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Computing the intersection of two splitting fields as $\mathbb Q$-vector spaces

Step-by-step mathematics solution: Computing the intersection of two splitting fields as $\mathbb Q$-vector spaces

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1. What the problem is asking (in plain language)

You are given two separable polynomials

[ f(x),\;g(x)\in\mathbb Q[x] ]

and you look at their splitting fields

[ K_f=\operatorname{Spl}{\mathbb Q}(f),\qquad K_g=\operatorname{Spl}{\mathbb Q}(g). ]

Both fields are finite Galois extensions of (\mathbb Q).
The question is:

Can we effectively produce a polynomial (h(x)\in\mathbb Q[x]) whose splitting field is exactly the intersection

[ L=K_f\cap K_g ? ]

Ideally the method should avoid having to compute the full Galois groups of (f) and (g); it would be nice if we could work only with linear algebra inside the compositum

[ K=K_fK_g=\operatorname{Spl}_{\mathbb Q}(fg). ]

2. Overview of the solution

Yes – an algorithm exists. The idea is to

  1. Build the compositum (K=K_fK_g). This is the splitting field of the product (fg); we can represent it as a simple extension (\mathbb Q[\theta]) for a primitive element (\theta).

  2. Identify the two subfields (K_f) and (K_g) inside (K). A root of (f) (resp. (g)) belonging to (K) will serve as a primitive element for the corresponding subfield.

  3. Compute the vector‑space intersection (L=K_f\cap K_g) inside the (\mathbb Q)-vector space (K) by solving a linear system.

  4. Produce a primitive element (\beta) of the intersection (a random (\mathbb Q)-linear combination of a basis of (L) works with probability  1).

  5. Return the minimal polynomial of (\beta) over (\mathbb Q). Its splitting field is precisely (L).

All steps are constructive and can be carried out with the standard algorithms for number fields (factorisation of polynomials over a number field, computing primitive elements, solving linear equations, etc.). No explicit computation of Galois groups is required, although the method can be phrased in group‑theoretic language if one wishes.

3. Detailed step‑by‑step algorithm

We write the algorithm in a way that could be implemented in a computer algebra system.

Input

Two separable polynomials

[ f(x),\;g(x)\in\mathbb Q[x]. ]

Output

A polynomial (h(x)\in\mathbb Q[x]) whose splitting field equals (K_f\cap K_g).

Step 0 – Preparations

  • Compute the degrees

    [ n_f=\deg f,\qquad n_g=\deg g, ] and verify separability (e.g. (\gcd(f,f’)=1) and (\gcd(g,g’)=1)).

  • Let

    [ n=[K_fK_g:\mathbb Q]=\deg\bigl(\operatorname{Spl}_{\mathbb Q}(fg)\bigr). ]

Step 1 – Construct the compositum (K=K_fK_g)

  1. Factor (f) and (g) over (\mathbb Q).
    Let (\alpha_1,\dots ,\alpha_{n_f}) be the (pairwise distinct) roots of (f) in an algebraic closure (\overline{\mathbb Q}); similarly (\beta_1,\dots , \beta_{n_g}) for (g).

  2. Pick a primitive element.
    By the primitive‑element theorem there exist rational numbers (c_1,c_2) such that

    [ \theta = \alpha_1 + c_1\beta_1 + c_2 ]

    generates the whole compositum (K).
    In practice one can try a few random integer pairs ((c_1,c_2)) until the minimal polynomial of (\theta) over (\mathbb Q) has degree (n). (The probability of success is (>1/2) for a random choice.)

  3. Compute the minimal polynomial of (\theta).
    Use the standard “resultant’’ or “modular’’ algorithms to obtain

    [ m_\theta (x)=\operatorname{MinPoly}_\mathbb{Q}(\theta)\in\mathbb Q[x]. ]

    The field (K) is now represented as

    [ K\;=\;\mathbb Q[\theta]\;\cong\;\mathbb Q[x]/(m_\theta (x)). ]

    All arithmetic in (K) will be performed with respect to the basis

    [ \mathcal B_K={1,\theta,\theta^2,\dots ,\theta^{n-1}}. ]

Step 2 – Embed the two splitting fields inside (K)

Factor the original polynomials in the ring (\mathbb Q[\theta]).

  1. Factor (f) over (K).
    Use the factorisation algorithm for polynomials over a number field to write

    [ f(x)=\prod_{i=1}^{n_f}(x-\alpha_i)\quad\text{in }K[x]. ]

    Pick any root, say (\alpha:=\alpha_1\in K).

  2. Set

    [ K_f = \mathbb Q[\alpha]\subseteq K. ]

    Since (f) is separable, (\alpha) generates the whole splitting field (K_f). Its degree (d_f=[K_f:\mathbb Q]) is the degree of the minimal polynomial of (\alpha) over (\mathbb Q) (computed by a resultant).

  3. Do the same for (g).
    Factor (g) in (K[x]), pick a root (\beta\in K) and set

    [ K_g = \mathbb Q[\beta]\subseteq K, \qquad d_g=[K_g:\mathbb Q]. ]

Thus we have concrete (\mathbb Q)-bases

[ \mathcal B_f={1,\alpha,\alpha^2,\dots ,\alpha^{d_f-1}},\qquad \mathcal B_g={1,\beta ,\beta^2,\dots ,\beta^{d_g-1}} ]

expressed as (\mathbb Q)-linear combinations of the basis (\mathcal B_K) of (K).

Step 3 – Compute the intersection (L = K_f\cap K_g) as a vector space

Write an arbitrary element of (K_f) as

[ x=\sum_{i=0}^{d_f-1}u_i\alpha^i,\qquad u_i\in\mathbb Q . ]

Express each power (\alpha^i) in the basis (\mathcal B_K); this gives a matrix

[ A_f\in\mathbb Q^{n\times d_f} ]

whose columns are the coordinates of (1,\alpha,\dots ,\alpha^{d_f-1}) in (\mathcal B_K).

Do the same for (K_g) and obtain (A_g\in\mathbb Q^{n\times d_g}).

An element (x) belongs to the intersection iff its coordinate vector (v=A_f\mathbf u) (with (\mathbf u\in\mathbb Q^{d_f})) also lies in the column space of (A_g). Therefore we must solve

[ A_f\mathbf u = A_g\mathbf w \quad\text{for }(\mathbf u,\mathbf w)\in \mathbb Q^{d_f}\times\mathbb Q^{d_g}. ]

This is a linear system over (\mathbb Q). Compute a basis of its solution space, then project the solutions onto the first (d_f) coordinates to obtain a basis ({\gamma_1,\dots ,\gamma_m}) of the intersection as a (\mathbb Q)‑vector space. (Equivalently, compute a basis of (\operatorname{im}A_f\cap\operatorname{im} A_g) inside (\mathbb Q^n).)

The dimension (m) equals ([L:\mathbb Q

Original question: Computing the intersection of two splitting fields as $\mathbb Q$-vector spaces on Mathematics Stack Exchange, licensed CC BY-SA.

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