Over algebraically closed field, we have the dictionaries between irreducible affine varieties and prime ideals. Different prime ideals define different varieties. But when the field is not algebraically closed, e.g. $\mathbf{R}$, we have polynomials $x^2+1$ and $x^2+2$ that define the same empty set.

  1. Are there more examples that different prime ideals define the same (non-empty) variety?

  2. Is is possible for a prime ideal to define a reducible algebraic set?

  • 2
    $\begingroup$ Yes, it's possible: $Y^2+X^2(X-1)^2$ over $\mathbb R$. $\endgroup$ – user26857 Nov 13 '16 at 20:41
  • 1
    $\begingroup$ I just realized that $<X^2+Y^2>$ and $<X,Y>$ define the same irreducible variety $\{(0,0)\}$. $\endgroup$ – ZQ Wan Nov 13 '16 at 22:00
  • $\begingroup$ (2) No, it is not; ever a prime ideal defines an irreducible algebraic set! $\endgroup$ – Armando j18eos Nov 14 '16 at 12:42

I guess you do not consider points of varieties that are not rational then (from the comments).

In that case, weird things are possible. For example, the answer to $2.$ is positive:

Consider e.g. a finite field $\mathbb{F}$. Then every subset of $\mathbb{A}^n_{\mathbb{F}}$ is closed (it is a finite union of closed points), and the only irreducible subsets are singletons (for the same reason). Thus, and ideal $I$ defines an irreducible subset iff $V(I)$ is a one-point set. So for example, $$I=(x_1, x_2, \dots, x_i), \;\;i <n$$ is a prime ideal in $\mathbb{F}[x_1, \dots, x_n],$ but $V(I)$ is not irreducible, since it contains more than one point.

For these reasons (which are connected to your previous question) it is preferable to define all your affine varieties in $\mathbb{A}^n_{\overline{\mathbb{F}}}$ instead of $\mathbb{A}^n_{\mathbb{F}}$ (where $\overline{\mathbb{F}}$ denotes the algebraic closure of $\mathbb{F}$) even when you consider polynomials only over $\mathbb{F}$. That is, an affine algebraic set (over $\mathbb{F}$) under this definition is a set of the form

$$V(S)=\{(a_1, a_2, \dots a_n) \in \overline{\mathbb{F}}^n\;|\; \forall f \in S: f(a_1, \dots, a_n)=0\},\;\; S \subseteq \mathbb{F}[x_1, x_2, \dots, x_n].$$

Then the correspondence works similarly as if $\mathbb{F}$ was alg. closed. The downside of this is that some singletons are no longer closed in the ($\mathbb{F}$-)Zariski topology (if you care about such property).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.