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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!

user26857
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user14242
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3 Answers3

101

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.

Xam
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Qiaochu Yuan
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  • Great!Just what I expected!Thank you very much! – user14242 Aug 11 '11 at 15:49
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    +1 Are there similar results for $K[x_1,x_2,....,x_n]$ ? – Amr Jul 28 '13 at 21:40
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    @Amr: nope. Things get much harder in the general case. – Qiaochu Yuan Jul 28 '13 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 $R = k[x]$? –  Sep 11 '16 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: https://math.stackexchange.com/questions/4043128/question-regarding-the-prime-and-maximal-ideals-of-kx-y – Jhon Doe Mar 01 '21 at 02:44
  • @QiaochuYuan I ask because in Dummit and Foote, for question 4 of section 15.5 requires $f$ to be irreducible in $R[Y]$ as well. – Jhon Doe Mar 01 '21 at 03:00
  • @QiaochuYuan I was not able to post this below the blog post so I am asking it here:Hi Mr. Yuan, thank you very much for writing this answer. I have a question in last line of the proof of the Proposition :

    "The prime ideals of $R[y]$ are precisely the zero ideal $(0)$, the ideals $(f(y))$ where f(y) is irreducible, and the maximal ideals (p, f(y)) where p $\in$ R is prime and f(y) is irreducible in $(R/p)[y]$."

    What contradiction is obtained after clearing the denominator?

     –  Oct 02 '23 at 19:48
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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.

Plop
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6

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, $n \geq 2$.

user237522
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