$M$ is maximal, $P$ is prime but not maximal

Question

If $R$ is commutative with $1 \in R$, then each maximal ideal of $R$ is also a prime.

The reverse doesn't hold.

For example, $R=K[x, y], P=\langle x \rangle, M=\langle x, y\rangle$.

Then it stands $$P=\langle x \rangle \subsetneq M=\langle x, y\rangle \subsetneq K[x, y]$$ $M$ is maximal, so also prime, but $P$ is prime but not maximal.

Can you explain how we know that $M$ is maximal and that $P$ is prime but not maximal?

$$$$

Edit:

To show that $P$ is prime do we do the following?

We consider the homomorphism $\phi: K[x, y] \rightarrow K[y]$ with $\phi(z)=z, \forall z \in K[y]$ and $\phi(x)=0$.

We consider a polynomial $p(x, y) \in K[x, y]$.

Applying the euclidean division of the polynomial and $x$, we get:

$$p(x,y)=x \cdot g(x, y)+r(x, y)$$

$$deg_x r(x, y)<deg(x)=1 \Rightarrow deg_x r(x, y)=0 \Rightarrow r(x, y)=r(y)$$

So $p(x, y)=x \cdot g(x, y)+r(y)$.

Let $p(x, y \in ker \phi$. Then $\phi ( p(x, y))=0 \Rightarrow r(y)=0$.

So, $p(x, y)=x \cdot g(x, y)$.

So, if $p(x, y) \in ker \phi$, then $p(x, y) \in \langle x \rangle$.

So, $ker \phi \subseteq \langle x \rangle$.

$x \in \langle x \rangle$

$\phi (x)=0$

So $x \in ker \phi$.

So, $\langle x \rangle \subseteq ker \phi$.

Therefore, $ker \phi =\langle x \rangle$.

We have that $Im \phi \subseteq K[y]$.

For each $z \in K[y]$, there is an element of $K[x, y]$ that gets mapped with $\phi$ to $z$.

So $K[y] \subseteq Im \phi$.

Therefore, $Im \phi =K$.

From the theorem of isomorphism we have that $$K[x, y] /\langle x \rangle \cong K[y]$$

$K[y]$ is an integral domain, so $K[x, y] /\langle x \rangle$ is an integral domain.

Therefore, $P=\langle x \rangle$ is a prime Ideal of $K[x, y]$.

$$$$

Can you check what I've tried?

Why is $K[y]$ an integral domain? Do we have to prove it?

Answer 1

A way to see this is that $K[x,y]/M \cong K$ is a field, while $K[x,y]/P \cong K[y]$ is not a field but an integral domain.

This assumes you know the criterion or definition that $I$ is a maximal ideal of $R$ if and only of $R/I$ is a field, and that $I$ is a prime ideal of $R$ if and only of $R/I$ is an integral domain.