Understanding of algebraic maps via number of inequalities in a semi-algebraic set Let $P_{(d,0,e)}(x) = dx^2 + e$, and $Q_{(a,b,c)}(x) = ax^2 + bx+c$ be two polynomials in $\mathbb{R}[x]$. Define an algebraic map $\phi:\mathbb{A}_{\mathbb{R}}^2 \times \mathbb{A}_{\mathbb{R}}^3 \to \mathbb{A}_{\mathbb{R}}^5$ that sends $((d,e),(a,b,c)) \in \mathbb{A}_{\mathbb{R}}^2 \times \mathbb{A}_{\mathbb{R}}^3$ to coefficents of multiplication of these two polynomials i.e. $(ad,bd,ae+cd,be,ce)\in \mathbb{A}_\mathbb{R}^5$. The image of this map is described by $$\mathcal{M}_{P,Q} = \{(A,B,C,D,E) \in \mathbb{A}_{\mathbb{R}}^5: AD^2+B^2E=BCD \textit{ and  } C^2 \ge 4AE\}.$$
In general, we might have $P_v$, and $Q_w$, of degree at most $m$ and $n$ respectively, where $v$ and $w$ are vectors that can represent the polynomials (Note that just like the above example, we might have some forced zeros in $v$ and $w$, but others are arbitrary numbers in $\mathbb{R}$.).
By Tarski-Seidenberg Theorem, we have that the image of the algebraic map $\phi$ is semi-algebraic i.e. it can be described by zeros of some polynomials and some inequalities. I'm wondering about the number of nonredundant inequalities that appear in $\mathcal{M}_{P,Q}$. Based on a few examples that I checked, this number was one, like the example above or zero if the image of $\phi$ is closed. Is this true for an arbitrary $P_v$ and $Q_w$?
$\textbf{Edit}$: I would like to add two more examples to this post that shows if there is no fixed zero in $v$, and $w$ then the problem is easier to handle.
Example: Let $P_{(d,e)}(x) = dx+e$, and $Q_{(a,b,c)}(x) = ax^2+bx+c$ then the map $\phi$ sends $$((d,e),(a,b,c)) \in \mathbb{A}_{\mathbb{R}}^2 \times \mathbb{A}_{\mathbb{R}}^3 \to (ad, bd+ae,cd+be,ce) \in \mathbb{A}_{\mathbb{R}}^4.$$ One can check that this map is surjective as every cubic polynomial in $\mathbb{R}$ has a real root corresponds to $dx+e$ and a degree 2 polynomial corresponds to $ax^2+bx+c$.
Example: Let $P_{(c,d)}(x) = cx+d$ and $Q_{(a,b)}(x)= ax+b$ then $\phi$ sends $$((c,d),(a,b)) \in \mathbb{A}_{\mathbb{R}}^2 \times \mathbb{A}_{\mathbb{R}}^2 \to (ac, ad+bc,bd) \in \mathbb{A}_{\mathbb{R}}^3.$$
Since not all degree 2 polynomials have real roots, we expect that this map is not surjective, so one can check that
$$\mathcal{M} = \{(A,B,C)\in \mathbb{A}_{\mathbb{R}}^3: B^2-4AC \ge 0\}.$$
These two examples illustrate that if we don't have $\textbf{forced}$ zeros in $v$ and $w$, then we are able to describe $\mathcal{M}$. To be more precise, if $v$ and $w$ don't have forced zero then the image of $\phi$ is $m+n+1$ dimensional (it is not hard to justify this via roots of the target polynomial) and two cases happen:

*

*If $m$ and $n$ are odd simultaneously then the decomposition $P_vQ_w$ implies having at least two real roots, so $\phi$ is not surjective. In this case, we expect that the discriminant to appear as an inequality.


*If $m$ and $n$ are not odd simultaneously then this decomposition exists in $\mathbb{R}[x]$. So, $\phi$ is surjective and we don't have any inequalities.
 A: 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.
