What are the all the prime ideals of $\mathbf{R}[x,y]$ and $\mathbf{C}[x, y]$? (Also, how do you prove that you've found all of them?) I'm trying to understand what the $\mathbf{R}$-algebraic vs $\mathbf{C}$-algebraic subsets of $\mathbf{C}^2$ are, defined as the zeros (in $\mathbf{C}^2$) of an ideal in $\mathbf{R}[x, y]$ and $\mathbf{C}[x,y]$, respectively.

Edit: I'm not that familiar with dimension theory, so I'd prefer to see an argument that does not rely on it.

  • $\begingroup$ How do you expect the answer to this to be? If $f\in C[x,y]$ is an irreducible polynomial, then $(f)$ is a prime ideal in $C[x,y]$ and there are maaaaany irreducible poynomials in that ring. Along with that, there is $(0)$ and the maximal ones, and that is that. Would you say that we "found all the prime ideals"? $\endgroup$ – Mariano Suárez-Álvarez Feb 7 '18 at 23:33
  • $\begingroup$ @MarianoSuárez-Álvarez I am interested in knowing how one proves the statement: the set of prime ideals of C[x,y] are precisely the zero ideal, the maximal ideal, and the ideals generated by an irreducible polynomial. In particular, I'm not familiar with Krull dimension, so an argument without passing to dimension theory, would be ideal. $\endgroup$ – Drew Brady Feb 7 '18 at 23:40
  • $\begingroup$ Mariano's comment to your question and how you've asked it is called an xy-problem(xyproblem.info), where you want to do X, but can't. You think the strange question Y should solve it, and you ask Y instead of X. $\endgroup$ – Alex Clark Feb 8 '18 at 0:11
  • $\begingroup$ Also, pun intended? (would be ideal). I would hardly call that 'passing to dimension theory', you can understand the definition of Krull dimension with standard ring theory you'd encounter in second or third year undergrad. $\endgroup$ – Alex Clark Feb 8 '18 at 0:14

Here's an answer for $\mathbf C[x,y]$: Hilbert's Nullstellensatz, asserts the maximal ideals have the form $(x-\alpha,y-\beta)$ for some $\;\alpha, \beta\in \mathbf C$.

On the other hand, $\mathbf C[x,y]$ is a U.F.D. of (Krull) dimension $2$. So the prime ideals of height $1$ are principal, generated by irreducible polynomials.

Add to this list the only ideal of height $0$, and you've finished.

  • 1
    $\begingroup$ How does one prove this without using dimension theory? $\endgroup$ – Drew Brady Feb 7 '18 at 23:40
  • $\begingroup$ No idea. That's only basic dimension theory, so it's quite possible it can de done by hand.. $\endgroup$ – Bernard Feb 7 '18 at 23:42
  • $\begingroup$ what about in the case when the field is not algebraically closed. Then what can we say, for example, about the maximal ideals of $\mathbf{R}[x, y]$? $\endgroup$ – Drew Brady Feb 8 '18 at 1:13
  • $\begingroup$ That is closely linked to real algebraic geometry, which I don't really know. $\endgroup$ – Bernard Feb 8 '18 at 9:30

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