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Let $V$ be an affine real algebraic set. That is, $V$ is the zero set of some polynomials in $\mathbb{R}^n$. I would like to show that there is not a proper algebraic subset $W\subset V$ which admits a surjective polynomial map $W\twoheadrightarrow V$. The plan to do this is to take the ordered list of dimensions of irreducible components of $V = V_1\cup \cdots \cup V_p$. For each irreducible component $V_i$, $W\cap V_i$ will equal $V_i$ or will have smaller dimension. Thus, the ordered list of dimensions of irreducible components of $W$ will be less than or equal to that of $V$ with respect to lexicographic order, with equality only if $W=V$. The same holds for images of polynomial maps which are algebraic sets, and hence the statement.

Thus, I am wondering if there is a name for the ordered list of dimensions of the irreducible components of an algebraic set? I would also be happy to know if this terminology exists for algebraic sets over algebraically closed fields (which is the usual setting for algebraic geometry). Maybe there is appropriate terminology in commutative algebra?

Also, if the above result is written down somewhere, that would be a helpful reference to have.

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  • $\begingroup$ I am just speaking for myself: I have not seen that before in the literature. Perhaps you could ask experts in applications of algebraic geometry to applied mathematics, since this is definitely relevant there (where you need to remove "extraneous" irreducible components that correspond to non-physical solutions). One such expert is Andrew Sommese at the University of Notre Dame. $\endgroup$
    – Jason Starr
    Commented Dec 31, 2021 at 12:45
  • $\begingroup$ @JasonStarr okay, thanks for the suggestion, I did consult my colleague bernd sturmfels who does applied algebraic geometry, but he didn’t have a suggestion. I might contact Andrew. $\endgroup$
    – Ian Agol
    Commented Dec 31, 2021 at 22:22
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    $\begingroup$ I'not aware of such a notion. However, instead of working with the ordered list of dimensions of the irreducible components, it is more common to work with the dimension function that assigns the dimension to each irreducible component. Such a dimension function behaves well with respect to morphisms and allows to prove the statement as well. $\endgroup$
    – Johannes Huisman
    Commented Jan 1, 2022 at 10:32
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    $\begingroup$ @JohannesHuisman Thanks for the comment, this makes sense. I was focused on the list of dimensions because it gives an invariant of the algebraic set, but I think your suggestion will give a streamlined proof. Do you have a reference for a paper or text in which this dimension function is defined or used? $\endgroup$
    – Ian Agol
    Commented Jan 1, 2022 at 18:25
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    $\begingroup$ For a Zariski-closed subset $X$ in $\mathbf{C}^n$ define the ordinal inductively $\ell(X)=\sup\{\ell(Y)+1\}$ where $Y$ ranges over proper closed subsets of $X$. This definition is valid because of noetherianity. Then for $X$ with $d_i$ $i$-dimensional components, we have $\ell(X)=\sum_i \omega^i\cdot d_i$ (sum in decreasing order). And $\ell$ is decreasing under taking surjective regular maps. $\endgroup$
    – YCor
    Commented Mar 30, 2022 at 21:12

1 Answer 1

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You can prove this without needing to keep track of dimensions.

Suppose that $W$ and $V$ are real algebraic sets with $W \subsetneq V$, and that there exists a surjective polynomial map $\phi: W \twoheadrightarrow V$. Then define a sequence of algebraic sets $W_i$ with $W_0 = W$ and $W_{i+1} = \phi^{-1}(W_i)$. Since $W \subsetneq V$ and $\phi$ is surjective, $W_1 = \phi^{-1}(W)$ must be a proper subset of $W_0$. Then $\phi|_{W_1}$ gives a surjection $W_1 \twoheadrightarrow W_0$, so by the same logic $W_2 \subsetneq W_1$.

By induction, the $W_i$ form a sequence of nested of algebraic sets $W_0\supsetneq W_1 \supsetneq \ldots$, which contradicts the Zariski topology on $\mathbb{R}^n$ being a Noetherian topological space.

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