# What do prime ideals in $k[x,y]$ look like?

Suppose that $k$ is an algebraically closed field. Then what do the prime ideals in the polynomial ring $k[x,y]$ look like?

As far as I know, the maximal ideals of $k[x,y]$ are of the form $(x-a,y-b)$ where $a,b\in k$. What can we say about the prime ideals? Are there similar results? And what about $k[x,y,z], k[x,y,z,w]$ and so on. Would someone be kind enough to give me some hints or referrence on this topic? Thank you very much!

• Do you know anything about algebraic geometry? This seems like a cruel question to ask out of context... Aug 11, 2011 at 14:55
• If you want an explicit description like the one for the maximal ones... you are not going to have much luck. Aug 11, 2011 at 15:02
• in $k[x,y],$ any non-maximal, nonzero prime is a principal ideal generated by an irreducible polynomial. Aug 11, 2011 at 15:08

For $k[x, y]$ this is not as bad as it sounds! The saving grace here is that $k[x]$ is a PID.

Proposition: Let $R$ be a PID. The prime ideals of $R[y]$ are precisely the ideals of the following form:

• $(0)$,
• $(f(y))$ where $f$ is an irreducible polynomial (recall that Gauss' lemma is valid over a UFD, so irreducibility over $R$ is equivalent to irreducibility over $\text{Frac}(R)$),
• $(p, f(y))$ where $p \in R$ is prime and $f(y)$ is irreducible in $(R/p)[y]$.

This is a nice exercise. If you get stuck, I prove it in in this blog post. The primes of the third type are maximal, so when $R = k[x]$ you've already listed them (by the weak Nullstellensatz). The only new prime ideals are those of the second type; they correspond to irreducible subvarieties of dimension $1$.

In general I'm not even sure what would count as a reasonable description, and I don't know enough algebraic geometry to comment.

• Great!Just what I expected!Thank you very much! Aug 11, 2011 at 15:49
• +1 Are there similar results for $K[x_1,x_2,....,x_n]$ ?
– Amr
Jul 28, 2013 at 21:40
• @Amr: nope. Things get much harder in the general case. Jul 28, 2013 at 21:44
• @QiaochuYuan could you please explain to me, why $(f)$, where $f \in R[y]$ is irreducible, cannot be maximal in $R[y]$ when $R = k[x]$?
– user128245
Sep 11, 2016 at 13:19
• @QiaochuYuan Hello, is possible to pick an $f\in R[Y]$ s.t f is also irreducible in $R[Y]$ as well as in $(R/p)[Y]$? I have asked in the following link. You are welcome to answer there: math.stackexchange.com/questions/4043128/… Mar 1, 2021 at 2:44

The prime ideals of $k[x,y]$ are $0$, the maximal ones, and $(P)$ where $P$ is any irreducible polynomial. This is because $k[x,y]$ has dimension two, and is a UFD. For higher-dimensional rings things are more complicated, and there is no explicit answer. However, many things can be said, for example about the minimal number of generators for prime ideals. Good references are Introduction to Commutative Algebra (Atiyah-Macdonald), Commutative algebra (Matsumura), Commutative algebra with a view toward algebraic geometry (Eisenbud). The last one is more complete and self-contained.

• Thank you for the referrences! Aug 11, 2011 at 15:48
• irreducible polynomials are degree $1$ over an algebraically closed field, no? Oct 31, 2018 at 23:40
• Irreducible polynomials in one variable are degree 1 over algebraically closed fields, but not necessarily for multivariate polynomials. See for example math.stackexchange.com/q/1208939/333148 Feb 19, 2019 at 7:53

For $$k[x,y]$$ there are excellent answers here; however, if one still wonders what happens in $$k[x_1,\ldots,x_n]$$, $$n>2$$ (as was commented), then it is worth to mention Theorem 2.6 of Miguel Ferrero's paper, which deals with prime ideals in polynomial rings in several variables, $$n \geq 2$$.