Finitely Generated Algebraic Extensions

The fact that finite extensions are algebraic and finitely generated — and its converse for finite generation — is the load-bearing wall underneath essentially all of classical field theory. It is the first step in proving that the algebraic numbers $\overline{\mathbb{Q}} \subseteq \mathbb{C}$ form a field (via the corollary), a fact silently used every time one manipulates expressions like $\sqrt2 + \sqrt[3]5$ as though they obviously satisfy polynomial equations. It underlies the construction of splitting fields of polynomials, where one adjoins finitely many roots and needs to know the resulting field is finite over the base — a prerequisite for defining and computing Splitting Fields and, from there, Galois groups. Iterating the finite-generation argument to an infinite process, replacing "finitely many algebraic generators" with "all algebraic elements over a base field," is exactly the construction behind the existence of an Algebraic Closure of any field. Beyond pure algebra, degree-counting via the Tower Law is the standard tool in classical constructibility problems: showing that doubling the cube or trisecting a general angle is impossible with straightedge and compass reduces to showing certain field extensions have degree not a power of $2$, using exactly the machinery here to compute and bound degrees of towers of quadratic extensions. In algebraic number theory, number fields are by definition finite extensions of $\mathbb{Q}$, and the theorem guarantees every number field is both algebraic (every algebraic number in it satisfies a polynomial with rational coefficients) and finitely generated (by a primitive element, in fact — foreshadowing the Primitive Element Theorem). In coding theory and cryptography, finite fields $\mathbb{F}_{p^n}$ arise as finite (hence algebraic, finitely generated) extensions of $\mathbb{F}_p$, and degree computations exactly like Example 1 determine the size and structure of the field used in Reed–Solomon codes and elliptic-curve cryptosystems.

:::quiz[If $[K:F] = 7$, which statement must be true for every $\alpha \in K$?]

• $[F(\alpha):F]$ is a power of 2
• $[F(\alpha):F] \in \{1, 7\}$ ✓
• $\alpha$ is transcendental over $F$
• $F(\alpha) = K$ for every choice of $\alpha$

:::quiz[$K = F(\alpha_1,\alpha_2)$ where $\deg m_{\alpha_1}^F = 3$ and $\deg m_{\alpha_2}^F = 5$. What can we conclude about $[K:F]$?]

• It equals exactly $15$
• It is finite and at most $15$ ✓
• It is infinite since $3$ and $5$ are coprime
• It equals $\gcd(3,5) = 1$

:::faq[Why does the bound $[K:F] \le \prod \deg m_{\alpha_i}^F$ sometimes fail to be an equality?] The bound comes from applying the Tower Law along a chain $F \subseteq F(\alpha_1) \subseteq F(\alpha_1,\alpha_2) \subseteq \cdots \subseteq K$, using at each step that $[E(\alpha_i):E] \le \deg m_{\alpha_i}^F(x)$. Equality at each step requires the minimal polynomial of $\alpha_i$ over $F$ to remain irreducible over the bigger field $E$ built from the previous generators. If $\alpha_i$'s minimal polynomial factors further once earlier generators are adjoined — for instance if $\alpha_2$ turns out to already lie in $F(\alpha_1)$, as in Example 2 with $\sqrt8 \in \mathbb{Q}(\sqrt2)$ — then $[E(\alpha_i):E]$ drops below $\deg m_{\alpha_i}^F$, and the overall product bound is strict.

Ask the tutor: How does this theorem set up the construction of a splitting field for an arbitrary polynomial?

Think of building a field extension the way you build a vector space: you start with $F$ and toss in new elements one at a time. Each time you adjoin an element $\alpha$ that is algebraic over your current field $E$, the resulting field $E(\alpha)$ is a finite-dimensional $E$-vector space, with basis $1, \alpha, \alpha^2, \dots, \alpha^{d-1}$ where $d$ is the degree of $\alpha$'s minimal polynomial over $E$. Adjoining finitely many algebraic elements one after another is like stacking finitely many finite-dimensional floors on top of each other in a tower — and the Tower Law says that the total height (degree) of the tower is the product of the individual floor heights, which is still finite. The subtlety, and the heart of the theorem, is that when we adjoin $\alpha_1, \dots, \alpha_n$ simultaneously — algebraic over the original $F$, not necessarily over each other's field yet — we must check that each $\alpha_i$ remains algebraic over the growing tower $F(\alpha_1,\dots,\alpha_{i-1})$. This is easy: any polynomial over $F$ killing $\alpha_i$ still lives in the bigger ring $F(\alpha_1,\dots,\alpha_{i-1})[x]$ and still kills $\alpha_i$. So the tower of simple extensions is forced to have finite floors all the way up, and finiteness of the whole tower follows from finiteness of each floor via the Tower Law — this is exactly the mechanism in the proof below. ```svg ```

Example 1: Bounding and computing a genuinely awkward degree. Let $F = \mathbb{Q}$ and consider $K = \mathbb{Q}(\sqrt{2}, \sqrt[3]{5})$. We compute $[K:\mathbb{Q}]$ using the finitely-generated-algebraic-extension machinery, and then verify the bound is tight here (it need not always be). The minimal polynomial of $\alpha_1 = \sqrt{2}$ over $\mathbb{Q}$ is $x^2 - 2$ (irreducible over $\mathbb{Q}$ by Eisenstein at $p=2$, or since $\sqrt2$ is irrational), of degree $2$. The minimal polynomial of $\alpha_2 = \sqrt[3]{5}$ over $\mathbb{Q}$ is $x^3-5$ (Eisenstein at $p=5$), of degree $3$. The theorem gives the a priori bound

and in particular $[K:\mathbb{Q}]$ is finite, which is exactly the substantive content we needed — a priori it was not obvious that a field generated by a square root and a cube root wouldn't be infinite-dimensional. To see the bound is achieved, use the Tower Law along $\mathbb{Q} \subseteq \mathbb{Q}(\sqrt2) \subseteq K$: we know $[\mathbb{Q}(\sqrt2):\mathbb{Q}]=2$. For the top floor, $\sqrt[3]{5}$ is a root of $x^3 - 5 \in \mathbb{Q}(\sqrt2)[x]$; if this cubic were reducible over $\mathbb{Q}(\sqrt2)$ it would have a root there, i.e. $\sqrt[3]{5} \in \mathbb{Q}(\sqrt2) = \{a+b\sqrt2 : a,b \in \mathbb{Q}\}$. But every element of $\mathbb{Q}(\sqrt2)$ has degree $1$ or $2$ over $\mathbb{Q}$, while $\sqrt[3]5$ has degree $3$ (since $x^3-5$ is its minimal polynomial), a contradiction. So $x^3-5$ is irreducible over $\mathbb{Q}(\sqrt2)$, giving $[K : \mathbb{Q}(\sqrt2)] = 3$, and

So here the theorem's upper bound of $6$ is achieved exactly. By the corollary, $\alpha = \sqrt2 + \sqrt[3]5 \in K$ is automatically algebraic over $\mathbb{Q}$ of degree dividing $[K:\mathbb{Q}] = 6$ (by Theorem 1's degree-divisibility statement); one can further show $\mathbb{Q}(\alpha) = K$ (a "primitive element" phenomenon), so in fact $[\mathbb{Q}(\alpha):\mathbb{Q}] = 6$ exactly, meaning $\alpha$ satisfies an irreducible sextic over $\mathbb{Q}$ — without our machinery, finding that sextic directly by eliminating radicals from $(\alpha - \sqrt2)^3 = 5$ is a genuine (if mechanical) algebra chore that the theorem lets us bypass entirely for the finiteness conclusion.

Example 2: A case where the bound is not tight, showing the inequality is genuinely an inequality. Let $F = \mathbb{Q}$, $\alpha_1 = \sqrt2$, and $\alpha_2 = \sqrt8 = 2\sqrt2$. Both have minimal polynomials of degree $2$ over $\mathbb{Q}$ (namely $x^2-2$ and $x^2-8$, the latter irreducible over $\mathbb{Q}$ since $\sqrt8=2\sqrt2\notin\mathbb{Q}$). The theorem's bound gives

But directly, $\sqrt8 = 2\sqrt2 \in \mathbb{Q}(\sqrt2)$ already, so $\mathbb{Q}(\sqrt2,\sqrt8) = \mathbb{Q}(\sqrt2)$, of degree exactly $2$ over $\mathbb{Q}$ — strictly less than the bound $4$. This illustrates that $\prod \deg m_{\alpha_i}$ is only an upper bound: the true degree $[E(\alpha_n):E]$ in the inductive step can be strictly smaller than $\deg m_{\alpha_n}^F$ whenever $\alpha_n$'s minimal polynomial over $F$ factors further, or already has a root, once we pass to the intermediate field $E$. This is precisely the gap tracked by inequality $(\ast\ast)$ in the proof: equality holds there exactly when $m_{\alpha_n}^F(x)$ remains irreducible over $E$, which fails in this example since $x^2-8 = (x-2\sqrt2)(x+2\sqrt2)$ splits completely already over $E = \mathbb{Q}(\sqrt2)$.