ and $k'$ are distinct (prove this, and don't over-think it).

Third, to see that $(x,y)$ is not principal, suppose $(x,y)=(f)$. Then there is some $p$ such that $x=fp$, and since $x$ has degree $1$, $f$ has either degree $1$ (in which case it is of the form $kx$ for $k\in K$) or is a unit. Similarly, $y=fq$ so $f$ has either degree $1$ (in which case it is of the form $k'y$ for $k'\in K$) or is a unit. Since $f$ cannot both have the form $kx$ and $k'y$, we conclude $f$ is a unit. But then $1\in (f)=(x,y)$, a contradiction (show this, again it is easy). Hence $(x,y)$ is not principal.

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Hint $\$ If $\rm\,(x,y) = (g)\,$ then $\rm\:g = gcd(x,y).\,$ But $\rm\:x\:$ is irreducible (by degree or primality), hence $\rm\,g= gcd(x,y) = x\,$ or $1,\,$ up to units. But $\rm\,g = 1\:$ $\Rightarrow$ $\rm\,(x,y) = (g) = (1),\,$ contra $\rm\,(x,y)\,$ maximal. Therefore $\rm\,g=x,\,$ hence $\rm\,x\:|\:y,\,$ which implies $\,0\:|\: 1\,$ by evaluating at $\rm\,x,y=0,1,\,$ contradiction.

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