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Let $\mathbb{R}[x_1, \ldots , x_n]$ denote the commutative ring of all polynomials in $n$ variables $x_1, \ldots, x_n$ with coefficients in $\mathbb{R}$.

Given a set with $k$ polynomials $\{f_1, . . . , f_k\}$ of $\mathbb{R}[x_1, \ldots, x_n]$, we consider the algebraic subset of $\mathbb{R}^n$, $V(f_1, . . . , f_k)$, which is defined as $$V(f_1, \ldots , f_k) = \{(a_1, . . . , a_n) \in \mathbb{R}^{n} : f_i(a_1, . . . , a_n) = 0 \mbox{ for all } 1 \leq i \leq k\}.$$ Analogously, we can consider the algebraic subset $Z(f_1, \ldots , f_k)$ of $\mathbb{C}^n$ as $$Z(f_1, \ldots , f_k) = \{(a_1, . . . , a_n) \in \mathbb{C}^{n} : f_i(a_1, \ldots , a_n) = 0 \mbox{ for all } 1 \leq i \leq k\}.$$

I want to learn about the relation between irreducible components of $V(f_1, \ldots , f_k)$ and $Z(f_1,\ldots, f_k)$. Is the number of irreducible components of $V(f_1, \ldots , f_k)$ less than or equal to the number of irreducible components of $Z(f_1, \ldots , f_k)$? For instance, if $n=2$, $k=1$ and $f_1(x_1,x_2) = x_1^2 + x_2^2$, we have $V(f_1) = {(0,0)}$ and $$Z(f_1) =\{(x,\sqrt{-1}x):x\in \mathbb{C}\} \cup \{(x,-\sqrt{-1}x):x\in \mathbb{C}\}.$$

KReiser
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Guillermo Duran
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  • I believe this is true. The irreducible components correspond to the minimal associated primes of the corresponding ideal in its primary decomposition. When passing to the algebraic closure, we should expect to see either the same amount or more minimal associated primes. I haven't really checked the details on the previous sentence, but I think this makes sense. – Dave Oct 19 '20 at 00:22

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Unfortunately, this is not true. Consider $f=(x^3-x)^2+y^2=0$. Then $f$ is irreducible over $\Bbb R$, because it factors as $(x^3-x+iy)(x^3-x-iy)$ over $\Bbb C$ and neither of these polynomials are in $\Bbb R[x,y]$. $Z(f)$ has two irreducible components, one per factor, but $V(f)$ is three distinct points and thus has three irreducible components.

If $V(I)$ doesn't have any irreducible components consisting of only singular points over $\Bbb R$, then the statement you want should hold - a smooth real point of $V(I)$ gives a smooth complex point of $Z(I)$, and smooth points are on one and only one irreducible component, which gives you that the number of irreducible components of $Z(I)$ is at least the number of irreducible components of $V(I)$. This is perhaps a bit restrictive, though.

This sort of counterexample is one reason to move to schemes - then the solution outlined in Dave's comment would be correct. Irreducible components of $V(I)$ correspond to minimal primes over $I$, and prime ideals can either remain prime or split in a field extension (they can't combine together). Alternatively, depending on the sort of problem you're looking to attack with this sort of thing, you may need to forgo schemes and look for a more semi-algebraic approach. If you're looking at getting in to real algebraic geometry, there are some text recommendations here but I don't recall any specific point where they get in to this exact question.

KReiser
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