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.