# Do different prime ideals define different irreducible varieties in general?

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?

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

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, $$I=(x_1, x_2, \dots, x_i), \;\;i is a prime ideal in $$\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).