Minimal polynomial of product, sum, etc., of two algebraic numbers

The standard proof, apparently due to Dedekind, that algebraic numbers form a field is quick and slick; it uses the fact that $$[F(\alpha) : F]$$ is finite iff $$\alpha$$ is algebraic, and entirely avoids the (to me, essential) issue that algebraic numbers are roots of some (minimal) polynomial. This seems to be because finding minimal polynomials is hard and largely based on circumstance.

There are more constructive proofs which, given algebraic $$\alpha$$, $$\beta$$, find an appropriate poly with $$\alpha \beta$$, $$\alpha + \beta$$, etc., as a root -- but of course these are not generally minimal.

You would of course want an algorithm to compute such min. polies, but assuming this is unfeasible (as it seems), my question is a bit different:

Every algebraic number $$\alpha$$, $$\beta$$, $$\alpha\beta$$, $$\alpha + \beta$$, etc., has a unique corresponding minimal polynomial, call it $$p_{\alpha}(x)$$, $$p_{\beta}(x)$$, $$p_{\alpha \beta}(x)$$, etc., and these polies have other roots, the conjugates, $$\alpha_1$$,...,$$\alpha_n$$, $$\beta_1$$,...,$$\beta_m$$, etc. Suppose I want to define an operation on this set of polies in the most naive way: $$p_{\alpha}(x) \star p_{\beta}(x) = p_{\alpha\beta}(x)$$. (Note that this is NOT the usual, direct multiplication of polies.)

But is this even well-defined? More specifically: suppose I swap $$\alpha$$ with one of its conjugates, $$\beta$$ with one of its conjugates, and multiply those together. Is the minimal polynomial of the new product the same as before? i.e. is this new product a conjugate of the old one? Meaning, I would need $$p_{\alpha_1 \beta_1}(x) = p_{\alpha_i \beta_k}(x)$$ for any combination of conjugates in order for this proposed operation to even make sense. And this seems unlikely -- that would be sort of miraculous right?

What about a similar operation for $$\alpha + \beta$$, $$\alpha - \beta$$, etc?

More broadly, given two algebraic numbers $$\alpha$$, $$\beta$$, I'm interested in the set of minimal polynomials corresponding to those algebraic numbers which can be generated by performing the field operations on $$\alpha$$, $$\beta$$ -- call this the "set of minimal polies attached to the number field" or something -- and if a similar field (or even just ring) structure can be put on these polies by defining appropriate operations on them. (Not the usual operations, which will clearly give you polies with roots outside of your number field.) I'm ultimately after questions like:

(1) How do the conjugates of $$\alpha \beta$$, $$\alpha + \beta$$, etc., relate to the conjugates of $$\alpha$$ and $$\beta$$?

(2) How do the coefficients of the min. polies of $$\alpha \beta$$, $$\alpha + \beta$$, etc., relate to those of the min. polies of $$\alpha$$ and $$\beta$$? Obviously, the algebraic integers form a ring; what else can be said?

(3) Degree?

It may be impractical to calculate any one such min. poly explicitly, but maybe interesting things can be said about the collection as a whole?

2 Answers

Neither operation is well defined.

For an example for multiplication, take $\alpha=\beta=\sqrt{2}$ which has minimal polynomial $x^2-2$. Now $\alpha\beta=2$ so this has minimal polynomial $x-2$ but if we took $\beta=-\sqrt{2}$, then $\alpha\beta=-2$ so has minimal polynomial $x+2$ which is obviously different. Notice how the degree reduces here but it could also go up or remain unchanged.

As for addition, consider the case where $\alpha=-\beta$. Then $\alpha + \beta=0$ so has minimal polynomial $x$. If we instead take $\alpha=\sqrt{2}$, $\beta=\sqrt{3}$ then the minimal polynomial of $\alpha + \beta$ has degree $4$. Hopefully this shows that very little can be said for addition; we don't even know whether the degree will increase or decrease or even stay the same (e.g. $\alpha=\beta$).

• multiplication is in just the same case as addition, it is not well-defined. Sep 16, 2015 at 15:33
• @mercio fixed now. I thought my first statement was true, but maybe I need some more conditions (which I can't think of right now) to impose it and exclude the trivial cases Sep 16, 2015 at 16:00
• Off the top of my head, I'm thinking that if $F=K(\alpha\beta)$ is an extension of both $F_1=K(\alpha)$ and $F_2=K(\beta)$ then the Galois automorphisms of $F$ (which of course then give your conjugates of $\alpha\beta$) will restrict down to $F_1, F_2$ and map $\alpha$, $\beta$ to their relevant conjugates and maybe this is all you need. Sep 16, 2015 at 16:07
• Check me, here: consider the cyclotomic polies $\Psi_3(x) = x^2 + x + 1$ and $\Psi_5(x) = x^4 + ... + 1$; their roots are of course primitive nth roots of unity, and multiplying any two of them gives a primitive 15th root of unity (right?), in fact I think every one is a unique such product. So everything works in this (very special) case, ya? Any other cases where it turns out OK? Sep 16, 2015 at 17:08
• Your idea will also work with cyclotomic polynomials $\Psi_i$ and $\Psi_j$ where $\gcd(i,j)=1$. Extending on this, if you take two Galois extensions $F=\mathbb{Q}(\alpha)$ and $K=\mathbb{Q}(\beta)$ with $F \cap K = \mathbb{Q}$ then it should turn out OK and is probably what I was originally thinking. (I think the technical condition may be linear disjointness but the conditions I've given imply that). Sep 16, 2015 at 18:13

Your operation $$p_\alpha\ast p_\beta=p_{\alpha+\beta}$$ (resp. $$p_\alpha\ast p_\beta=p_{\alpha\beta}$$ etc.) is not well defined at the level of particular cases because of the fall of dimensions of $$[F(\alpha+\beta):F]$$ (resp. $$[F(\alpha\beta):F]$$) w.r.t. $$[F(\alpha):F][F(\beta):F]$$ but will be such at the level or "universal formulas" or "in general position".

A) Firstly, we remark that the minimal polynomial of $$\alpha$$ (resp. $$\beta$$), $$p_\alpha$$ (resp. $$p_\beta$$) in an algebraically closed extension reads $$p_\alpha(X)=\prod_{a\in A}(X-a)^{m(a)}\ ;\ p_\beta(X)=\prod_{b\in B}(X-b)^{m(b)}$$ Let us solve the general problem (embracing $$x+y$$ and $$xy$$) of a polynomial operation with integer coefficients $$T(x,y)\in \mathbb{Z} [x,y]$$ (then, it can be sum $$T(x,y):=x+y$$, product, $$T(x,y):=xy$$, difference $$T(x,y):=x-y$$ or any law as $$T(x,y)=x+y+xy$$ etc.). So, one has that $$p_\alpha\star_T p_\beta:=p_{T(\alpha,\beta)}$$ divides
$$\prod_{a\in A,\, b\in B}(X-T(a,b))^{m(a)m(b)}\qquad (1)$$ And, in case $$[F(T(\alpha,\beta)):F]=[F(\alpha):F][F(\beta):F]$$ (i.e. $$F(\alpha), F(\beta)$$ are linearly disjoint) we have equality.

One can prove (exercise, but I can elaborate), using similar methods as in here (in fact, use $$T(A\otimes I,I\otimes B)$$) that the polynomial (1) depends uniquely on the coefficients of $$p_\alpha$$ and $$p_\beta$$, so one can define $$p_\alpha\bullet_T p_\beta :=\prod_{a\in A,\, b\in B}(X-T(a,b))^{m(a)m(b)}$$ B) In the general case, let $$R$$ be a ring, $$X$$ an indeterminate and $$R_1[X]$$ be the set of monic polynomials, we get a (more rigorously defined) law $$\bullet_T$$ defined on monic polynomials and derived from $$T\in \mathbb{Z} [x,y]$$ as follows

• Let $$P,Q$$ in $$R_1[X]$$ of degrees $$p,q$$ respectively
• Take two disjoint finite alphabets $$A,B$$ of cardinality $$p,q$$
• Form, in $$\mathbb{Z}[A\cup B][X]$$, the products $$U=\prod_{a\in A}(X-a)=X^{p}+\sum_{k=1}^{p} \alpha_k\,X^{p-k}$$ $$V=\prod_{b\in B}(X-b)=X^{q}+\sum_{k=1}^{q} \beta_k\,X^{q-k}$$ (elementary symmetric functions with signs $$\pm 1$$) and $$W=\prod_{a\in A,\, b\in B}(X-T(a,b))=X^{pq}+\sum_{k=1}^{pq} \gamma_k\,X^{pq-k}\ .$$ The coefficients $$\gamma_k$$ depend uniquely on $$\alpha_i,\beta_j$$, so we set $$W:=U\bullet_T V$$. As the coefficients are all in $$\mathbb{Z}$$, one can specialize this to $$R_1[X]$$, compute $$P\bullet_T Q$$ and recover what was said in (A).

Late edit After having read all the comments, let us check the method with $$T(x,y)=xy$$ and cyclotomic polynomials. We consider two primes $$p,q$$ (different) and $$F$$, a field with no primitive $$p$$th (res. $$q$$th) root of unity (call it $$\alpha$$ - resp. $$\beta$$ - in some extension). Take any extension $$\Omega$$ with $$pq$$th roots of unity then, with (A) $$p_\alpha(X)=\sum_{k=0}^{p-1}X^k\ ;\ p_\beta(X)=\sum_{k=0}^{q-1}X^k\ ;\ p_\alpha(X)\bullet_T p_\beta(X)=\sum_{k=0}^{pq-1}X^k=p_{\alpha\beta}(X)$$ By (B), we get the same result by specialization.

Remarks i) If $$T$$ is associative (as $$x+y,\ xy,\ x+y+xy$$ etc.), so is $$\bullet_T$$ and we can recover the law you seemed to wish.

ii) (For Late edit) This law being generic it does not, however, take into account the dimension fall when $$p,q$$ are composite and have a non trivial common factor (this is the cost of being nice). In fact, it has contact with the law the MO required in the following sense $$gcd(p_{\alpha}(X),p_{\beta}(X))=1\Longrightarrow p_\alpha(X)\bullet_{xy} p_\beta(X)=p_{\alpha\beta}(X)$$

iii) Even in the case of addition ($$T(x,y):=x+y$$), the coefficients of the law $$\bullet_{x+y}$$ are not easy to compute. To appreciate this, the interested reader can have a look there : A. Lascoux ("Classes de Chern d'un produit tensoriel", C. R. Acad. Sci. Paris Sér. A-B 286 (1978), no. 8, A385–A387)