Uniqueness of Splitting Fields

The uniqueness of splitting fields, and the isomorphism extension theorem behind it, is the mechanism that makes "the Galois group of $f$ over $F$" a well-defined object rather than merely a group attached to one arbitrary construction. It underlies the construction of the Algebraic Closure of any field, since an algebraic closure can be built as a "splitting field of all polynomials at once," and its uniqueness up to isomorphism uses the same root-by-root matching argument. In coding theory and cryptography, finite fields $\mathbb{F}_{p^n}$ are constructed as splitting fields of $x^{p^n}-x$ over $\mathbb{F}_p$, and their uniqueness (any two finite fields of the same order are isomorphic) is a direct instance of this theorem, essential for the consistency of finite-field arithmetic used in AES encryption and Reed–Solomon error-correcting codes. In computer algebra systems, when a program constructs a splitting field symbolically (e.g. to compute a Galois group), the correctness of comparing outputs from different algorithms — say, different choices of which root to adjoin first — rests on exactly this uniqueness statement. The theorem's separable-case root count, $[E:F]$ automorphisms exactly, is the starting point for the fundamental correspondence between subgroups and subfields central to The Galois Group of an Extension and the Fundamental Theorem of Galois Theory.

:::quiz[If $E$ and $E'$ are splitting fields of the same polynomial $f\in F[x]$ over $F$, what can we conclude?]

• $E$ and $E'$ are equal as subsets of some ambient field
• $E$ and $E'$ are isomorphic as abstract rings, with no further guarantee
• There is an isomorphism $E \to E'$ restricting to the identity on $F$ ✓
• $E$ and $E'$ must have different degrees over $F$ if constructed differently

:::quiz[When is the number of extensions of $\sigma$ to the splitting field exactly equal to $[E:F]$, rather than merely at most $[E:F]$?]

• Always, regardless of any hypothesis
• Only when $F$ has characteristic zero
• When $f$ has no repeated irreducible factors, i.e. is separable over $F$ ✓
• When $E = F$

Ask the tutor: Can you show me how the root-count in the separable case fails, step by step, for $f(x) = x^p - t$ over $\mathbb{F}_p(t)$?

Think of building a splitting field as laying railroad track one root at a time. Starting from $F$, we pick an irreducible factor $p(x)$ of $f(x)$ in $F[x]$, form $F(\alpha) = F[x]/(p(x))$ for a root $\alpha$, and repeat with the leftover factors of $f$ over the new bigger field. Each step is forced up to isomorphism: adjoining a root of an irreducible polynomial $p(x)$ always produces the ring $F[x]/(p(x))$, regardless of which root of $p(x)$, in whichever splitting field you happen to be sitting inside, you chose to adjoin. The isomorphism extension theorem says this track-laying process can be matched, step by step, between two different constructions. If someone else adjoined a different root of the same irreducible polynomial $p(x)$, the two intermediate fields $F(\alpha)$ and $F(\alpha')$ are still isomorphic via a map sending $\alpha \mapsto \alpha'$ and fixing $F$. The heart of the proof is that this local, single-step isomorphism can always be pushed forward to the rest of the tower by induction, because the "amount of field left to build" — measured by degree — strictly decreases at each step.

Example 1: Matching two different constructions of the splitting field of $x^3 - 2$ over $\mathbb{Q}$. Let $f(x) = x^3 - 2 \in \mathbb{Q}[x]$, irreducible over $\mathbb{Q}$ by Eisenstein at $p=2$. Its roots in $\mathbb{C}$ are $\alpha = \sqrt[3]{2}$ (real cube root), $\omega\alpha$, and $\omega^2\alpha$, where $\omega = e^{2\pi i/3}$ is a primitive cube root of unity. Construction A: $E = \mathbb{Q}(\alpha, \omega\alpha, \omega^2\alpha) = \mathbb{Q}(\alpha,\omega)$, built by first adjoining $\alpha$ (degree 3 over $\mathbb{Q}$), then $\omega$ (degree 2 over $\mathbb{Q}(\alpha)$, since $\mathbb{Q}(\alpha)\subset \mathbb{R}$ while $\omega\notin\mathbb{R}$, and $\omega$ satisfies $x^2+x+1=0$). So $[E:\mathbb{Q}] = 3\cdot 2 = 6$. Construction B: someone else builds a splitting field $E'$ by first adjoining the different root $\beta = \omega\alpha$: set $F_1' = \mathbb{Q}(\beta)$, also degree 3 over $\mathbb{Q}$ since $\beta$ has the same minimal polynomial $x^3-2$ (as $\beta^3 = \omega^3\alpha^3 = 2$, and $x^3-2$ is irreducible so it is the minimal polynomial of any of its roots). Then $E' = F_1'(\alpha) = \mathbb{Q}(\beta,\alpha)$, and one checks $E' = \mathbb{Q}(\alpha,\omega)$ as well since $\omega = \beta/\alpha \in E'$ — but let's not assume the constructions coincide as subsets of $\mathbb{C}$, and instead run the theorem's algorithm abstractly, taking $E, E'$ as abstract splitting fields. Apply the theorem with $\sigma = \mathrm{id}_\mathbb{Q}$. Step 1: $p(x) = x^3-2$ is our chosen irreducible factor of $f$. In $E$, pick the root $\alpha$; in $E'$, pick the root $\beta$. By the single-step lemma there is an isomorphism

Concretely $\sigma_1(a + b\alpha + c\alpha^2) = a + b\beta + c\beta^2$ for $a,b,c \in \mathbb{Q}$; this is well-defined and a ring isomorphism because both sides are $\mathbb{Q}[x]/(x^3-2)$ in disguise. Step 2: now $E = \mathbb{Q}(\alpha)(\omega,\ldots)$ is the splitting field of $f$ over $\mathbb{Q}(\alpha)$ (all three roots $\alpha,\omega\alpha,\omega^2\alpha$ lie in $E$ and generate it over $\mathbb{Q}(\alpha)$, since $\omega = (\omega\alpha)/\alpha \in E$), and $E'$ is the splitting field of $\sigma_1(f) = x^3-2$ over $\mathbb{Q}(\beta)$. Over $\mathbb{Q}(\alpha)$, $f$ factors as $(x-\alpha)(x^2+\alpha x + \alpha^2)$; the quadratic factor is irreducible over $\mathbb{Q}(\alpha) \subset \mathbb{R}$ (its roots $\omega\alpha,\omega^2\alpha$ are non-real). We adjoin a root of this quadratic, say $\omega\alpha$, in $E$; in $E'$ we must send it to some root of $\sigma_1(x^2+\alpha x+\alpha^2) = x^2+\beta x + \beta^2$ in $E'$, and $E'$ contains such a root since $E'$ splits $\sigma_1(f)$. By the base case ($n=1$ for this last step, since adjoining a root of a quadratic over $\mathbb{Q}(\alpha)$ already reaches all of $E$), we extend $\sigma_1$ to a full isomorphism $\tau: E \to E'$ with $\tau(\alpha) = \beta$, $\tau(\omega\alpha) \mapsto$ a chosen root of $x^2+\beta x+\beta^2$. Since $[E:\mathbb{Q}] = 6$ and $x^3-2$ is separable (its three complex roots are distinct — check $\gcd(x^3-2, 3x^2)=1$ over $\mathbb{Q}$), the theorem guarantees exactly $6$ such $\tau$'s in total: $3$ choices for where $\alpha$ goes (any of the 3 roots of $x^3-2$ in $E'$) times $2$ choices for where the adjoined quadratic root goes at the second stage. This matches $|\mathrm{Gal}(E/\mathbb{Q})| = 6$, foreshadowing The Galois Group of an Extension.

Example 2: A base isomorphism that is not the identity, over a finite field. Let $F = \mathbb{F}_3(t)$, the field of rational functions in one variable $t$ over $\mathbb{F}_3 = \mathbb{Z}/3\mathbb{Z}$, and let $F' = \mathbb{F}_3(u)$ similarly. Let $\sigma: F \to F'$ be the field isomorphism determined by $\sigma(t) = u$ (and $\sigma|_{\mathbb{F}_3} = \mathrm{id}$); this is a genuine, non-identity-flavored isomorphism between two different (though isomorphic) fields — a useful stress test since $F\ne F'$ literally as sets. Let $f(x) = x^2 - t \in F[x]$. This is irreducible over $F$: if it factored, it would have a root $r(t) \in \mathbb{F}_3(t)$ with $r(t)^2 = t$, but comparing degrees of numerator/denominator (writing $r = p/q$ in lowest terms, $p^2 = tq^2$ forces $\deg p^2$ even and $\deg(tq^2)$ odd unless $q$ is a unit, and then $p^2=tq^2$ with $q\in\mathbb{F}_3^\times$ still forces $t$ to be a perfect square in $\mathbb{F}_3[t]$, impossible by comparing degrees mod 2) — no such rational function exists. Then $\sigma(f)(x) = x^2 - u \in F'[x]$, also irreducible over $F'$ by the identical argument. A splitting field of $f$ over $F$ is $E = F(\sqrt t\,) = \mathbb{F}_3(t)(\sqrt t)$, and here we should be careful: $\mathrm{char}(F) = 3 \ne 2$, so $x^2 - t$ is separable (its two roots $\pm\sqrt t$ are distinct since $2\sqrt t \ne 0$ in characteristic $3$). Likewise $E' = F'(\sqrt u) = \mathbb{F}_3(u)(\sqrt u)$ is a splitting field of $\sigma(f)$ over $F'$, with $[E':F'] = 2$. By the theorem, $\sigma$ extends to an isomorphism $\tau: E \to E'$, and since $f$ is separable of degree $2$ there are exactly $[E:F] = 2$ such extensions. Explicitly they are

both restricting to $\sigma$ on $F$ (sending $t\mapsto u$). One checks directly that $\tau_1$ is a well-defined field homomorphism: every element of $E$ is $a(t) + b(t)\sqrt t$ with $a,b \in F$, and we set $\tau_1(a(t)+b(t)\sqrt t) = \sigma(a)(u) + \sigma(b)(u)\sqrt u$; this respects addition and multiplication because $(\sqrt t)^2 = t \mapsto u = (\sqrt u)^2$ under $\sigma$, matching the relation $(\sqrt u)^2 = u$ in $E'$. This example shows the theorem working with genuinely different base fields, not merely two copies of the same field with the identity map — the mechanism is purely about matching irreducible-factor structure across $\sigma$.