Simple Extensions and F(alpha)

The construction $F(\alpha) \cong F[x]/(m(x))$ is the single most-used computational device in Field & Galois Theory and far beyond it. In algebraic number theory, number fields like $\mathbb{Q}(\sqrt[3]{2})$ or $\mathbb{Q}(\zeta_n)$ (cyclotomic fields) are defined and computed with exactly this representation, and computer algebra systems (PARI/GP, Magma, Sage) store algebraic numbers internally as polynomials modulo a minimal polynomial. In coding theory and cryptography, finite fields $\mathbb{F}_{p^n} \cong \mathbb{F}_p[x]/(m(x))$ for irreducible $m$ of degree $n$ are the arithmetic backbone of Reed–Solomon codes, AES (built over $\mathbb{F}_{2^8} = \mathbb{F}_2[x]/(x^8+x^4+x^3+x+1)$), and elliptic-curve cryptography over binary fields. In computer algebra and symbolic computation, algebraic numbers are manipulated as polynomial residues, with exact arithmetic performed by polynomial division — precisely the reduction step used in the worked examples above. In Galois theory itself, the theorem is the base case for the Tower Law (which shows how degrees multiply when you adjoin several elements in stages) and for the construction of Splitting Fields (which are built by repeatedly adjoining one root at a time, exactly as in Worked Example 2, until a polynomial factors completely into linear factors).

:::quiz[Why is $F[x]/(m(x))$ a field rather than merely a ring, when $m(x)$ is the minimal polynomial of $\alpha$?]

• Because $F[x]$ is always a field
• Because $m(x)$ is monic
• Because $m(x)$ is irreducible, making $(m(x))$ a maximal ideal ✓
• Because $\alpha$ is transcendental

:::faq[Why must $\alpha$ be algebraic for $F(\alpha)\cong F[x]/(m(x))$ to hold with $m$ a nonzero polynomial?] The isomorphism comes from the First Isomorphism Theorem applied to the evaluation map $\varphi_\alpha: F[x]\to E$, $f\mapsto f(\alpha)$. The theorem always says $F[x]/\ker\varphi_\alpha \cong F[\alpha]$, but this is only useful when $\ker\varphi_\alpha$ is a nonzero ideal, which happens exactly when some nonzero polynomial vanishes at $\alpha$ — the definition of algebraic. If $\alpha$ is transcendental, $\ker\varphi_\alpha = (0)$, so the "quotient" is really just $F[x]$ itself, with no finite-degree structure and no minimal polynomial $m(x)$ to speak of.

:::faq[Is $F[\alpha]$ always equal to $F(\alpha)$?] Only when $\alpha$ is algebraic over $F$. In that case, the theorem shows the ring $F[\alpha]$ is already a field (because $(m(x))$ is maximal), so it must equal the smallest field containing $F$ and $\alpha$, which is $F(\alpha)$. When $\alpha$ is transcendental, $F[\alpha]$ is only an integral domain (isomorphic to the polynomial ring $F[x]$), and $F(\alpha)$ is strictly larger, isomorphic to the field of rational functions $F(x)$, obtained by formally adjoining inverses of nonzero elements of $F[\alpha]$.

:::faq[Does the isomorphism $F(\alpha)\cong F[x]/(m(x))$ depend on which root $\alpha$ of $m(x)$ we pick, if $m$ has several roots in some bigger field?] The abstract isomorphism class does not depend on the choice: for any root $\alpha$ of an irreducible $m(x)$, $F(\alpha) \cong F[x]/(m(x))$, so all such $F(\alpha)$ are isomorphic to each other via isomorphisms fixing $F$. However, as subsets of a fixed ambient field $E$, different roots $\alpha \ne \alpha'$ of $m(x)$ typically give different (though isomorphic) subfields $F(\alpha)$ and $F(\alpha')$ of $E$. This subtlety — same abstract field, different embeddings — is exactly the phenomenon that Galois theory studies via automorphisms permuting roots.

:::faq[Is it possible for $F[x]/(m(x))$ to fail to be a field even when $m$ is the minimal polynomial of some element?] No — if $m(x)$ genuinely is the minimal polynomial of an algebraic element $\alpha$ over $F$, then $m$ is automatically irreducible (proved in Step 4 of the proof, by a contradiction argument using that $E$ has no zero divisors), so $F[x]/(m(x))$ is automatically a field. The only way to get a non-field quotient $F[x]/(f(x))$ is to quotient by a reducible nonzero polynomial $f$, but such an $f$ can never arise as the minimal polynomial of an actual field element, precisely because minimal polynomials are irreducible by construction.

Ask the tutor: Can you show me step-by-step how to find the minimal polynomial and degree of $\mathbb{Q}(\sqrt{2}+\sqrt{3})$ over $\mathbb{Q}$?

Think of $F[x]$ as the world of formal polynomial expressions in a symbol $x$, and think of evaluation at $\alpha$ as a machine that takes a formal expression and produces an actual number (or field element) $f(\alpha) \in E$. The evaluation map $\varphi_\alpha : F[x] \to E$, $f(x) \mapsto f(\alpha)$, is a ring homomorphism. Its kernel records exactly the "coincidences" — the polynomial relations that $\alpha$ happens to satisfy. If $\alpha$ is algebraic, this kernel is nonzero, generated by the minimal polynomial $m(x)$; the First Isomorphism Theorem then says the image of $\varphi_\alpha$, which is precisely $F[\alpha]$, is isomorphic to $F[x]/(m(x))$. The subtlety — and the whole point of irreducibility — is that a generic quotient $F[x]/(f(x))$ is only a ring, with zero divisors if $f$ factors. It becomes a field exactly when $f$ is irreducible, because then $(f)$ is a maximal ideal in the PID $F[x]$. So the fact that $m(x)$ is irreducible (forced by minimality: any factorization would produce a smaller-degree polynomial vanishing at $\alpha$) is precisely the mechanism that upgrades the ring $F[\alpha]$ into the field $F(\alpha)$. This is why $F[\alpha] = F(\alpha)$ whenever $\alpha$ is algebraic: no further field operations (inverses) need to be adjoined, they are already present. Geometrically, picture $F(\alpha)$ as a vector space stacked in layers over $F$: the layers are $1, \alpha, \alpha^2, \dots, \alpha^{n-1}$, and the relation $m(\alpha) = 0$ tells you exactly how to collapse $\alpha^n$ and all higher powers back down into this finite list of layers.

Example 1: an algebraic number field, $\mathbb{Q}(\sqrt[3]{2}\,)$ inside $\mathbb{R}$. Let $F = \mathbb{Q}$, and let $\alpha = \sqrt[3]{2} \in \mathbb{R}$, a real root of $f(x) = x^3 - 2$. We claim $f$ is the minimal polynomial of $\alpha$ over $\mathbb{Q}$. First, $f(x) = x^3 - 2$ is irreducible over $\mathbb{Q}$ by Eisenstein's criterion at the prime $p=2$: the leading coefficient $1$ is not divisible by $2$, the coefficients $0, 0, -2$ of $x^2, x, 1$ are all divisible by $2$, and $-2$ is not divisible by $2^2 = 4$. So $f$ is irreducible over $\mathbb{Q}$ (equivalently over $\mathbb{Z}$, by Gauss's Lemma, since $f$ is monic). Since $f(\alpha) = (\sqrt[3]2)^3 - 2 = 2-2=0$ and $f$ is monic irreducible, $f$ is exactly the minimal polynomial $m_\alpha(x)$ of $\alpha$ over $\mathbb{Q}$. By the theorem, $\deg m_\alpha = 3$, so

with basis $\{1, \alpha, \alpha^2\} = \{1, \sqrt[3]{2}, \sqrt[3]{4}\}$, and

Let's verify the arithmetic really does reduce mod $m(x)$, as the theorem promises. Compute $\alpha^{-1}$ (its existence as an element of this explicit form is itself a non-obvious fact the theorem guarantees). We want $a,b,c \in \mathbb{Q}$ with $(a+b\alpha+c\alpha^2)\alpha = 1$, i.e.

Since $\alpha^3 = 2$, this is $2c + a\alpha + b\alpha^2 = 1$. Matching coefficients against the basis (using linear independence, part 2 of the theorem, to justify that matching coefficients is legitimate): $2c=1$, $a=0$, $b=0$. So $c = 1/2$, and

Check: $\sqrt[3]2 \cdot \tfrac12\sqrt[3]4 = \tfrac12 \sqrt[3]{8} = \tfrac12 \cdot 2 = 1$. Correct — and this inverse was produced entirely by polynomial reduction mod $x^3-2$, exactly as the isomorphism $F[x]/(m) \cong F(\alpha)$ predicts, with no appeal to any a priori field structure beyond that guaranteed by the theorem.

Example 2: a finite field, $\mathbb{F}_2(\alpha)$ from an irreducible cubic. Let $F = \mathbb{F}_2 = \{0,1\}$, and let $m(x) = x^3 + x + 1 \in \mathbb{F}_2[x]$. We check irreducibility: a cubic over a field is irreducible iff it has no root in that field (any factorization of a cubic must have a linear factor). Testing: $m(0) = 0+0+1=1 \ne 0$, and $m(1) = 1+1+1 = 1 \ne 0$ (arithmetic mod 2). So $m(x)$ has no roots in $\mathbb{F}_2$, hence is irreducible over $\mathbb{F}_2$. Let $\alpha$ be a root of $m(x)$ in some extension field $E \supseteq \mathbb{F}_2$ (its existence, which we take as given here, is the subject of the Splitting Fields construction — precisely one of the things this theorem foreshadows). By the theorem, $m(x)$ is the minimal polynomial of $\alpha$ (it's already monic irreducible and vanishes at $\alpha$), $\deg m = 3$, and

with basis $\{1,\alpha,\alpha^2\}$ over $\mathbb{F}_2$, so $[\mathbb{F}_2(\alpha):\mathbb{F}_2] = 3$ and $|\mathbb{F}_2(\alpha)| = 2^3 = 8$. Let's compute a nontrivial product using only the relation $\alpha^3 = \alpha + 1$ (from $m(\alpha)=0$, rearranged; note $-1=1$ in $\mathbb{F}_2$). Take $\beta = 1+\alpha$ and $\gamma = \alpha^2+\alpha$, both nonzero elements of $\mathbb{F}_2(\alpha)$. Then

Now reduce using $\alpha^3 = \alpha+1$:

So $\gamma = \beta^{-1}$: the element $\alpha^2+\alpha$ is the multiplicative inverse of $1+\alpha$ in this 8-element field, computed purely by polynomial reduction mod $x^3+x+1$ — again exactly the content of the theorem, now in the finite-field setting that underlies coding theory and cryptography.