Why $(x)$ is a prime non-maximal ideal in $K[x,y]$ where $K$ is a field?

Question

The ring of poynomials $K[x,y]$, $K$ a field is not a principal ideal domain. I want to show that in this ring non-zero prime ideals need not be maximal.

I am having difficulties in showing that $(x)$ is prime but it is not maximal.

Let $f(x)$, $g(x)$ be two elements of $K[x,y]$ such that $f(x)g(x)$ is in $(x)$. Then there exists $h(x)$ in $K[x,y]$ such that $f(x)g(x)=xh(x)$. So, $x$ divides the product $f(x)g(x)$.

How can I arrive in the conclusion that either $f(x)$ or $g(x)$ is in $(x)$?

Would you help me, please?

Answer 1

The ring homomorphism $K[x,y] \to K[y], f(x,y) \mapsto f(0,y)$ induces an isomorphism $K[x,y]/(x) \cong K[y]$. Now $K[y]$ is a domain, but not a field, thus $(x)$ is prime, but not maximal.

Answer 2

The answer of MatheiBoulomenos exploits the equivalence "$P$ prime $\iff$ $A/P$ domain" and is certainly correct.

But you can prove that $(x)$ is prime directly from the definition as follows.

Let $P(x,y)$ and $Q(x,y)$ be two polynomials with $$ P(x,y)Q(x,y)\in(x).\qquad\qquad(*) $$

Write $P(x,y)=p(y)+xP_1(x,y)$ and $Q(x,y)=q(y)+xQ_1(x,y)$ separating the monomials that do not contain $x$. Then $$ P(x,y)Q(x,y)=p(y)q(y)+x[P_1(x,y)q(y)+Q_1(x,y)p(y)+xP_1(x,y)Q_1(x,y)] $$ and the hypothesis $(*)$ implies that $p(y)q(y)=0$. But since $K[y]$ is a domain this implies that, for instance, $p(y)=0$ which means that $P(x,y)\in(x)$.

Answer 3

An alternate argument to show that $(x)$ is not maximal:

$$(x)\subsetneq (x,y)\subsetneq K[x,y]$$

Answer 4

Hint $ $ Compare the residues $K[x,y]/I$ of the ideals $I$ in the chain below.

$\begin{align} \color{#b0f}{(x)}&\subseteq \, (x,y)\subseteq (1)\quad\ \rm [ideals]\\[.3em] \color{#b0f}{K[y]} &\supset \ \ \ K\ \,\supset\ (0)\quad\ \rm [residues]\\[.5em] {\rm Hence}\ \ K[x,y]/(x)=\color{#b0f}{K[y]}&{\rm\,\ is\ a\ \ \color{#c00}{nonfield}\ \ \color{#0a0}{domain}}\\[.3em] \Rightarrow\ \color{#b0f}{(x)}&{\rm\,\ is\ a\ \color{#c00}{nonmaximal}\ \color{#0a0}{prime}}\end{align}$

Answer 5

In a comment you wrote that you seek a direct proof, not using quotient rings. Below is one.

If $D$ is ring and $f\in D[x],\,$ then dividing $f\,$ by $x$

yields $\, f = x\, g + f_0\, $ for $\,f_0 = f(0)\,$ and $\,g\in D[x]$

Therefore $\,x\mid f\iff x\mid f_0 \iff 0 = f_0\,\ $ (by eval at $\,x=0)$

$\quad\ \begin{align} {\rm thus}\ \ x\ {\rm is\ prime\ in }\ D[x] \ &\iff\ x\,\mid\, f\, g\ \ \ \Rightarrow\ \ x\mid f\ \ \ {\rm or}\,\ \ x\mid g\\[.3em] &\iff \ 0\!=\! f_0\: g_0\,\Rightarrow\, 0 =\!f_0\ \ {\rm or}\ \ 0 = g_0\\[.3em] &\iff \ D\ \text{is a domain}\ [\!\iff {\rm \color{#c00}{0\ is\ prime}\ in} \ D] \end{align}$

Applying this to $D = K[y]$ yields that $\,x\,$ is prime in $K[y][x] \cong K[y,x]$

Remark $\ $ Thus the definition of $x$ is prime in $D[x],$ when evaluated at $x=0$, ends up being equivalent to $\,\color{#c00}{0\rm\ is\ prime}\,$ in $D$ (via $\,0\mid d\!\iff\! 0 = d),$ an equivalent form of $D$ is a domain.

Note $\,(x)$ prime $\iff x$ prime follows by comparing the definitions using "divides = contains" for principal ideals, i.e. $\ x\mid f\iff (x)\supseteq (f)\iff (x) \ni f$

When you learn about quotient rings and evaluation homs you will find it insightful to study how the above elementwise proof relates to the equivalent structural form.