Field Extensions and Subfields

The vector-space viewpoint on field extensions is the load-bearing wall of essentially all of Galois theory: the entire tower law for degrees, the classification of straightedge-and-compass constructible numbers, and the impossibility of trisecting a general angle or doubling the cube all rest on treating $K/F$ as a vector space and computing its dimension. In algebraic number theory, rings of integers in number fields ($K/\mathbb{Q}$ finite extensions) are studied precisely via this vector space structure combined with integrality conditions, underlying the arithmetic of class groups and unit groups. In coding theory and cryptography, finite field extensions $\mathbb{F}_{p^n}/\mathbb{F}_p$ are the vector-space backbone of Reed–Solomon codes, AES's finite-field arithmetic, and elliptic-curve cryptography over extension fields, where elements of $\mathbb{F}_{p^n}$ are literally represented as coordinate vectors relative to a chosen $\mathbb{F}_p$-basis. In algebraic geometry and complex analysis, function field extensions such as $\mathbb{C}(t)/\mathbb{C}(t^n)$ model branched covers of the Riemann sphere, with the extension degree computing the number of sheets. Even in computer algebra systems, symbolic computation with algebraic numbers (like $\sqrt[3]{2}$) is implemented by fixing a vector space basis of the extension field, exactly as in the worked example above, and reducing all field arithmetic to matrix and vector operations on coordinate tuples.

:::quiz[For the extension $\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q}$, which set is a basis for $\mathbb{Q}(\sqrt[3]{2})$ as a $\mathbb{Q}$-vector space?]

• $\{1, \sqrt[3]{2}\}$
• $\{\sqrt[3]{2}, \sqrt[3]{4}, \sqrt[3]{8}\}$
• $\{1, \sqrt[3]{2}, \sqrt[3]{4}\}$ ✓
• $\{1\}$

Ask the tutor: How does the vector-space dimension $[K:F]$ let us decide whether an element $\alpha \in K$ is algebraic or transcendental over $F$?

Think of a field as a self-contained number system: you can add, subtract, multiply, and divide (by nonzero elements) and never leave the system. A subfield is a smaller self-contained system living inside a larger one, using the exact same rules of arithmetic. The extension $K/F$ records that $F$'s arithmetic is a faithful restriction of $K$'s arithmetic — nothing new is introduced when you restrict attention to $F$-elements. The crucial shift in perspective is to forget that $K$ has its own multiplication and division except insofar as we allow multiplying $K$-elements by $F$-elements. What remains is exactly the data of a vector space: vectors are elements of $K$, scalars are elements of $F$, vector addition is $K$'s addition, and scalar multiplication is $K$'s multiplication restricted to $F \times K$. All eight vector space axioms (associativity and commutativity of addition, existence of zero and additive inverses, distributivity of scalar multiplication over vector and scalar addition, associativity of scalar multiplication, and the unit law $1 \cdot v = v$) are simply special cases of the field axioms of $K$ that happen to only involve elements of $F$ in the scalar slot. This is why the proof below is largely a matter of extracting the right instances of the field axioms of $K$. Picture $\mathbb{Q}(\sqrt{2}) = \{a + b\sqrt{2} : a, b \in \mathbb{Q}\}$ sitting over $\mathbb{Q}$. Every element is uniquely a $\mathbb{Q}$-linear combination of $1$ and $\sqrt{2}$, so as a vector space $\mathbb{Q}(\sqrt{2})$ looks exactly like $\mathbb{Q}^2$, with basis $\{1, \sqrt{2}\}$. The field multiplication is extra structure layered on top of this 2-dimensional vector space; it is not visible to the vector space alone, but it is what let us build the field in the first place (by demanding $\sqrt{2}\cdot\sqrt{2}=2$).

Example 1: $K = \mathbb{Q}(\sqrt[3]{2})$ over $F = \mathbb{Q}$. Let $F = \mathbb{Q}$ and let $K = \mathbb{Q}(\sqrt[3]{2}) = \{a + b\sqrt[3]{2} + c\sqrt[3]{4} : a,b,c \in \mathbb{Q}\} \subseteq \mathbb{R}$, where $\sqrt[3]{2}$ is the real cube root of $2$ and $\sqrt[3]{4} = (\sqrt[3]{2})^2$. We check $F$ is a subfield of $K$ and exhibit the $F$-vector space structure concretely. First, $\mathbb{Q} \subseteq K$ since every rational $a$ can be written as $a + 0\cdot\sqrt[3]{2} + 0 \cdot \sqrt[3]{4}$. The field operations of $K$ (addition, multiplication, inversion, inherited from $\mathbb{R}$) restrict on $\mathbb{Q}$ to the usual field operations of $\mathbb{Q}$, and $\mathbb{Q}$ is closed under these since sums, products, negatives, and inverses of rationals are rational. So $\mathbb{Q}$ is a subfield of $K$ and $K/\mathbb{Q}$ is a field extension. By the theorem, $K$ is a $\mathbb{Q}$-vector space, with vector addition

and scalar multiplication for $q \in \mathbb{Q}$:

The set $\{1, \sqrt[3]{2}, \sqrt[3]{4}\}$ spans $K$ over $\mathbb{Q}$ by definition of $K$, and it is $\mathbb{Q}$-linearly independent (a standard consequence of the irreducibility of $x^3 - 2$ over $\mathbb{Q}$, to be justified rigorously once we develop Minimal Polynomials). Granting this, $\{1,\sqrt[3]{2},\sqrt[3]{4}\}$ is a basis, so

Let us verify a genuinely non-symmetric linear combination behaves correctly under the vector space axioms. Take vectors $u = 2 - \sqrt[3]{2} + 5\sqrt[3]{4}$ and $v = -3 + 4\sqrt[3]{2} + \sqrt[3]{4}$ in $K$, and scalars $a = \tfrac{1}{2}, b = -3 \in \mathbb{Q}$. Compute

Then

On the other hand $a\cdot(u) + b\cdot(v)$ should equal, by distributivity axioms (V5)–(V7) established in the theorem, the vector obtained by first forming $\tfrac{1}{2}u - 3v$ symbolically and then simplifying — which is exactly the same computation, confirming consistency (this is not a separate check but an illustration that the abstract axioms proved above reduce to ordinary polynomial arithmetic in $\sqrt[3]{2}$, exactly as guaranteed by Step 3–5 of the proof). Notice this computation never needed to multiply $u$ by $v$; multiplying two vectors in $K$ is possible (since $K$ is a field) but is extra structure not used by the vector space axioms — a point the proof made precise in Step 2.

Example 2: $K = \mathbb{F}_2(t)$ over $F = \mathbb{F}_2(t^2)$, a positive-characteristic, infinite, function-field example. Let $\mathbb{F}_2 = \{0,1\}$ be the field with two elements, and let $\mathbb{F}_2(t)$ be the field of rational functions in an indeterminate $t$ with coefficients in $\mathbb{F}_2$, i.e. the field of fractions of the polynomial ring $\mathbb{F}_2[t]$. Let

This is the subfield of $K$ consisting of all rational functions expressible using only even powers of $t$. Checking $F$ is a subfield. The element $0 = 0/1$ and $1 = 1/1$ both lie in $F$ (take $p(x) = 0$ or $p(x)=1$, $q(x)=1$). If $f = p_1(t^2)/q_1(t^2)$ and $g = p_2(t^2)/q_2(t^2)$ are in $F$, then

and both numerator and denominator are again polynomials evaluated at $t^2$ (since sums and products of such polynomials, re-expressed, are still polynomials in $t^2$: e.g. if $p_1(x) = x+1, q_2(x) = x^2$ then $p_1(t^2)q_2(t^2) = (t^2+1)t^4$ is a polynomial in $t^2$). Hence $f+g, fg \in F$. Negation is trivial in characteristic $2$ since $-f = f$. If $f \ne 0$ then $f^{-1} = q_1(t^2)/p_1(t^2) \in F$. So $F$ is closed under all required operations, and one checks the field axioms (associativity, distributivity, etc.) are inherited automatically since $F \subseteq K$ and $K$ is a field — closure is the only thing that needs checking beyond what $K$ already guarantees. Thus $F$ is a subfield and $K/F$ is a field extension. The vector space structure. By the theorem, $K = \mathbb{F}_2(t)$ is an $F$-vector space. Here is a genuinely non-trivial, asymmetric instance of this structure. Consider the two vectors

and the scalars $a = \dfrac{t^2}{t^2+1} \in F$ and $b = t^2+1 \in F$ (note $t^2+1 = (t^2) + 1$ is manifestly a polynomial in $t^2$, so it lies in $F$; likewise $t^2/(t^2+1) \in F$). We may form the $F$-linear combination

Simplify the first term: $t^3 + t = t(t^2+1)$, so $\dfrac{t^3+t}{t^2+1} = t$, and hence the first term is $\dfrac{t^2}{t^2+1}\cdot t = \dfrac{t^3}{t^2+1}$. So

Putting over a common denominator $(t^2+1)(t^4+t^2+1)$:

This element of $K$ is not generally in $F$ — indeed $u = t \notin F$ (since $t$ is not a rational function of $t^2$: over $\mathbb{F}_2$, $[K:F] = 2$ with $\{1,t\}$ a basis, a fact from the theory of inseparable extensions in characteristic $p$ foreshadowed here but proved properly once degrees are developed). This example emphasizes: (i) $F$ can be an infinite field and $K/F$ can still be a finite-dimensional vector space extension ($\dim_F K = 2$ here, with basis $\{1,t\}$, since every rational function in $t$ can be split into its "even part" plus $t$ times its "odd part," both even parts being genuine elements of $F$); (ii) characteristic $2$ causes no breakdown of the vector space axioms proved in the theorem, which used only general field axioms valid in any characteristic.