The tower law

Theorem

The tower law

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Statement

Let \(F \subseteq K \subseteq L\) be a tower of fields, i.e. \(K\) is a subfield of \(L\) containing \(F\) as a subfield, so that \(L\) is a vector space over \(K\), \(K\) is a vector space over \(F\), and \(L\) is a vector space over \(F\). Then \[ [L:F] \;=\; [L:K]\,[K:F], \] where \([L:F] = \dim_F L\) denotes the degree of the extension \(L/F\), and the equality holds in the extended natural numbers \(\mathbb{N} \cup \{\infty\}\): in particular \([L:F]\) is finite if and only if both \([L:K]\) and \([K:F]\) are finite, and in that case if \(\{a_i\}_{i \in I}\) is an \(F\)-basis of \(K\) and \(\{b_j\}_{j \in J}\) is a \(K\)-basis of \(L\), then \(\{a_i b_j\}_{(i,j) \in I \times J}\) is an \(F\)-basis of \(L\).

Why it matters

The tower law is the single load-bearing computational tool of finite field extension theory. Almost every degree calculation in Galois theory — showing \(\sqrt[3]{2}\) is not expressible using only square roots, bounding the degree of a splitting field, proving the classical impossibility of squaring the circle, trisecting an angle, or doubling the cube with straightedge and compass — proceeds by building a tower of intermediate fields and multiplying degrees along it.

It also converts an additive-looking problem (adjoining generators one at a time) into a multiplicative one (degrees), which is what makes divisibility arguments possible: if \([L:F]\) is prime, no intermediate field can exist except \(F\) and \(L\) themselves, because any intermediate degree would have to divide the prime.

Hypotheses

\(F, K, L\) are fields (not merely rings) with \(F \subseteq K \subseteq L\). If \(K\) is replaced by a ring that is not a field — e.g. \(F = \mathbb{Q}\), \(K = \mathbb{Q}[x]\), \(L = \mathbb{Q}(x)\) — then "\([K:F]\)" is still meaningful (it is infinite, since \(\mathbb{Q}[x]\) is an infinite-dimensional \(\mathbb{Q}\)-vector space), but the basis-tensoring argument used in the proof breaks down as soon as one allows non-field intermediate rings with zero divisors, because "basis of a module over a ring with zero divisors" behaves very differently from "basis of a vector space"; the theorem as stated is a statement about vector space dimensions and needs \(K\) a field so that \(L\) is a \(K\)-vector space in the first place. The inclusions \(F \subseteq K \subseteq L\) are field homomorphisms fixing the operations (i.e. genuine subfields, not just abstract field maps composed without a compatible tower). Without an actual containment \(F \subseteq K \subseteq L\) making \(L\) simultaneously an \(F\)-algebra and a \(K\)-algebra compatibly, there is no single vector space structure on \(L\) over both \(F\) and \(K\) for the bases to interact on, and the identity \([L:F]=[L:K][K:F]\) is not even well-posed. The degree \([L:K]\) is interpreted as a cardinal number when \(L/K\) is infinite, and the product \([L:K][K:F]\) is cardinal multiplication (with the convention \(\kappa \cdot \infty = \infty\) for \(\kappa \geq 1\)). If one insists on a naive "finite or infinite" dichotomy without cardinal arithmetic, subtler phenomena are lost: e.g. \([\mathbb{R}:\mathbb{Q}] = [\mathbb{C}:\mathbb{Q}] = 2^{\aleph_0}\) while \([\mathbb{C}:\mathbb{R}]=2\), and the law \(2^{\aleph_0} = 2 \cdot 2^{\aleph_0}\) is a genuine (if easy) cardinal-arithmetic fact, not something visible from finite bookkeeping alone.

Proof

Fix an \(F\)-basis \(\{a_i\}_{i \in I}\) of \(K\) and a \(K\)-basis \(\{b_j\}_{j \in J}\) of \(L\).

Every vector space over a field has a basis (this uses the axiom of choice via Zorn's Lemma when the space is infinite-dimensional); apply this to \(K\) as an \(F\)-vector space and to \(L\) as a \(K\)-vector space. A

Define \(S = \{a_i b_j : (i,j) \in I \times J\} \subseteq L\). We claim \(S\) is an \(F\)-basis of \(L\); this claim, once proved, gives \(|I \times J| = |I|\cdot|J|\) as cardinalities of bases, i.e. \([L:F]=[L:K][K:F]\).

Dimension of an \(F\)-vector space is by definition the cardinality of any basis (well-defined since all bases of a vector space have the same cardinality). It remains to prove \(S\) spans \(L\) over \(F\) and is \(F\)-linearly independent. B

Spanning: let \(\ell \in L\) be arbitrary. Since \(\{b_j\}_{j\in J}\) spans \(L\) over \(K\), \(\ell = \sum_{j \in J} k_j b_j\) for some \(k_j \in K\), almost all zero.

Definition of a spanning set / basis of the \(K\)-vector space \(L\): every element is a finite \(K\)-linear combination of basis vectors. A

Since \(\{a_i\}_{i \in I}\) spans \(K\) over \(F\), each \(k_j \in K\) can be written \(k_j = \sum_{i \in I} f_{ij} a_i\) with \(f_{ij} \in F\), almost all zero. Substituting, \[ \ell = \sum_{j\in J} \Big(\sum_{i \in I} f_{ij} a_i\Big) b_j = \sum_{(i,j) \in I \times J} f_{ij}\,(a_i b_j). \]

Substitution and distributivity/associativity of the ring multiplication on \(L\) (a field, hence a commutative ring, so scalar multiplication by elements of \(K \subseteq L\) distributes over the sum); the double sum is finite since only finitely many \(f_{ij}\) and \(k_j\) are nonzero. A

Hence every \(\ell \in L\) is an \(F\)-linear combination of elements of \(S\), so \(S\) spans \(L\) as an \(F\)-vector space.

Definition of spanning set, applied to the expression obtained in Step 4. A

Linear independence: suppose \(\sum_{(i,j)\in I\times J} f_{ij}\,(a_i b_j) = 0\) with \(f_{ij} \in F\), almost all zero. Regroup as \[ \sum_{j \in J} \Big(\underbrace{\sum_{i\in I} f_{ij} a_i}_{=: \,k_j \,\in\, K}\Big) b_j = 0. \]

Regrouping a finite sum by the distributive law; each inner sum \(k_j := \sum_i f_{ij}a_i\) lies in \(K\) because \(K\) is closed under \(F\)-linear combinations of its own elements \(a_i\) (\(F \subseteq K\)). B

Since \(\{b_j\}_{j \in J}\) is \(K\)-linearly independent and \(\sum_j k_j b_j = 0\) with \(k_j \in K\), every \(k_j = 0\).

Definition of linear independence of the \(K\)-basis \(\{b_j\}\) of \(L\). A

For each fixed \(j\), \(k_j = \sum_{i \in I} f_{ij} a_i = 0\) with \(f_{ij} \in F\); since \(\{a_i\}_{i \in I}\) is \(F\)-linearly independent, \(f_{ij} = 0\) for every \(i \in I\). As \(j\) was arbitrary, \(f_{ij}=0\) for all \((i,j) \in I \times J\).

Definition of linear independence of the \(F\)-basis \(\{a_i\}\) of \(K\), applied separately for each \(j \in J\) (this is legitimate because the vanishing \(k_j = 0\) established in Step 7 holds for every individual \(j\), not merely in aggregate). C

Hence \(S = \{a_ib_j\}\) is \(F\)-linearly independent, and combined with Step 5 it spans \(L\) over \(F\); so \(S\) is an \(F\)-basis of \(L\).

Definition of a basis: spanning \(+\) linear independence. A

The map \(I \times J \to S,\ (i,j) \mapsto a_ib_j\) is a bijection (injective because distinct pairs give distinct basis elements of a linearly independent, hence pairwise-distinct-and-nonzero, family; surjective by definition of \(S\)), so \(|S| = |I \times J| = |I|\cdot|J|\). Therefore \[ [L:F] = |S| = |I|\cdot|J| = [K:F]\cdot[L:K], \] with the product read as ordinary multiplication if both factors are finite and as cardinal multiplication (giving \(\infty\) whenever a factor is infinite and the other is nonzero) in general.

Dimension is a cardinality invariant of a vector space (invariance of basis cardinality, via the Steinitz exchange lemma in the finite-dimensional case and a Zorn's-lemma argument in general); cardinal arithmetic for the infinite case. C

Result

[L:F] \;=\; [L:K]\,[K:F] \qquad \text{for } F \subseteq K \subseteq L

Reading. Stack an extension \(L\) over an intermediate field \(K\) over a base field \(F\); the size of the whole extension, measured in dimension over \(F\), is exactly the product of the sizes of the two steps. Concretely, an \(F\)-basis of the top field \(L\) is built by taking all products of a \(K\)-basis of \(L\) with an \(F\)-basis of \(K\).

Scope. Holds for any tower of fields \(F \subseteq K \subseteq L\), with no finiteness assumption required for the statement itself (only for the ordinary-arithmetic reading of the product); it holds equally for towers of arbitrary length by induction, and is the field-theoretic instance of the general "rank is multiplicative in towers of free modules" phenomenon.

Corollaries & converses

Divisibility. If \([L:F]\) is finite, then \([K:F]\) and \([L:K]\) both divide \([L:F]\), for any intermediate field \(K\).
Primality obstruction. If \([L:F]\) is prime, the only intermediate fields between \(F\) and \(L\) are \(F\) and \(L\) themselves — there is no room for a proper intermediate field, since its degree over \(F\) would have to be a proper divisor of a prime other than \(1\).
Finite generation by towers. If \(L = F(\alpha_1,\dots,\alpha_n)\) with each \(\alpha_k\) algebraic over \(F(\alpha_1,\dots,\alpha_{k-1})\), then \([L:F] = \prod_{k=1}^n [F(\alpha_1,\dots,\alpha_k) : F(\alpha_1,\dots,\alpha_{k-1})]\), by iterating the tower law along the chain of successive simple extensions; this is the standard tool for computing degrees like \([\mathbb{Q}(\sqrt2,\sqrt3):\mathbb{Q}]\).
Induction to \(n\)-step towers. For \(F_0 \subseteq F_1 \subseteq \cdots \subseteq F_n\), \([F_n:F_0] = \prod_{k=1}^n [F_k:F_{k-1}]\), by repeated application of the two-step law.
Converse fails. Knowing only \([L:F]\) does not determine \([K:F]\) or \([L:K]\) individually, nor does knowing the product \([L:K][K:F]\) determine which intermediate field \(K\) was used: different intermediate fields of a given extension can realize the same product in different ways (e.g. inside \(\mathbb{Q}(\sqrt2,\sqrt3)/\mathbb{Q}\), which has degree \(4\), both \(K=\mathbb{Q}(\sqrt2)\) and \(K=\mathbb{Q}(\sqrt3)\) give \([K:\mathbb{Q}]=[L:K]=2\), yet they are different fields \(K\) sitting inside the same tower).
No converse divisibility-to-existence statement. A divisor \(d\) of \([L:F]\) need not be realized as \([K:F]\) for an actual intermediate field \(K\); e.g. for a non-Galois or inseparable extension, or simply because not every divisor of the degree corresponds to a subfield (the correspondence between intermediate fields and divisors is controlled by the Galois correspondence only in the Galois case, and even divisibility of degree by \(d\) does not by itself construct a subfield of that degree in general).

Fails without

Drop "\(K\) is a field." Take \(F = \mathbb{Q}\), \(R = \mathbb{Q}[x]\) (a ring, not a field), \(L = \mathbb{Q}(x)\). Then \(R\) is not a \(K\)-vector-space-compatible intermediate object in the sense required: \(L\) is not a finite-dimensional \(\mathbb{Q}[x]\)-module in the vector-space sense needed (indeed \(\mathbb{Q}[x]\) is not a field so "\([L:\mathbb{Q}[x]]\)" as a vector space dimension is not even the right invariant — one would need module rank over a ring with zero divisors, a wholly different and much less rigid theory), so the clean multiplicative identity has no counterpart.
Drop the containment \(F \subseteq K \subseteq L\) (use unrelated fields instead). Let \(F=\mathbb{Q}\), \(K = \mathbb{F}_2\), \(L=\mathbb{Q}(\sqrt2)\). There is no field homomorphism \(\mathbb{Q}\to\mathbb{F}_2\) failing... rather, more simply: without an actual tower (a chosen embedding making \(F \subseteq K \subseteq L\) literally true), \([K:F]\) is not even defined, since \(K\) is not an \(F\)-vector space at all unless \(F\) embeds in \(K\). The identity presupposes the tower; discard it and the two sides of the equation are not simultaneously meaningful.
Confuse "finite" with "well-ordered towers only," ignoring cardinal arithmetic. With \(F=\mathbb{Q} \subseteq K = \mathbb{Q}(x) \subseteq L = \overline{\mathbb{Q}(x)}\) (algebraic closure), \([K:F]\) and \([L:K]\) are both infinite; the naive statement "finite times finite is finite, otherwise all bets are off" is not wrong here, but if one tries to *compute* \([L:F]\) from finite sub-cases without cardinal arithmetic, one loses the ability to distinguish, e.g., countably infinite from continuum-many, which matters for finer classification questions (transcendence degree considerations sit on top of exactly this kind of bookkeeping).

Common errors

Adding degrees instead of multiplying. Students often write \([L:F] = [L:K]+[K:F]\) by false analogy with additivity of dimension for direct sums; the tower law is a statement about composing bases via a Cartesian product \(I \times J\), which multiplies cardinalities, not a direct-sum decomposition.
Forgetting that intermediate fields must actually be fields. Attempting to apply the law with \(K\) an arbitrary subring (e.g. \(\mathbb{Z}[\sqrt2]\) sitting inside \(\mathbb{Q}(\sqrt2)\)) and expecting the same clean multiplicative bookkeeping, when \(\mathbb{Z}[\sqrt2]\) is not a field over which \(\mathbb{Q}(\sqrt2)\) is even naturally a vector space in the required sense.
Assuming every divisor of \([L:F]\) gives an intermediate field. E.g. assuming a degree-6 extension must have a degree-2 and a degree-3 intermediate field just because \(6=2\cdot3\); this need not happen unless the extension has enough structure (such as being Galois with a subgroup lattice matching those divisors).
Sloppy indexing in the basis-product proof. Writing \(\{a_ib_j\}\) as a basis without checking that \(i\) ranges over the basis of \(K/F\) and \(j\) over the basis of \(L/K\) in the correct order, then miscounting the size as \(|I|+|J|\) rather than \(|I|\cdot|J|\) when both index sets are finite and explicit (e.g. believing a degree-2-over-degree-3 tower has \(5\) basis elements rather than \(6\)).
Misapplying the law to non-field extensions of rings in commutative algebra, e.g. assuming an analogous "rank multiplies in towers" for general ring extensions \(A \subseteq B \subseteq C\) without checking that \(B\) and \(C\) are free/finite modules of the right kind; the field-theoretic tower law relies essentially on vector spaces always having bases, which fails for general modules.

Discussion

The tower law is often the very first genuinely structural theorem a student meets in field theory, because it converts the vague idea of "how big is this extension" into an honest arithmetic invariant that composes predictably. Its proof is elementary linear algebra — nothing beyond bases and the distributive law — yet its consequences reach immediately into classical antiquity: the three Greek construction problems (doubling the cube, trisecting a general angle, squaring the circle) are all resolved by showing that a constructible number must lie at the top of a tower of degree-2 steps over \(\mathbb{Q}\), so \([L:\mathbb{Q}]\) must be a power of \(2\); since \(\sqrt[3]{2}\) generates a degree-3 extension of \(\mathbb{Q}\), and \(3 \nmid 2^n\) for any \(n\), doubling the cube is impossible. This is the tower law wearing an applied hat.

Historically the result predates the modern axiomatic definition of a field extension as a vector space; it is implicit in the 19th-century work on algebraic numbers (Kummer, Dedekind, Kronecker) long before Steinitz's 1910 paper Algebraische Theorie der Körper gave field theory its modern, basis-and-dimension footing. Once dimension was recognised as the right invariant, the tower law became essentially a triviality of linear algebra dressed in field-theoretic language — which is precisely why it is so robust: it needs almost no hypotheses beyond "these are fields, nested."

Conceptually the tower law is the degree-1 shadow of a much more general phenomenon: rank is multiplicative in towers of free modules, and more generally there are "multiplicativity of index" statements throughout algebra (index of subgroups in a tower, \([G:H][H:K]=[G:K]\) for a chain of subgroups, itself provable by an essentially identical coset-counting argument). Recognising the tower law as an instance of this pattern — rather than a fact peculiar to fields — is one of the marks of mathematical maturity in a first course on algebraic structures.

A genuine subtlety, easy to miss on a first pass: the theorem as usually taught silently assumes all extensions are algebraic or otherwise "nice," but the proof given above uses only the existence of bases, valid for *any* field extension, algebraic or transcendental, finite or infinite. The common misconception is that the tower law is somehow a theorem "about algebraic extensions"; it is not — it is a theorem about vector space dimension, and algebraicity plays no role in the proof at all. The connection to algebraic extension theory (e.g. that finite extensions are exactly the finitely generated algebraic ones) is a separate fact, proved independently and often confused with the tower law itself.

Worked examples

Compute \([\mathbb{Q}(\sqrt2,\sqrt3):\mathbb{Q}]\).

Set up the tower \(\mathbb{Q} \subseteq \mathbb{Q}(\sqrt2) \subseteq \mathbb{Q}(\sqrt2,\sqrt3)\). A

\([\mathbb{Q}(\sqrt2):\mathbb{Q}] = 2\), since \(x^2-2\) is irreducible over \(\mathbb{Q}\) (Eisenstein at \(p=2\)) and is the minimal polynomial of \(\sqrt2\), so \(\{1,\sqrt2\}\) is a \(\mathbb{Q}\)-basis of \(\mathbb{Q}(\sqrt2)\).

Degree of a simple algebraic extension equals the degree of the minimal polynomial. A

Over \(K=\mathbb{Q}(\sqrt2)\), the element \(\sqrt3\) satisfies \(x^2-3=0\); this polynomial remains irreducible over \(K\), because if \(\sqrt3 \in \mathbb{Q}(\sqrt2)\) then \(\sqrt3 = p+q\sqrt2\) for \(p,q\in\mathbb{Q}\), and squaring gives \(3 = p^2+2q^2 + 2pq\sqrt2\); since \(\sqrt2\) is irrational this forces \(pq=0\), and checking both cases (\(p=0\) or \(q=0\)) against \(3=p^2+2q^2\) yields no rational solution, a contradiction.

Irreducibility check by direct contradiction, using that \(\{1,\sqrt2\}\) is a \(\mathbb{Q}\)-basis so the representation \(p+q\sqrt2\) is unique. B

Hence \([\mathbb{Q}(\sqrt2,\sqrt3):\mathbb{Q}(\sqrt2)] = 2\), with basis \(\{1,\sqrt3\}\) over \(K\). By the tower law, \[ [\mathbb{Q}(\sqrt2,\sqrt3):\mathbb{Q}] = [\mathbb{Q}(\sqrt2,\sqrt3):\mathbb{Q}(\sqrt2)]\cdot[\mathbb{Q}(\sqrt2):\mathbb{Q}] = 2\cdot 2 = 4, \] with explicit \(\mathbb{Q}\)-basis \(\{1,\sqrt2,\sqrt3,\sqrt6\}\) obtained as the products \(a_ib_j\).

Direct application of the theorem, with basis constructed exactly as in Step 2 of the proof. A

[\mathbb{Q}(\sqrt2,\sqrt3):\mathbb{Q}] = 4, \quad \text{basis } \{1,\sqrt2,\sqrt3,\sqrt6\}

Reading. Two independent square-root adjunctions each double the degree, and the tower law confirms the doubling compounds multiplicatively rather than additively.

Show that \(\sqrt[3]{2} \notin \mathbb{Q}(\sqrt2)\), using degree considerations rather than a direct algebraic manipulation.

Strategy: compute \([\mathbb{Q}(\sqrt2):\mathbb{Q}]\) and \([\mathbb{Q}(\sqrt[3]2):\mathbb{Q}]\) and use the divisibility corollary of the tower law. B

\([\mathbb{Q}(\sqrt2):\mathbb{Q}]=2\) as in Example 1. Also \([\mathbb{Q}(\sqrt[3]2):\mathbb{Q}]=3\), since \(x^3-2\) is irreducible over \(\mathbb{Q}\) by Eisenstein at \(p=2\), and is the minimal polynomial of \(\sqrt[3]2\).

Degree of a simple algebraic extension equals the degree of the minimal polynomial (Eisenstein's criterion supplies irreducibility). A

Suppose, for contradiction, \(\sqrt[3]2 \in \mathbb{Q}(\sqrt2)\). Then \(\mathbb{Q}(\sqrt[3]2) \subseteq \mathbb{Q}(\sqrt2)\), giving a tower \(\mathbb{Q} \subseteq \mathbb{Q}(\sqrt[3]2) \subseteq \mathbb{Q}(\sqrt2)\).

If an element lies in a field, the field it generates over \(\mathbb{Q}\) is contained in that field, since \(\mathbb{Q}(\sqrt2)\) is closed under the field operations and contains \(\mathbb{Q}\) and \(\sqrt[3]2\), hence contains every element of \(\mathbb{Q}(\sqrt[3]2)\), the smallest such field. B

By the tower law, \([\mathbb{Q}(\sqrt2):\mathbb{Q}] = [\mathbb{Q}(\sqrt2):\mathbb{Q}(\sqrt[3]2)]\cdot[\mathbb{Q}(\sqrt[3]2):\mathbb{Q}]\), i.e. \(2 = [\mathbb{Q}(\sqrt2):\mathbb{Q}(\sqrt[3]2)]\cdot 3\). The right side is a multiple of \(3\), but \(2\) is not a multiple of \(3\) — contradiction.

Direct application of the tower law, then elementary arithmetic (divisibility). A

\sqrt[3]{2} \notin \mathbb{Q}(\sqrt2)

Reading. A degree-3 element cannot live inside a degree-2 extension, because 3 does not divide 2; this is the tower law converting a "which numbers are expressible" question into a one-line divisibility check.

Problems

Compute \([\mathbb{Q}(\sqrt2,\sqrt[3]2):\mathbb{Q}]\), justifying irreducibility of the relevant minimal polynomials at each stage.
Solution

\([\mathbb{Q}(\sqrt2):\mathbb{Q}]=2\) and \([\mathbb{Q}(\sqrt[3]2):\mathbb{Q}]=3\) as before. By the divisibility corollary, \([\mathbb{Q}(\sqrt2,\sqrt[3]2):\mathbb{Q}]\) is a multiple of both \(2\) and \(3\) (since it factors through the towers over each of \(\mathbb{Q}(\sqrt2)\) and \(\mathbb{Q}(\sqrt[3]2)\) separately), hence a multiple of \(\mathrm{lcm}(2,3)=6\). On the other hand, using the tower \(\mathbb{Q}\subseteq\mathbb{Q}(\sqrt2)\subseteq\mathbb{Q}(\sqrt2,\sqrt[3]2)\), \(\sqrt[3]2\) satisfies \(x^3-2\) over \(\mathbb{Q}(\sqrt2)\), a degree-3 polynomial, so \([\mathbb{Q}(\sqrt2,\sqrt[3]2):\mathbb{Q}(\sqrt2)] \leq 3\); by the tower law \([\mathbb{Q}(\sqrt2,\sqrt[3]2):\mathbb{Q}] \leq 3\cdot2=6\). Combining the multiple-of-6 lower bound with the \(\leq 6\) upper bound gives \([\mathbb{Q}(\sqrt2,\sqrt[3]2):\mathbb{Q}]=6\).
Let \([L:F]=15\) with \(F \subseteq K \subseteq L\). List the possible values of \([K:F]\), and explain why \(K\) can never be a proper nontrivial intermediate field of the "wrong" size, e.g. degree \(4\), over \(F\).
Solution

Since \([K:F]\) divides \([L:F]=15=3\cdot5\), the only possible values are the positive divisors of \(15\): \(1, 3, 5, 15\). A value of \(4\) is impossible because \(4 \nmid 15\): if \([K:F]=4\) then by the tower law \(15=[L:K]\cdot4\), forcing \([L:K]=15/4\), not an integer, contradicting that \([L:K]\) is a well-defined (nonnegative integer) vector space dimension.
Prove, using the tower law, that if \([L:F]\) is finite and \(p\) is a prime with \(p \nmid [L:F]\), then no element of \(L\setminus F\) can generate over \(F\) a simple extension of degree \(p\).
Solution

Suppose \(\alpha \in L\) has \([F(\alpha):F]=p\). Since \(F \subseteq F(\alpha) \subseteq L\), the tower law gives \([L:F] = [L:F(\alpha)]\cdot[F(\alpha):F] = [L:F(\alpha)]\cdot p\), so \(p \mid [L:F]\), contradicting \(p \nmid [L:F]\). Hence no such \(\alpha\) exists.
Let \(F \subseteq K \subseteq L\) be an infinite tower with \([K:F]=\aleph_0\) (countably infinite) and \([L:K]=2\). Determine \([L:F]\) and justify using cardinal arithmetic.
Solution

By the tower law (cardinal form), \([L:F]=[L:K]\cdot[K:F] = 2\cdot\aleph_0 = \aleph_0\), using the standard cardinal arithmetic fact that for any infinite cardinal \(\kappa\) and finite \(n\geq1\), \(n\cdot\kappa=\kappa\). So \([L:F]=\aleph_0\): adjoining one further quadratic step on top of a countably infinite-dimensional extension does not change its (still countably infinite) cardinality.
A student claims: "since \([\mathbb{Q}(\sqrt2,\sqrt3):\mathbb{Q}]=4\) and \(4=2\cdot2\), the only intermediate fields are \(\mathbb{Q}(\sqrt2)\) and \(\mathbb{Q}(\sqrt3)\), because those are the only ways to factor \(4\) as \(2\times2\)." Identify the logical gap in this reasoning and, without doing a full Galois-theoretic classification, name one further intermediate field it misses.
Solution

The gap: the tower law tells you that any intermediate field's degree over \(\mathbb{Q}\) must divide \(4\) (giving candidates \(1,2,4\)), but it says nothing about how many degree-2 intermediate fields exist, nor does it claim the factorization \(4=2\cdot2\) is realized in only one way. In fact \(\mathbb{Q}(\sqrt6)\) is a third intermediate field of degree \(2\) over \(\mathbb{Q}\) (since \((\sqrt6)^2=6\), minimal polynomial \(x^2-6\), and \(\sqrt6 = \sqrt2\cdot\sqrt3 \in \mathbb{Q}(\sqrt2,\sqrt3)\)), distinct from \(\mathbb{Q}(\sqrt2)\) and \(\mathbb{Q}(\sqrt3)\). The tower law constrains possible degrees; it does not by itself enumerate or bound the number of intermediate fields of a given degree — that finer question belongs to the Galois correspondence.