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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?

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    $\begingroup$ Yes, it's possible: $Y^2+X^2(X-1)^2$ over $\mathbb R$. $\endgroup$
    – user26857
    Nov 13, 2016 at 20:41
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    $\begingroup$ I just realized that $<X^2+Y^2>$ and $<X,Y>$ define the same irreducible variety $\{(0,0)\}$. $\endgroup$
    – ZQ Wan
    Nov 13, 2016 at 22:00
  • $\begingroup$ (2) No, it is not; ever a prime ideal defines an irreducible algebraic set! $\endgroup$ Nov 14, 2016 at 12:42

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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).

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