Defining polynomial of compositum of splitting fields

Question

Let $L_1,\dotsc,L_n/K$ be finite separable field extensions. Then the compositum extension $L:=L_1\cdot\dotsb\cdot L_n/K$ is also finite and separable. Thus by the primitive element theorem, there are $\theta_i$ and $\theta$ in some fixed algebraic closure $\overline{K}$ of $K$ such that $L_i=K(\theta_i)$ and $L=K(\theta)$.

Suppose the $f_i(x)$ are the minimal polynomials of the $\theta_i$ over $K$, so that $L_i\cong K[x]/\langle f_i(x)\rangle$. Assuming that the $f_i$ are known, is there an algorithm to compute a generating polynomial for $L/K$, i.e. an irreducible polynomial $f\in K[x]$ so that $L=L_1\cdot\dotsb\cdot L_n\cong K[x]/\langle f(x)\rangle$, and $f$ can be computed from the $f_i$ in some way?

After some searching, I think the answer is related to the resultant $$\operatorname{Res}_y(f_i(y),f_j(x-y)),$$

which is mentioned in section 2 of [1] and section 3.2 of [2]; [1] states that the above resultant does the trick when $n=2$ (albeit without a formal proof), while [2] explains that the resultant has roots which are precisely the sums of a root of $f_i$ and a root of $f_j$ (not hard to see), but I don't see when we get irreducibility.

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[1] Awtrey, C., Cesarski, T., & Jakes, P. (2017). Determining Galois Groups of Reducible Polynomials Via Discriminants and Linear Resolvents. JP Journal of Algebra, Number Theory and Applications, 39(5), 685–702.

[2] Soicher, L. (1981). The Computation of Galois Groups [Master’s Thesis, Concordia University]. https://spectrum.library.concordia.ca/id/eprint/4916/1/MK49693.pdf

Answer 1

For monic polynomials $f,g$ define $f*g=\mathrm{Res}_y(f(y),g(x-y))$. This is an associative and commutative operation on univariate monic polynomials with neutral element $x$. If $f=\prod_{i=1}^m(x-\alpha_i),\,g=\prod_{j=1}^k(x-\beta_j)$ then $f*g=\prod_{i=1}^m\prod_{j=1}^k(x-\alpha_i-\beta_j)$.

In your setup, if $L_1,\ldots,L_n$ are linearly disjoint (i.e. $L_1\otimes\cdots\otimes L_n$ is a field) then $f_1*\cdots*f_n$ is irreducible and one of its roots generates $L$. This doesn't work in general, even in the case $n=2$: consider $K=\mathbb Q$ and let $\alpha_1,\alpha_2,\alpha_3$ be the roots of $x^3-2$. Set $L_i=\mathbb Q(\alpha_i)$ and $f_i=x^3-2$ ($1\le i\le 3$). Then the roots of $f_1*f_2$ are $-\alpha_i,2\alpha_i,\,1\le i\le 3$ and neither of these (taken alone) generates $L_1L_2$ which has degree 6 over $\mathbb Q$.

What one can do in general is take $f=\frac 1{c_1^{\deg f_1}}f_1(c_1x)*\cdots*\frac 1{c_n^{\deg f_n}}f(c_nx)$ for some $c_1,\ldots,c_n\in K^\times$. This polynomial is not always irreducible, but if it happens to be separable (which happens for a generic choice of $c_1,\ldots,c_n$ in the sense of being in a nonempty Zariski open subset) then one of the roots of $f$ generates $L_1\cdots L_n$.