Understanding of algebraic maps via number of inequalities in a semi-algebraic set

Solution 1:

Short answer : I think you're misunderstanding the Tarski-Seidelberg theorem and asking too much of it. As you say, it shows that there is a description with equalities and inequalities, but there is no "canonical" description. The theorem says that a variety is a finite union of components where each component is exactly defined by a finite system of equalities and inequalities. But there is no "natural,obvious" set of components (typically the proof uses a so-called cylindrical decomposition which requires to arbitrarily choose a preliminary ordering of the coordinates, and even that does not suffice to make the decomposition unique). So when you speak of "the number of non-redundant inequalities", while it is true that one can define the minimal number of inequalities involved in a decomposition, you have to realize that the inequalities in question will often be "unrelated" in the sense that they hold on different components, and further there might be several decompositions achieving the minimum number of inequalities.

All the examples you give in the OP are far too simple and not at all representative of the general case. They are all single-component cases.

Detailed answer : Consider the multiplication of a polynomial of degree at most 1 with a polynomial of degree at most 3, producing a polynomial of degree at most 4 :

$$ (a_0+a_1x)(b_0+b_1x+b_2x^2+b_3x^3)=c_0+c_1x+c_2x^2+c_3x^3+c_4x^4 $$

(thus $c_0=a_0b_0$, $c_4=a_1b_3$ etc). This defines a map $\phi : {\mathbb R}^2 \times {\mathbb R}^4 \to {\mathbb R}^5$, by $\phi((a_1,a_0),(b_3,\ldots,b_0))=(c_4,\ldots,c_0)$.

Then we have the following :

Theorem. Both the image $I$ of $\phi$ and its complement $I^c$ are Zariski-dense in ${\mathbb R}^5$ (and in particular, any non-strict polynomial inequality true on either of them will be true on all of ${\mathbb R}^5$, and thus be trivial).

Proof of theorem. It will suffice to show that for any $(c_4,c_3,c_2,c_1)\in ({\mathbb R} \setminus \lbrace 0 \rbrace)\times {\mathbb R}^3$, there are infinite subsets $X_{-},X_{+}$ of $\mathbb R$ such that $(c_4,c_3,c_2,c_1,c_0)$ is in $I$ if $c_0\in X_{-}$ and in $I^{c}$ if $c_0\in X^{+}$.

Now, when $c_4\neq 0$, $c=(c_4,c_3,c_2,c_1,c_0)$ is in $I$ (or $I^c$) iff $\frac{1}{c_4}c=(1,\frac{c_3}{c_4},\ldots,\frac{c_0}{c_4})$ is, so it suffices to find infinite subsets $X_{-},X_{+}$ of $\mathbb R$ such that $(1,c_3,c_2,c_1,c_0)$ is in $I$ if $c_0\in X_{-}$ and in $I^{c}$ if $c_0\in X^{+}$.

Now, $(1,c_3,c_2,c_1,c_0)\in I$ iff the polynomial $C=x^4+c_3x^3+c_2x^2+c_1x+c_0$ has a real root. If we consider the truncated polynomial $D=C-c_0=x^4+c_3x^3+c_2x^2+c_1x$, then $D$ has even degree and so has a global minimum on $\mathbb R$ which we will call $\mu$. Then we can take $X_{-}=(-\infty,\mu)$ and $X_{+}=(\mu,+\infty)$, which finishes the proof.