Proving which ideals of $K[x,y]$ are primes or maximals.

Question

Let $K$ be a field. Which of these ideals of $K[x,y]$ are primes? And maximals? $$I_1=(x+y,y-3), I_2=(x^2+y)$$ I know what is a prime ideal, a maximal ideal and a field, and I still can't see how to solve this problem. These are just 2 of a large list of ideals, I would appreciate if someone could solve it so I can use these as an example to solve the others.

Answer 1

Try to figure out what the rings $K[x,y]/I_1$ and $K[x,y]/I_2$ are. If it's a domain, then you have a prime idea. If it's a field, then you have a maximal ideal.

For example, in $K[x,y]/I_1$ we will have $x\equiv -y$ and $y\equiv 3$, which means that $K[x,y]/I_1\simeq K$, which is a field. In $K[x,y]/I_2$ we have $y\equiv -x^2$, which means that $f(x,y)\in K[x,y]$ is congruent to $f(x,-x^2)$ in $K[x,y]/I_2$. You should then be able to convince yourself that $K[x,y]/I_2\simeq K[x]$, which is a domain.

Answer 2

You have $I_1 = (x+3,y-3)$, can you see how $K[x,y]/I_1 \cong K$ and conclude that $I_1$ is maximal and hence a prime?

For the second one you have $K[x,y]/I_2 \cong K[x]$. This follows, as you have $y \equiv -x^2 \pmod {I_2}$. Hence we have that $I_2$ is prime, but not maximal, as $K[x]$ is an integral domain, but not a field.

Answer 3

$K[x,y]/(x+y,y-3)\cong K[x]/ (x+3)\cong K$, so $I_1$ is a maximal ideal, because $K$ is a field.

$K[x,y]/(x^2+y)\cong K[x,-x^2]\cong K[x]$, so $I_2$ is a prime ideal, because $K[x]$ is a domain integral. .