A term only composed of the following operatings shall henceforth be called a radix term because I don't know how these terms are called.
A radix term $t$ is either an integer or
- a sum of two radix terms,
- a product of two radix terms,
- or a radix term raised to the $q$-th power where $q\in\mathbb{Q}$
Is there an algorithm that can construct a polynomial with integer coefficients for which the value of a given radix term is a root? Is it possible to construct a pair of polynomials with integer coefficients such that the intersection of the sets of roots of these two polynomials contains only a given radix term? If that is not possible, is it possible to provide a set of finitely many polynomials with integer coefficients such that the intersection of all their root sets contains only a given radix term? If all these things are not possible, what restrictions do we have to make to radix terms such that such an algorithm can be provided.
I'm interested in explicit algorithms to construct such polynomials if possible.
Yes, but it can result in fairly horrible polynomials. It's slightly easier to work with monic polynomials with rational coefficients, but of course you can clear denominators to get a integer polynomial with the given root afterwards if you prefer integer coefficients.
The algorithm works by induction on the structure of the term, of course. The case for a root is actually the easy one -- if $\alpha$ is a root of the polynomial $p(X)$, then $\sqrt[n]\alpha$ is a root of $p(X^n)$.
Sums and products need a more involved process, but in principle it's not difficult. Suppose $\alpha$ is a root of $p(X)$ of degree $n$ and $\beta$ is a root of $q(X)$ of degree $k$. Then consider the $\mathbb Q$-vector space $V$ spanned by $\alpha^i\beta^j$ for $0\le i<n$ and $0\le j<k$. It is closed under both addition and multiplication, because whenever we get a term of degree $n$ or more in $\alpha$ or degree $k$ or more in $\beta$ we can use the fact that $p(\alpha)=0$ and $q(\beta)=0$ to rewrite those higher degrees to a combination of our basis vectors.
Now compute reduced expressions for $(\alpha\beta)^m$ (or $(\alpha+\beta)^m$ for addition) for $0\le m\le nk$. Since there are here $nk+1$ elements of an at most $nk$-dimensional vector space, they are linearly dependent. We can use standard linear algebra techniques to find a rational combination of them that is zero. But such a rational combination is exactly a rational polynomial that has $\alpha\beta$ (or $\alpha+\beta$) as a root, and then we're done.
No, and not with more than two polynomials either. For example, any integer polynomial that has $\sqrt2$ as a root will also have $-\sqrt2$ as a root. This is because the set $\mathbb Q[\sqrt2]$ of numbers of the form $a+b\sqrt2$ with $a,b\in\mathbb Q$ is a subring of $\mathbb R$, and the function $\varphi(a+b\sqrt2)=a-b\sqrt2$ is a ring homomorphism from $\mathbb Q[\sqrt2]$ to itself. Since the $\varphi$ maps integers to themselves, applying $\varphi$ to both sides of the equation $P(\sqrt2)=0$ yields $P(\varphi(\sqrt2))=\varphi(0)$ or in other words $P(-\sqrt2)=0$.
In general, algebraic numbers that have the same primitive polynomial cannot be distinguished by (integer) polynomials.