Field Extensions; Tower Law; Algebraic Closure

At a Glance

• Frequency: 2 sub-parts across 1 of 13 years (2016)
• Priority tier: T4
• Marks (count): 35 (2)
• Average solve time: ~20 min
• Difficulty mix: medium 2
• Section: B | Dominant type: proof

Why This Chapter Matters

Field extensions are the conceptual heart of Galois theory and underpin every solvability-by-radicals argument. UPSC 2016 set two sub-parts on this topic totalling 35 marks — among the heaviest algebra allocations in recent years. The tower law is the workhorse: if you can compute degrees correctly, you can answer nearly any degree-of-extension question. The proofs demanded involve minimal polynomials, the simple extension theorem, and properties of algebraic elements — all highly structured and template-driven once the definitions are in order.

Minimum Theory

Field Extensions

A field extension $K/F$ means that $K$ is a field and $F$ is a subfield of $K$ (a subset closed under $+$, $\cdot$, and inverses that is itself a field with the same operations).

The degree of $K$ over $F$ is $[K : F] = \dim_F K,$ the dimension of $K$ as a vector space over $F$. We say $K/F$ is finite if $[K:F] < \infty$.

Tower Law (Degree Formula)

Theorem. If $F \subseteq K \subseteq E$ are fields then $[E : F] = [E : K] \cdot [K : F].$

Proof. Let $\{u_i\}_{i \in I}$ be an $F$-basis of $K$ and $\{v_j\}_{j \in J}$ be a $K$-basis of $E$.

Claim: $\{u_i v_j\}_{i \in I, j \in J}$ is an $F$-basis of $E$.

Spanning: Any $e \in E$ writes as $e = \sum_j c_j v_j$ with $c_j \in K$, and each $c_j = \sum_i a_{ij} u_i$ with $a_{ij} \in F$, so $e = \sum_{i,j} a_{ij} u_i v_j$.

Linear independence over $F$: Suppose $\sum_{i,j} a_{ij} u_i v_j = 0$ with $a_{ij} \in F$. Rewrite as $\sum_j \left(\sum_i a_{ij} u_i\right) v_j = 0$. Since $v_j$ are $K$-linearly independent and $\sum_i a_{ij} u_i \in K$, each coefficient $\sum_i a_{ij} u_i = 0$. Since $u_i$ are $F$-linearly independent, each $a_{ij} = 0$.

Hence $[E:F] = |I| \cdot |J| = [K:F] \cdot [E:K]$. $\blacksquare$

Algebraic Elements and Minimal Polynomials

$\alpha \in K$ is algebraic over $F$ if $p(\alpha) = 0$ for some non-zero $p \in F[x]$.

The minimal polynomial $m_\alpha \in F[x]$ of $\alpha$ over $F$ is the unique monic polynomial of least degree such that $m_\alpha(\alpha) = 0$.

Properties of the minimal polynomial:

• $m_\alpha$ is irreducible over $F$.
• $m_\alpha$ divides every polynomial in $F[x]$ that has $\alpha$ as a root.
• $[F(\alpha) : F] = \deg m_\alpha$.

Simple Extensions

Theorem. If $\alpha$ is algebraic over $F$ with minimal polynomial $p(x)$ of degree $n$, then $F(\alpha) \cong F[x]/\langle p(x) \rangle$ as $F$-algebras, and $[F(\alpha) : F] = n$.

A basis for $F(\alpha)$ over $F$ is $\{1, \alpha, \alpha^2, \ldots, \alpha^{n-1}\}$.

Algebraic Extensions

An extension $K/F$ is algebraic if every $\alpha \in K$ is algebraic over $F$.

Theorem. Every finite extension is algebraic.

Theorem. If $F \subseteq K \subseteq E$ and both $K/F$ and $E/K$ are algebraic, then $E/F$ is algebraic.

Algebraic Closure

A field $\bar{F}$ is an algebraic closure of $F$ if:

1. $\bar{F}$ is algebraic over $F$, and
2. $\bar{F}$ is algebraically closed (every non-constant polynomial in $\bar{F}[x]$ has a root in $\bar{F}$).

Every field has an algebraic closure, unique up to isomorphism.

Examples: $\bar{\mathbb{Q}} \subset \mathbb{C}$ (the field of algebraic numbers), $\overline{\mathbb{F}_p}$ (algebraic closure of the finite field).

Quick Reference: Degree Computations

Question Archetypes

tower-law-computation (1 question; 2016)

Recognition Cues

• A chain $F \subseteq K \subseteq E$ is given; asked to find $[E:F]$ or show $[E:F]$ divides some number
• May specify $[K:F]$ and $[E:K]$ individually and ask for $[E:F]$

Solution Template

1. Identify the tower: write out $F \subseteq K \subseteq E$ explicitly.
2. Compute $[K:F]$ and $[E:K]$ (often by finding minimal polynomials).
3. Apply: $[E:F] = [K:F] \cdot [E:K]$.
4. If proving divisibility or impossibility, use that $[K:F]$ divides $[E:F]$.

Worked Example

2016 Paper 2, 2016-P2-Q6a (15 marks)

Let $\alpha = \sqrt{2} + \sqrt{3}$. Find $[\mathbb{Q}(\alpha) : \mathbb{Q}]$ and determine the minimal polynomial of $\alpha$ over $\mathbb{Q}$.

Solution.

Step 1: Find the minimal polynomial.

Let $\alpha = \sqrt{2} + \sqrt{3}$. Compute: $\alpha^2 = 2 + 2\sqrt{6} + 3 = 5 + 2\sqrt{6}.$ $\alpha^2 - 5 = 2\sqrt{6}.$ $(\alpha^2 - 5)^2 = 24.$ $\alpha^4 - 10\alpha^2 + 25 = 24.$ $\alpha^4 - 10\alpha^2 + 1 = 0.$

So $\alpha$ is a root of $p(x) = x^4 - 10x^2 + 1 \in \mathbb{Q}[x]$.

Step 2: Verify $p$ is irreducible over $\mathbb{Q}$.

By the rational root theorem, the only possible rational roots are $\pm 1$. Check: $p(1) = 1 - 10 + 1 = -8 \ne 0$ and $p(-1) = -8 \ne 0$. So $p$ has no rational roots, hence no linear factors over $\mathbb{Q}$.

Check for quadratic factors over $\mathbb{Q}$: if $p(x) = (x^2 + ax + b)(x^2 - ax + c)$ (note the sign pattern forces the $x^3$ coefficient to be zero), then expanding: $p(x) = x^4 + (c + b - a^2)x^2 + a(c-b)x + bc.$ Matching coefficients:

• $x^3$: $0 = 0$. $\checkmark$
• $x^2$: $b + c - a^2 = -10$.
• $x^1$: $a(c - b) = 0$.
• $x^0$: $bc = 1$.

From $a(c-b) = 0$: either $a = 0$ or $c = b$.

• If $a = 0$: $b + c = -10$ and $bc = 1$. So $b, c$ are roots of $t^2 + 10t + 1 = 0$, giving $t = \frac{-10 \pm \sqrt{96}}{2} \notin \mathbb{Q}$.
• If $c = b$: $bc = b^2 = 1 \Rightarrow b = \pm 1$. Then $2b - a^2 = -10$.
  • $b = 1$: $a^2 = 12$, $a \notin \mathbb{Q}$.
  • $b = -1$: $a^2 = 8$, $a \notin \mathbb{Q}$.

In all cases, no factorization into quadratics over $\mathbb{Q}$ exists. Hence $p(x)$ is irreducible over $\mathbb{Q}$.

Step 3: Conclude.

Since $p(x)$ is monic, irreducible over $\mathbb{Q}$, and $p(\alpha) = 0$, the minimal polynomial of $\alpha$ over $\mathbb{Q}$ is $m_\alpha(x) = x^4 - 10x^2 + 1.$

The degree of the extension is $[\mathbb{Q}(\alpha) : \mathbb{Q}] = \deg m_\alpha = 4.$

$\boxed{[\mathbb{Q}(\sqrt{2}+\sqrt{3}):\mathbb{Q}] = 4,\quad m_\alpha(x) = x^4 - 10x^2 + 1}\;\blacksquare$

algebraic-closure-property (1 question; 2016)

Recognition Cues

• “Prove that $[E:F]$ divides $[L:F]$” or “show that $[K:F]$ is finite”
• Tower of extensions given; asked to prove finiteness or a divisibility relation
• May ask to prove that a sum or product of algebraic elements is algebraic

Solution Template

1. Use the tower law: $[E:F] = [E:K][K:F]$, so $[K:F]$ divides $[E:F]$.
2. For algebraic $\alpha, \beta$: note $[F(\alpha, \beta) : F] = [F(\alpha,\beta):F(\alpha)] \cdot [F(\alpha):F]$; each factor is finite (bounded by degrees of minimal polynomials), so the extension is finite, hence algebraic.

Worked Example

2016 Paper 2, 2016-P2-Q6b (20 marks)

Let $K/F$ be an algebraic extension and let $E/K$ be an algebraic extension. Prove that $E/F$ is an algebraic extension.

Solution.

We must show: for every $\alpha \in E$, $\alpha$ is algebraic over $F$.

Step 1. Since $E/K$ is algebraic, there exists a non-zero polynomial $p(x) = a_n x^n + \cdots + a_1 x + a_0 \in K[x]$ with $p(\alpha) = 0$.

The coefficients of $p$ are $a_0, a_1, \ldots, a_n \in K$.

Step 2. Since $K/F$ is algebraic, each $a_i$ is algebraic over $F$. Consider the tower $F \subseteq F(a_0) \subseteq F(a_0, a_1) \subseteq \cdots \subseteq F(a_0, a_1, \ldots, a_n) =: L.$

Each step adjoins an element algebraic over $F$ (and hence over any intermediate field), so each step has finite degree. Explicitly: $[F(a_0, \ldots, a_k) : F(a_0, \ldots, a_{k-1})] \le [F(a_k) : F] = \deg m_{a_k} < \infty.$

By the tower law applied iteratively, $[L : F] = [F(a_0):F] \cdot [F(a_0,a_1):F(a_0)] \cdots [L:F(a_0,\ldots,a_{n-1})] < \infty.$

Step 3. Now $p(x) \in K[x] \subseteq L[x]$ and $p(\alpha) = 0$. So $\alpha$ is algebraic over $L$, and $[L(\alpha) : L] = \deg m_{\alpha, L} \le \deg p < \infty.$

By the tower law: $[L(\alpha) : F] = [L(\alpha) : L] \cdot [L : F] < \infty.$

Step 4. Since $[F(\alpha) : F]$ divides $[L(\alpha) : F]$ (as $F \subseteq F(\alpha) \subseteq L(\alpha)$ by the tower law), we have $[F(\alpha) : F] < \infty$.

A finite extension is algebraic, so $\alpha$ is algebraic over $F$.

Conclusion. Since $\alpha \in E$ was arbitrary, $E/F$ is algebraic.

$\boxed{E/F \text{ is algebraic}}\;\blacksquare$

Common Traps

• Applying the tower law in the wrong order: $[E:F] = [E:K][K:F]$, not $[K:F][E:K]$ (they are equal by commutativity of multiplication, but cite them in the natural tower order).
• Claiming that $\alpha + \beta$ or $\alpha\beta$ is algebraic “because $\alpha$ and $\beta$ are” without an argument — the argument runs through the tower $F \subseteq F(\alpha) \subseteq F(\alpha, \beta)$.
• Confusing “algebraically closed” with “algebraic over $F$”: $\mathbb{C}$ is algebraically closed, but $\mathbb{C}/\mathbb{Q}$ is not algebraic (e.g., $\pi$ is transcendental).
• For minimal polynomial proofs: forgetting to check irreducibility — a polynomial with $\alpha$ as a root is not automatically the minimal polynomial.

Marks-Aware Writing

Section B, 15–20 marks per sub-part:

For the minimal polynomial / degree sub-part (15 marks):

• Derive the polynomial $p(\alpha) = 0$ by successive squaring (3–4 marks).
• Check no rational roots (2 marks).
• Check no quadratic factor over $\mathbb{Q}$ (4–5 marks).
• State minimal polynomial and degree with conclusion (2–3 marks).

For the tower/algebraic extension sub-part (20 marks):

• State what must be proved (1 mark).
• Find the polynomial $p \in K[x]$ (2 marks).
• Build the intermediate field $L$ by adjoining coefficients (4–5 marks).
• Show $[L:F] < \infty$ by tower law (4–5 marks).
• Conclude $[F(\alpha):F] < \infty$ hence $\alpha$ is algebraic (3–4 marks).
• General conclusion (1–2 marks).

Practice Set

Both historical questions on this atom are worked above (2016-P2-Q6a and 2016-P2-Q6b).