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University Mathematics — Year 1 · স্নাতক প্রথম বর্ষ

19Finite Dimension

A space spanned by finitely many vectors carries a well-defined dimension — the common size of all its bases. The proofs below all flow from one combinatorial engine, the exchange lemma: a free family can never outnumber a generating one. With dimension come the tools used everywhere afterwards: the incomplete basis theorem, the rank of a family, Grassmann’s formula.

19.1 Existence of bases

Definition 19.1

$E$ is finite-dimensional when it has a finite generating family. (Otherwise infinite-dimensional: so is $K[X]$ , whose finite families only span polynomials of bounded degree.)

Theorem 19.2 (Exchange lemma)

Let $(g_1, \dots, g_n)$ generate $E$ and $(f_1, \dots, f_p)$ be free in $E$ . Then $p \leq n$ .

Proof. We prove by induction on $k \leq p$ : after renumbering the $g_j$ , the family $(f_1, \dots, f_k, g_{k+1}, \dots, g_n)$ generates $E$ — which forces $k \leq n$ at each stage, and $p \leq n$ at the end.

$k = 0$ : the hypothesis. Step: assume it for $k - 1$ ; then $f_k$ is a combination of $(f_1, \dots, f_{k-1},$ $g_k, \dots, g_n)$ . In this combination some $g_j$ ( $j \geq k$ ) has a nonzero coefficient — otherwise $f_k$ would be a combination of $f_1, \dots, f_{k-1}$ , contradicting freeness. Renumber so that $j = k$ , and solve for $g_k$ : $g_k$ is a combination of $(f_1, \dots, f_k, g_{k+1}, \dots, g_n)$ . Every vector of $E$ , expressed through the $(k-1)$ -family, can then be re-expressed through the $k$ -family: it generates. (If $k - 1 = n$ , no $g$ remains and $f_k$ would be a combination of the $f_i$ ’s alone: impossible; so $k \leq n$ .) ∎

Example 19.3 (The exchange, watched once)

In $\R^2$ , take the generating family $\bigl(g_1, g_2\bigr) = \bigl((1,0), (0,1)\bigr)$ and the free family $\bigl(f_1, f_2\bigr) = \bigl((1,2), (3,4)\bigr)$ . Step $1$ : $f_1 = 1\cdot g_1 + 2\cdot g_2$ ; the coefficient of $g_2$ is nonzero, so exchange $g_2$ for $f_1$ : the family $\bigl(f_1, g_1\bigr)$ still generates ( $g_2 = \frac12(f_1 - g_1)$ ). Step $2$ : $f_2 = (3,4) = 2\,f_1 + 1\cdot g_1$ ; the coefficient of the remaining $g_1$ is nonzero (it must be: $f_2$ is not a multiple of $f_1$ ), so exchange again: $\bigl(f_1, f_2\bigr)$ generates $\R^2$ . Had there been a third free vector $f_3$ , no $g$ would remain to absorb it — which is precisely how the lemma forbids $3$ free vectors in $\R^2$ . The proof above is this bookkeeping done in general.

Theorem 19.4 (Bases in finite dimension)

Let $E \neq \{0\}$ be finite-dimensional.

From every finite generating family one can extract a basis.
(Incomplete basis theorem) Every free family extends to a basis, using vectors of any chosen generating family.
All bases of $E$ are finite, with the same number of elements: the dimension $\dim E$ . (Convention: $\dim\{0\} = 0$ .)

Proof. (1) Discard, one at a time, any vector that is a combination of the others; the family stays generating, since in any expression using the discarded vector one may substitute its combination of the survivors. The process terminates — each step shrinks a finite family by one — and it halts exactly when no remaining vector is a combination of the others. The final family is still generating, and it is free: a nontrivial null combination would carry some nonzero coefficient, and dividing by it would solve for the corresponding vector in terms of the others, making it discardable after all — contradicting that the process had halted.

(2) Let $(f_1, \dots, f_p)$ be free, $(g_1, \dots, g_n)$ generating. Run through $g_1, \dots, g_n$ , appending $g_j$ to the current family whenever it is not already in its span (Proposition 18.19 (2) keeps the family free). The final family is free, and generating: every $g_j$ lies in its span — either it was appended, or it was already a combination.

(3) Two bases are each free and each generating: the exchange lemma gives both inequalities between their cardinalities. (Finiteness: a basis is free, hence by exchange no larger than a finite generating family.) ∎

Example 19.5

$\dim K^n = n$ (canonical basis); $\dim K_n[X] = n + 1$ (monomials); $\dim_\R \C = 2$ ; the solution space of $y'' + ay' + by = 0$ has dimension $2$ (Theorem 5.10: the solutions are parametrized bijectively and linearly by $(\lambda, \mu) \in K^2$ ).

Example 19.6 (The same set, two dimensions)

The set $\C^2$ of pairs of complex numbers is a $\C$ -vector space of dimension $2$ (canonical basis $e_1, e_2$ ) — and an $\R$ -vector space of dimension $4$ , with basis

(1, 0),\quad (\iu, 0),\quad (0, 1),\quad (0, \iu):

every $(z, w) = (a + \iu b,\ c + \iu d)$ has real coordinates $(a, b, c, d)$ , uniquely. Dimension is not a property of the set of vectors alone: it counts the degrees of freedom relative to the allowed scalars, and halving the scalar supply from $\C$ to $\R$ doubles the count. (The weekend problem exploits the extreme case of this sensitivity, with scalars shrunk all the way to $\Q$ .)

Example 19.7 (Running the completion algorithm)

Complete the free family $\bigl((1,1,1)\bigr)$ into a basis of $\R^3$ using the canonical vectors. Run the proof of Theorem 19.4 (2) on the generating family $(e_1, e_2, e_3)$ : is $e_1 \in \operatorname{Vect}(1,1,1)$ ? No (multiples of $(1,1,1)$ have equal coordinates) — append it. Is $e_2 \in \operatorname{Vect}\bigl((1,1,1), e_1\bigr)$ ? A combination $\alpha(1,1,1) + \beta e_1$ has equal second and third coordinates, and $e_2$ does not — append it. The family $\bigl((1,1,1), e_1, e_2\bigr)$ is free with $3$ vectors: stop, it is a basis (Proposition 19.8 will make this reflex official). Note the outcome depends on the order in which the $g_j$ are scanned: completion is an algorithm, not a formula.

Proposition 19.8 (The two-out-of-three rule)

Let $\dim E = n$ and $\mathcal{F}$ a family of exactly $n$ vectors of $E$ . Then

\mathcal{F} \text{ free} \iff \mathcal{F} \text{ generating} \iff \mathcal{F} \text{ basis}.

Moreover any free family has $\leq n$ vectors, any generating family $\geq n$ .

Proof. The cardinality bounds are the exchange lemma against a basis. If $\mathcal{F}$ (size $n$ ) is free but not generating, some $x$ lies outside its span; appending $x$ gives a free family of $n + 1$ vectors: impossible. If $\mathcal{F}$ is generating but not free, extracting a basis (Theorem 19.4 (1)) gives a basis of $< n$ vectors: impossible. ∎

Example 19.9 (Two-out-of-three, saving half the work)

Is $\bigl((1,1,0), (0,1,1), (1,0,1)\bigr)$ a basis of $\R^3$ ? Count: three vectors, dimension three — so freeness alone decides. A null combination gives $a + c = 0$ , $a + b = 0$ , $b + c = 0$ ; adding all three, $2(a + b + c) = 0$ , and subtracting each original equation from $a + b + c = 0$ leaves $b = c = a = 0$ : free, hence a basis, with the generating half of the verification supplied by the theorem for free. Compare Example 18.18, where the same double verification had to be done by hand — one chapter of theory converts into exactly that saving, on every basis check for the rest of the book.

Method 19.10 (Computing a dimension)

Three standard routes, in decreasing order of frequency.

Parametrize, then read off a basis. Solve the defining constraints, express the general element linearly in the surviving parameters, and check that the vectors multiplying the parameters are free: the dimension is the number of parameters. (Run below on a concrete subspace of $\R^4$ .)
Exhibit a bijective linear parametrization. When the elements are determined by finitely many values — initial conditions of a recurrence (Exercise 19.10), coefficients of a solution formula (Example 19.5) — the dimension is the number of those values.
Use the formulas. Grassmann for intersections and sums, rank for spans, and later rank–nullity for kernels and images: dimensions are usually computed, not guessed.

In all three routes, the two-out-of-three rule is the finisher: once the count matches, freeness or generation alone concludes.

Example 19.11 (Route 1, in full)

Dimension of $H = \{(x, y, z, t) \in \R^4 : x + y + z + t = 0 \text{ and } x = t\}$ . Solve: $t = x$ and $y + z = -2x$ , so $z = -2x - y$ with $x, y$ free:

(x,\ y,\ -2x - y,\ x) = x\,(1, 0, -2, 1) + y\,(0, 1, -1, 0) .

The two vectors are free (look at the first two coordinates: $(x, y) = (0,0)$ ), so they form a basis of $H$ and $\dim H = 2$ . The count was predictable — two independent linear constraints in $\R^4$ should each eat one dimension — but the parametrization proves it and hands over a basis, which the prediction alone never does; Exercise 19.9 makes the “each equation eats at most one dimension” slogan into a theorem.

Example 19.12 (Route 2, in full)

Dimension of $W = \{P \in \R_n[X] : P(1) = P(2) = 0\}$ (for $n \geq 2$ ). By the factor theorem applied twice (Theorem 8.7; the roots $1$ and $2$ are distinct), $P \in W$ exactly when $P = (X - 1)(X - 2)\,Q$ with $\deg Q \leq n - 2$ . The correspondence $Q \mapsto (X-1)(X-2)Q$ is linear, reaches all of $W$ , and is injective (a product is zero only if $Q = 0$ ): $W$ is parametrized bijectively and linearly by $\R_{n-2}[X]$ , so

\dim W = \dim \R_{n-2}[X] = n - 1 .

A basis comes with the parametrization: the images of the monomials, $\bigl((X-1)(X-2),\ (X-1)(X-2)X,\ \dots,\ (X-1)(X-2)X^{n-2}\bigr)$ . Each new evaluation constraint at a fresh point costs exactly one dimension — the counting backbone of Lagrange interpolation (Theorem 8.23).

Example 19.13 (An integral constraint costs one dimension too)

Dimension of $H = \{P \in \R_2[X] : \int_0^1 P = 0\}$ . Writing $P = a + bX + cX^2$ , the constraint reads $a + \frac b2 + \frac c3 = 0$ ; solve for $a$ and parametrize:

P = b\Bigl(X - \frac12\Bigr) + c\Bigl(X^2 - \frac13\Bigr),

so $H = \operatorname{Vect}\bigl(X - \frac12,\ X^2 - \frac13\bigr)$ , dimension $2$ (the two polynomials have distinct degrees: free). One linear condition — whether an evaluation, an integral, or any other linear recipe — removes at most one dimension, and exactly one as soon as the condition is not identically zero. Chapter 20 will name such recipes linear forms and their solution sets hyperplanes; the basis vectors found here reappear in Chapter 23 as the start of the Legendre family.

19.2 Subspaces, rank, Grassmann

Theorem 19.14 (Subspaces)

Let $E$ be finite-dimensional and $F$ a subspace. Then $F$ is finite-dimensional, $\dim F \leq \dim E$ , with equality if and only if $F = E$ . Moreover every subspace has a supplementary subspace.

Proof. Free families of $F$ have at most $\dim E$ vectors (exchange lemma in $E$ : a free family of $F$ is in particular free in $E$ , and $E$ has a finite generating family). Among the free families of $F$ pick one of maximal size $p$ — possible since the sizes are integers bounded by $\dim E$ . It generates $F$ : otherwise some $x \in F$ would lie outside its span, and appending $x$ would give a free family of $F$ of size $p + 1$ (Proposition 18.19 (2)), contradicting maximality. Being free and generating, it is a basis of $F$ , and $\dim F = p \leq \dim E$ . If $p = \dim E = n$ : a free family of $n$ vectors of $E$ is a basis of $E$ (Proposition 19.8), so $F \supseteq \operatorname{span} = E$ . Supplementary: complete a basis $(f_1, \dots, f_p)$ of $F$ into a basis $(f_1, \dots, f_p, g_{p+1}, \dots, g_n)$ of $E$ (incomplete basis theorem); then $G = \operatorname{Vect}(g_{p+1}, \dots, g_n)$ satisfies $E = F \oplus G$ (existence and uniqueness of decompositions = coordinates in the big basis). ∎

Definition 19.15 (Rank of a family)

The rank of a finite family of vectors is the dimension of its span: $\operatorname{rk}(x_1, \dots, x_p) = \dim \operatorname{Vect}(x_1, \dots, x_p) \leq \min(p, \dim E)$ , with equality to $p$ iff the family is free.

Example 19.16 (Computing a rank by elimination)

Rank of $\bigl((1,2,3), (2,3,4), (3,4,5), (1,1,1)\bigr)$ in $\R^3$ . The span is unchanged when one subtracts from a vector a combination of the others (both families span the same combinations): replace $(2,3,4)$ by $(2,3,4) - (1,2,3) = (1,1,1)$ and $(3,4,5)$ by $(3,4,5) - (1,2,3) = (2,2,2)$ . The span is now $\operatorname{Vect}\bigl((1,2,3), (1,1,1), (2,2,2), (1,1,1)\bigr) = \operatorname{Vect}\bigl((1,2,3), (1,1,1)\bigr)$ , and these two vectors are not proportional: the rank is $2$ . This subtract-and-discard procedure is systematized as Gaussian elimination in Chapter 22.

Example 19.17 (Summing by concatenation)

Take, in $\R^3$ ,

F = \operatorname{Vect}\bigl((1,2,3),\ (1,1,1)\bigr), \qquad G = \operatorname{Vect}\bigl((2,3,4)\bigr) .

The sum $F + G$ is spanned by the concatenated family of all three generators, and

(2, 3, 4) = (1, 2, 3) + (1, 1, 1)

shows the third is redundant: $F + G = F$ , of dimension $2$ — equivalently $G \subseteq F$ , which the relation displays. Grassmann confirms: $\dim(F \cap G) = 2 + 1 - 2 = 1 = \dim G$ . Sums are computed by concatenating generators and then reducing the pile by the rank algorithm; no new technique is ever needed.

Theorem 19.18 (Grassmann’s formula)

For finite-dimensional subspaces $F, G$ of $E$ :

\dim(F + G) = \dim F + \dim G - \dim(F \cap G) .

In particular $F + G$ is direct iff $\dim(F + G) = \dim F + \dim G$ .

Proof. Start from a basis $(e_1, \dots, e_r)$ of $F \cap G$ ; complete it into a basis $(e_1, \dots, e_r, f_1, \dots, f_s)$ of $F$ and into a basis $(e_1, \dots, e_r, g_1, \dots, g_t)$ of $G$ (Theorem 19.4 (2)). We claim

\mathcal{B} = (e_1, \dots, e_r, f_1, \dots, f_s, g_1, \dots, g_t)

is a basis of $F + G$ ; the formula follows by counting: $(r + s) + (r + t) - r = r + s + t$ .

$\mathcal{B}$ generates $F + G$ : any $u + v$ ( $u \in F$ , $v \in G$ ) expands through it. Freeness: suppose $\sum \alpha_i e_i + \sum \beta_j f_j + \sum \gamma_k g_k = 0$ . The vector $w = \sum \gamma_k g_k = -\sum\alpha_i e_i - \sum\beta_j f_j$ lies in $G \cap F$ , so it expands on $(e_i)$ alone; but $w$ also expands on $(g_k)$ , and in the basis of $G$ these two expressions must coincide: all $\gamma_k = 0$ (and the $e$ -coordinates match). The relation reduces to $\sum\alpha_i e_i + \sum\beta_j f_j = 0$ , a relation in the basis of $F$ : all remaining coefficients vanish. ∎

Example 19.19

Two distinct planes $F, G$ (dimension $2$ ) of $\R^3$ satisfy $F + G = \R^3$ (their sum strictly contains a plane), so $\dim(F \cap G) = 2 + 2 - 3 = 1$ : they always intersect along a line — no “parallel planes” through the origin.

Example 19.20 (Grassmann in action, in $\R^4$ )

Let $F = \operatorname{Vect}\bigl(e_1,\ e_2,\ (1,1,1,0)\bigr)$ and $G = \operatorname{Vect}(e_3, e_4)$ in $\R^4$ . Dimensions: $\dim F = 3$ (the third generator has a nonzero third coordinate, outside $\operatorname{Vect}(e_1, e_2)$ ) and $\dim G = 2$ . Sum: $F + G$ contains $e_1, e_2, e_4$ and $e_3 = (1,1,1,0) - e_1 - e_2$ : it is all of $\R^4$ . Grassmann then computes the intersection’s size with no elimination at all:

\dim(F \cap G) = 3 + 2 - 4 = 1 .

To identify the line, look inside $G$ : a vector $(0, 0, c, d)$ lies in $F$ exactly when it is $a e_1 + b e_2 + \lambda(1,1,1,0)$ , forcing $\lambda = -a = -b$ and $d = 0$ : the intersection is $\operatorname{Vect}(e_3)$ — consistent, since $e_3$ was exhibited in $F$ above and lies in $G$ by definition. Typical division of labor: Grassmann predicts how much to look for, the linear system then finds what.

Example 19.21 (Intersecting by equations)

When both subspaces come as solution sets, intersecting is just stacking the equations. In $\R^3$ : $F = \{x + y + z = 0\}$ and $G = \{x = y\}$ give

F \cap G = \{x = y,\ 2x + z = 0\} = \{(x,\ x,\ -2x)\} = \operatorname{Vect}(1, 1, -2),

a line. Cross-check by Grassmann: $F + G = \R^3$ (the planes are distinct, so their sum strictly contains a plane), hence $\dim(F\cap G) = 2 + 2 - 3 = 1$ . The two descriptions of a subspace — by equations, by generators — each make one operation trivial: equations intersect by stacking, generators sum by concatenating; converting between them is exactly what solving a linear system means (Chapter 22).

Remark 19.22 (Common pitfalls)

Dimensions do not add along sums unless the sum is direct: two planes of $\R^3$ have $\dim(F + G) = 3$ , not $4$ ; always correct by the intersection term (Grassmann). An inclusion needs the dimension and the inclusion: $\dim F = \dim G$ alone never gives $F = G$ (two distinct lines of $\R^2$ ); the equality case of Theorem 19.14 requires $F \subseteq G$ first. Counting parameters is not yet a proof: “two equations in $\R^4$ , so dimension $2$ ” fails when the equations are dependent ( $x + y = 0$ and $2x + 2y = 0$ leave dimension $3$ ); only an actual parametrization or a rank computation decides. Do not speak of the dimension of a non-subspace: solution sets of inhomogeneous systems miss $0$ ; their “dimension” is that of the associated homogeneous solution space (Chapter 22 makes this precise). Infinite dimension exists: $K[X]$ contains free families of every size (the monomials), so no finite generating family can exist — statements like the two-out-of-three rule are strictly finite-dimensional and fail badly on $K[X]$ (Corollary 20.9 will show the same for injectivity/surjectivity). Rank is about the span, not the list: repeating a vector, reordering, or rescaling by nonzero constants leaves the rank unchanged, and $\operatorname{rk} = p$ (the number of vectors) is a property to prove — it is exactly freeness. A family of $5$ vectors of rank $2$ carries three vectors’ worth of redundancy, which elimination (Example 19.16) locates explicitly.

Remark 19.23 (Where dimension goes to work)

Dimension is the book’s favorite counting argument from here on. Chapter 20 proves the rank–nullity theorem, the functional version of Grassmann’s formula; Chapter 21 computes ranks by row reduction; Chapter 22 turns “ $n$ vectors of $K^n$ form a basis” into one number being nonzero. The weekend problem below shows dimension doing arithmetic: counting dimensions over the field $\Q$ proves irrationality statements that look untouchable by hand. In the Year 3 volume the same dimension counts, refined by group theory, decide which classical construction problems are solvable — that story is Galois theory.

Remark 19.24 (Perspectives inside Book 3)

Dimension is the conserved quantity of the rest of this volume, and it is worth naming the conservation laws in advance. Chapter 20 proves $\dim E = \dim\ker u + \operatorname{rk} u$ : what a linear map crushes plus what it keeps always totals the source. Chapter 22 refines this into the structure of solution sets: $p$ unknowns minus $\operatorname{rk} A$ pivots leaves the dimension of the solution space, which Gaussian elimination exhibits as free parameters. Chapter 23 splits $\dim E = \dim F + \dim F^\perp$ orthogonally, and the weekend problem of Chapter 25 spends exactly this budget: $n$ data points, $2$ parameters fitted, $n - 2$ dimensions of residual. Whenever a count refuses to balance in a later chapter, the error is a forgotten kernel or a non-direct sum — come back to Grassmann’s formula first.

19.3 Exercises

Exercise 19.1 ★

Give a basis and the dimension of:

$F = \{(x,y,z) \in \R^3 : x + y + z = 0\}$ ;
$G = \{(x,y,z,t) \in \R^4 : x = y,\ z = 2t\}$ ;
$H = \{P \in \R_3[X] : P(1) = P'(1) = 0\}$ .

Solution

Solution of Exercise 19.1.

$z = -x - y$ : $F = \{(x, y, -x-y)\} = \operatorname{Vect}\bigl((1,0,-1), (0,1,-1)\bigr)$ ; the two vectors are free (coordinates): $\dim F = 2$ .
$G = \{(x, x, 2t, t)\} = \operatorname{Vect}\bigl((1,1,0,0), (0,0,2,1)\bigr)$ : free, $\dim G = 2$ .
$P(1) = P'(1) = 0$ means $(X-1)^2 \mid P$ (Proposition 8.11): $P = (X-1)^2(aX + b)$ . Basis $\bigl((X-1)^2, X(X-1)^2\bigr)$ , dimension $2$ .

Exercise 19.2 ★

Compute the rank of the family $\bigl((1,1,1), (1,2,3), (3,5,7), (0,1,2)\bigr)$ in $\R^3$ , and extract a basis of its span.

Solution

Solution of Exercise 19.2.

$(3,5,7) = (1,1,1) + 2(1,2,3)$ and $(0,1,2) = (1,2,3) - (1,1,1)$ : both are combinations of the first two, which are free (not proportional). Rank $2$ ; basis of the span: $\bigl((1,1,1), (1,2,3)\bigr)$ .

Exercise 19.3 ★

Complete the free family $\bigl((1,1,0,0), (0,0,1,1)\bigr)$ into a basis of $\R^4$ using canonical vectors, and justify.

Solution

Solution of Exercise 19.3.

Try appending $e_1 = (1,0,0,0)$ and $e_3 = (0,0,1,0)$ . The family $\bigl((1,1,0,0), (0,0,1,1), e_1, e_3\bigr)$ is free: a null combination $\alpha(1,1,0,0) + \beta(0,0,1,1) + \gamma e_1 + \delta e_3 = 0$ reads $(\alpha + \gamma, \alpha, \beta + \delta, \beta) = 0$ , so $\alpha = \beta = 0$ , then $\gamma = \delta = 0$ . Four free vectors in dimension $4$ : a basis (Proposition 19.8).

Exercise 19.4 ★

Prove that $\bigl(1, X, X(X-1), X(X-1)(X-2)\bigr)$ is a basis of $\R_3[X]$ , and find the coordinates of $X^3$ in it.

Solution

Solution of Exercise 19.4.

Degrees $0, 1, 2, 3$ pairwise distinct: free (Proposition 18.19), four vectors in dimension $4$ : basis. For $X^3$ : expand downward,

X(X-1)(X-2) = X^3 - 3X^2 + 2X, \qquad X(X-1) = X^2 - X,

so $X^3 - X(X-1)(X-2) = 3X^2 - 2X$ ; and $3X^2 - 2X = 3(X^2 - X) + X = 3\,X(X-1) + X$ . Hence

X^3 = X(X-1)(X-2) + 3\,X(X-1) + 1\cdot X + 0\cdot 1 :

coordinates $(0, 1, 3, 1)$ on $\bigl(1, X, X(X-1), X(X-1)(X-2)\bigr)$ . (These are Stirling numbers in disguise.)

Exercise 19.5 ★★

Let $F$ and $G$ be subspaces of dimensions $4$ and $5$ of a space $E$ with $\dim E = 7$ . What are the possible values of $\dim(F \cap G)$ ? Give an instance realizing each value with $E = \R^7$ .

Solution

Solution of Exercise 19.5.

Grassmann: $\dim(F \cap G) = 4 + 5 - \dim(F + G)$ , and $F + G$ is a subspace of $E$ containing $G$ : $5 \leq \dim(F+G) \leq 7$ . Hence $\dim(F \cap G) \in \{2, 3, 4\}$ . Realizations in $\R^7$ with canonical basis $(e_1, \dots, e_7)$ , taking $G = \operatorname{Vect}(e_1, \dots, e_5)$ :

$F = \operatorname{Vect}(e_1, e_2, e_6, e_7)$ : $F + G = \R^7$ , intersection $\operatorname{Vect}(e_1, e_2)$ , dimension $2$ ;
$F = \operatorname{Vect}(e_1, e_2, e_3, e_6)$ : intersection of dimension $3$ ;
$F = \operatorname{Vect}(e_1, e_2, e_3, e_4) \subseteq G$ : dimension $4$ .

Exercise 19.6 ★★

Let $H = \{P \in \R_n[X] : P(1) = 0\}$ . Prove that $H$ is a hyperplane of $\R_n[X]$ (a subspace of dimension $n$ ), exhibit a basis of $H$ (think of the factor theorem: $P = (X-1)Q$ ), and give a supplementary line.

Solution

Solution of Exercise 19.6.

$H$ is a subspace (Exercise 18.1 (4) verbatim). By the factor theorem (Theorem 8.7), $P \in H \iff P = (X-1)Q$ with $\deg Q \leq n - 1$ : the map $Q \mapsto (X-1)Q$ is a linear bijection from $\R_{n-1}[X]$ onto $H$ , so a basis of $H$ is

\bigl((X-1),\ (X-1)X,\ (X-1)X^2,\ \dots,\ (X-1)X^{n-1}\bigr), \qquad \dim H = n .

A supplementary line: $\operatorname{Vect}(1)$ (constants). Indeed $H \cap \operatorname{Vect}(1) = \{0\}$ (a nonzero constant does not vanish at $1$ ) and dimensions add up to $n + 1$ : by Grassmann, $H \oplus \operatorname{Vect}(1) = \R_n[X]$ .

Exercise 19.7 ★★

Let $u_1, \dots, u_p$ be vectors of rank $r$ . Prove that removing one vector yields a family of rank $r$ or $r - 1$ , and that appending one vector yields rank $r$ or $r + 1$ . Deduce that rank changes by at most $1$ under any single insertion or deletion.

Solution

Solution of Exercise 19.7.

Deletion: removing $u_p$ , the span can only shrink; and it shrinks by at most one dimension, since adding $u_p$ back to a basis of the smaller span gives a generating family of the larger with at most one extra vector. Symmetrically, appending a vector $v$ : the new span contains the old with at most one extra generator, so its dimension is $r$ (if $v$ was already in the span) or $r + 1$ (otherwise, by Proposition 18.19 (2) a basis extends). Both statements together give the “rank is $1$ -Lipschitz” conclusion.

Exercise 19.8 ★★★

Let $F_1 \subseteq F_2 \subseteq \dots \subseteq F_k$ be subspaces of $E$ ( $\dim E = n$ ) with $F_i \neq F_{i+1}$ for all $i$ . Prove $k \leq n + 1$ . Deduce that a strictly increasing chain of subspaces of $\R^n$ has length at most $n + 1$ , and exhibit one of maximal length.

Solution

Solution of Exercise 19.8.

Along a strictly increasing chain, dimensions strictly increase ( $F_i \subseteq F_{i+1}$ , $F_i \neq F_{i+1}$ and Theorem 19.14: equality of dimensions would force equality of spaces). So $\dim F_1 < \dim F_2 < \dots < \dim F_k$ is a strictly increasing sequence of integers in $\intint{0}{n}$ : at most $n + 1$ values, $k \leq n + 1$ . Maximal chain in $\R^n$ :

\{0\} \subsetneq \operatorname{Vect}(e_1) \subsetneq \operatorname{Vect}(e_1, e_2) \subsetneq \dots \subsetneq \R^n ,

of length exactly $n + 1$ .

Exercise 19.9 ★★★

Let $E$ be of dimension $n$ and $F$ , $G$ two hyperplanes (dimension $n - 1$ ), $F \neq G$ . Compute $\dim(F \cap G)$ . Generalize: the intersection of $k$ hyperplanes has dimension $\geq n - k$ .

Solution

Solution of Exercise 19.9.

$F + G$ strictly contains $F$ (since $G \not\subseteq F$ ), so $\dim(F + G) = n$ and Grassmann gives $\dim(F \cap G) = (n-1) + (n-1) - n = n - 2$ .

General claim, by induction on $k$ : the intersection $I_k$ of $k$ hyperplanes has $\dim I_k \geq n - k$ . True for $k = 1$ . Step: $I_{k+1} = I_k \cap H_{k+1}$ , and Grassmann inside $E$ :

\dim(I_k \cap H_{k+1}) = \dim I_k + (n - 1) - \dim(I_k + H_{k+1}) \geq \dim I_k + (n-1) - n \geq n - k - 1 . \qedhere

Exercise 19.10 ★★

Let $E$ be the set of real sequences satisfying $u_{n+2} = 3u_{n+1} - 2u_n$ for all $n$ .

Show that $E$ is a subspace of the space of sequences, and that a sequence of $E$ is entirely determined, linearly, by the pair $(u_0, u_1)$ ; deduce $\dim E = 2$ .
Check that the constant sequence $(1)$ and the geometric sequence $(2^n)$ lie in $E$ and form a basis of $E$ .
Find the sequence of $E$ with $u_0 = 0$ , $u_1 = 1$ .

Solution

Solution of Exercise 19.10.

The condition $u_{n+2} - 3u_{n+1} + 2u_n = 0$ is linear and satisfied by the zero sequence: $E$ is a subspace. By induction, $u_0$ and $u_1$ determine every $u_n$ , and the dependence is linear (each step is a linear combination of the two previous values); conversely every pair $(a, b)$ arises from exactly one sequence of $E$ (define $u_n$ by the recurrence). As in Example 19.5, $E$ is parametrized bijectively and linearly by $(u_0, u_1) \in \R^2$ : $\dim E = 2$ .
Constants: $3\cdot1 - 2\cdot1 = 1$ . Geometric: $3\cdot 2^{n+1} - 2\cdot 2^n = (6 - 2)2^n = 2^{n+2}$ . Both lie in $E$ . Freeness: $a\cdot 1 + b\cdot 2^n = 0$ for all $n$ gives, at $n = 0$ and $n = 1$ : $a + b = 0$ , $a + 2b = 0$ , so $a = b = 0$ . Two free vectors in dimension $2$ : a basis (Proposition 19.8).
Solve $a + b = 0$ , $a + 2b = 1$ : $b = 1$ , $a = -1$ , so $u_n = 2^n - 1$ (the Mersenne sequence).

Exercise 19.11 ★★

Let $E$ be of dimension $n$ .

If $F, G$ are subspaces with $\dim F + \dim G > n$ , prove $F \cap G \neq \{0\}$ . Illustrate: two subspaces of dimensions $51$ and $50$ of $\R^{100}$ always share a nonzero vector.
If $H$ is a hyperplane and $F$ a subspace with $F \cap H = \{0\}$ , prove $\dim F \leq 1$ .

Solution

Solution of Exercise 19.11.

Grassmann: $\dim(F \cap G) = \dim F + \dim G - \dim(F + G) \geq \dim F + \dim G - n > 0$ , so $F \cap G \neq \{0\}$ . With $n = 100$ : $51 + 50 - 100 = 1 > 0$ , the intersection contains a line.
If $F \cap H = \{0\}$ , the sum is direct and $\dim F + (n - 1) = \dim(F \oplus H) \leq n$ , so $\dim F \leq 1$ . (Conversely a line not contained in $H$ does satisfy this: hyperplanes miss almost nothing.)

Exercise 19.12 ★★★

(Common supplementary) Let $F, G$ be subspaces of $E$ (finite dimension) with $\dim F = \dim G$ . Prove that $F$ and $G$ admit a common supplementary: there is a subspace $S$ with $E = F \oplus S = G \oplus S$ . (Induct downward on $\dim F$ : if $F \neq G \neq E$ , pick $x \notin F \cup G$ — Exercise 18.12 allows it — and consider $F \oplus \operatorname{Vect}(x)$ and $G \oplus \operatorname{Vect}(x)$ .)

Solution

Solution of Exercise 19.12.

Downward induction on $d = \dim F = \dim G$ , from $d = n$ to $d = 0$ . If $d = n$ : $F = G = E$ and $S = \{0\}$ works. Suppose the statement holds for subspace pairs of dimension $d + 1 \leq n$ , and let $\dim F = \dim G = d < n$ .

If $F = G$ : take for $S$ any supplementary of $F$ (Theorem 19.14). If $F \neq G$ : both are proper, so by Exercise 18.12 there exists $x \notin F \cup G$ . The sums $F' = F \oplus \operatorname{Vect}(x)$ and $G' = G \oplus \operatorname{Vect}(x)$ are direct ( $x \notin F$ , $x \notin G$ ) and have dimension $d + 1$ ; by the induction hypothesis they admit a common supplementary $S'$ : $E = F' \oplus S' = G' \oplus S'$ . Set $S = \operatorname{Vect}(x) \oplus S'$ — direct, because $S' \cap \operatorname{Vect}(x) \subseteq S' \cap F' = \{0\}$ .

Then $F + S = F + \operatorname{Vect}(x) + S' = F' + S' = E$ , and $F \cap S = \{0\}$ : if $f = \lambda x + s'$ with $f \in F$ , $s' \in S'$ , then $s' = f - \lambda x \in F' \cap S' = \{0\}$ , so $f = \lambda x$ , forcing $\lambda = 0$ ( $x \notin F$ ) and $f = 0$ . Hence $E = F \oplus S$ , and symmetrically $E = G \oplus S$ .

19.4 Problem: Dedekind’s tower law

Problem 19.1

Nothing in Chapters 18–19 used anything about the scalars beyond the field axioms (Definition 7.22): one may therefore take $K = \Q$ and measure sets of real numbers with the yardstick of $\Q$ -dimension. This problem computes the dimension of $\Q(\sqrt2, \sqrt3)$ , proves Dedekind’s tower law $\dim_\Q M = \dim_\Q K \cdot \dim_K M$ , and harvests irrationality theorems by pure dimension counting — no $\varepsilon$ , no decimals, just bases.

Part I — Rational scalars.

Check that $\Q$ is a field and that every definition and proof of Chapters 18–19 uses only the field axioms of the scalars; conclude that $\R$ is a $\Q$ -vector space and that the exchange lemma, the basis theorems and Grassmann’s formula hold over $\Q$ . Point out the one step of the proof of Theorem 19.2 where division by a nonzero scalar is performed.
Show that $(1, \sqrt2)$ is free over $\Q$ but linked over $\R$ . Set $\Q(\sqrt2) = \operatorname{Vect}_\Q(1, \sqrt2) = \{a + b\sqrt2 : a, b \in \Q\}$ ; what is $\dim_\Q \Q(\sqrt2)$ ?
Show that $\Q(\sqrt2)$ is stable under multiplication. For $u = a + b\sqrt2$ set $\sigma(u) = a - b\sqrt2$ and $N(u) = u\,\sigma(u) = a^2 - 2b^2$ . Show $\sigma(uv) = \sigma(u)\sigma(v)$ , deduce $N(uv) = N(u)N(v)$ , and show $N(u) \neq 0$ whenever $u \neq 0$ .
Deduce that every nonzero $u \in \Q(\sqrt2)$ has its inverse in $\Q(\sqrt2)$ , namely $u^{-1} = \sigma(u)/N(u)$ : $\Q(\sqrt2)$ is a subfield of $\R$ . Compute $\dfrac1{3 + 2\sqrt2}$ and $\dfrac1{1 + \sqrt2}$ .

Part II — Adjoining $\sqrt3$ .

Prove that $\sqrt6 \notin \Q$ (compare the exponent of $2$ on both sides of $6q^2 = p^2$ , as in Exercise 6.7), then that $\sqrt3 \notin \Q(\sqrt2)$ (square $\sqrt3 = a + b\sqrt2$ and discuss the cases $ab \neq 0$ , $b = 0$ , $a = 0$ ).
Let $M = \operatorname{Vect}_\Q(1, \sqrt2, \sqrt3, \sqrt6)$ . Show that $M$ is stable under multiplication (a table of the products of basis vectors suffices).
Write $K = \Q(\sqrt2)$ . Show that $(1, \sqrt3)$ is free over the field $K$ , and deduce that $M = K + K\sqrt3$ is a $K$ -vector space of dimension $2$ with basis $(1, \sqrt3)$ .
Prove that $(1, \sqrt2, \sqrt3, \sqrt6)$ is free over $\Q$ , hence $\dim_\Q M = 4$ . (Group a null relation as $(a + b\sqrt2) + (c + d\sqrt2)\sqrt3 = 0$ and apply questions 7 then 2.)

Part III — The tower law. Let $\Q \subseteq K \subseteq M \subseteq \R$ where $K$ and $M$ are subfields, $(e_1, \dots, e_m)$ is a basis of $K$ as a $\Q$ -vector space, and $(f_1, \dots, f_n)$ a basis of $M$ as a $K$ -vector space.

Show that the $mn$ products $(e_i f_j)$ generate $M$ over $\Q$ .
Show that the family $(e_i f_j)$ is free over $\Q$ . (Reorganize a null $\Q$ -combination as $\sum_j \bigl(\sum_i \lambda_{ij} e_i\bigr) f_j$ , whose inner coefficients live in $K$ .)
Conclude with Dedekind’s tower law:

$\dim_\Q M \;=\; \dim_\Q K \,\cdot\, \dim_K M ,$

and check it on $\Q \subseteq \Q(\sqrt2) \subseteq M$ against questions 7 and 8.
(Fields for free) Let $A \subseteq \R$ be a finite-dimensional $\Q$ -subspace containing $1$ and stable under multiplication, and let $u \in A$ , $u \neq 0$ . Show that if $(a_1, \dots, a_d)$ is a basis of $A$ , then $(u a_1, \dots, u a_d)$ is again a basis of $A$ ; deduce that $u$ has an inverse in $A$ : $A$ is a subfield of $\R$ . Which earlier questions does this recover?
Show that for every $x \in A$ (as in question 12, $\dim_\Q A = d$ ) the family $(1, x, x^2, \dots, x^{d})$ is linked: every element of $A$ is a root of a nonzero polynomial with rational coefficients, of degree at most $d$ .

Part IV — One number generates everything. Set $s = \sqrt2 + \sqrt3$ .

Compute the coordinates of $s^2$ , $s^3$ and $s^4$ in the basis $(1, \sqrt2, \sqrt3, \sqrt6)$ of $M$ .
Show that $(1, s, s^2, s^3)$ is free over $\Q$ , and deduce $\operatorname{Vect}_\Q(1, s, s^2, s^3) = M$ : every element of $M$ is a rational polynomial in $s$ . Express $\sqrt2$ and $\sqrt3$ as such polynomials.
Verify $s^4 - 10s^2 + 1 = 0$ , and show that $X^4 - 10X^2 + 1$ is the monic polynomial of least degree vanishing at $s$ . Determine its four real roots.
Deduce: $1/s = 10s - s^3$ ; identify this number. Show that $a + b\sqrt2 + c\sqrt3 + d\sqrt6$ ( $a,b,c,d \in \Q$ ) is rational if and only if $b = c = d = 0$ ; in particular $\sqrt2 + \sqrt3 + \sqrt6$ is irrational (strengthening Exercise 10.7).

Part V — The cube root stays outside. Set $t = 2^{1/3}$ .

Show that $X^3 - 2$ has no rational root (the rational-root criterion of Exercise 8.5, or a valuation count); deduce that $(1, t)$ is free over $\Q$ .
Show that $(1, t, t^2)$ is free over $\Q$ . (If some nonzero $P \in \Q[X]$ of degree $\leq 2$ kills $t$ , take one of least degree and divide $X^3 - 2$ by it, Theorem 8.3; conclude that $X^3 - 2$ would have a rational root.) Deduce that $A = \operatorname{Vect}_\Q(1, t, t^2)$ has dimension $3$ , is stable under multiplication, and is a subfield of $\R$ .
Prove that $t \notin M$ : otherwise $M$ would be a vector space over the field $A$ , and the tower law would force $3 \mid 4$ . So $2^{1/3}$ is not a rational combination of $1, \sqrt2, \sqrt3, \sqrt6$ .
Deduce that $t \notin \Q(\sqrt2)$ , that $t + \sqrt2$ is irrational, and that no rationals $a, b$ satisfy $2^{1/3} = a + b\sqrt3$ .

Part VI — All the subfields, and synthesis.

Prove the divisibility of degrees: if $\Q \subseteq A \subseteq B \subseteq \R$ are subfields with $\dim_\Q B$ finite, then $\dim_\Q A$ divides $\dim_\Q B$ . What are the possible dimensions of subfields of $M$ ?
Determine all subfields of $M$ of dimension $2$ over $\Q$ . (Reduce to finding the $w = a + b\sqrt2 + c\sqrt3 + d\sqrt6 \in M \setminus \Q$ with $w^2 \in \Q$ : expand $w^2$ and annihilate the three irrational coordinates.)
Express $\dfrac1{1 + \sqrt2 + \sqrt3}$ in the basis $(1, \sqrt2, \sqrt3, \sqrt6)$ (multiply by well-chosen conjugates).
Synthesis, in four sentences: why $\Q$ -dimension is an arithmetic invariant of a set of reals; what finiteness of the dimension forces about every element (question 13); how the two-out-of-three rule produced inverses out of thin air (question 12); and what the tower law forbids (question 20). Name the theorem proved in Part III.

Solution

Solution of Problem 19.1.

1. $\Q$ contains $0 \neq 1$ , is stable under addition, multiplication and opposites, and every nonzero rational has a rational inverse: a field (Definition 7.22). The definitions of span, freeness, basis and the proofs of Chapters 18–19 use only vector addition, distributivity and scalar arithmetic in a field — never an absolute value, an order or a limit. So $\R$ , with its own addition and the multiplication $\Q \times \R \to \R$ , is a $\Q$ -vector space, and all the theorems apply. Division by a scalar occurs once in the proof of Theorem 19.2: to “solve for $g_k$ ” one divides by its nonzero coefficient.

2. If $a + b\sqrt2 = 0$ with $a, b \in \Q$ not both zero: $b \neq 0$ would give $\sqrt2 = -a/b \in \Q$ , contradicting the irrationality of $\sqrt2$ (Exercise 6.7 with $p = 2$ ); so $b = 0$ , then $a = 0$ : free over $\Q$ . Over $\R$ the relation $\sqrt2\cdot 1 + (-1)\cdot\sqrt2 = 0$ is nontrivial: linked. Hence $\dim_\Q \Q(\sqrt2) = 2$ , with unique coordinates $(a, b)$ .

3. $(a + b\sqrt2)(c + d\sqrt2) = (ac + 2bd) + (ad + bc)\sqrt2 \in \Q(\sqrt2)$ . Then

\sigma(uv) = (ac + 2bd) - (ad + bc)\sqrt2 = (a - b\sqrt2)(c - d\sqrt2) = \sigma(u)\sigma(v),

so $N(uv) = uv\,\sigma(uv) = u\sigma(u)\,v\sigma(v) = N(u)N(v)$ . If $N(u) = a^2 - 2b^2 = 0$ with $u \neq 0$ : $b \neq 0$ would give $(a/b)^2 = 2$ , a rational square root of $2$ ; so $b = 0$ , then $a = 0$ and $u = 0$ : contradiction. Hence $N(u) \neq 0$ for $u \neq 0$ .

4. $u \cdot \dfrac{\sigma(u)}{N(u)} = \dfrac{N(u)}{N(u)} = 1$ , and $\sigma(u)/N(u) \in \Q(\sqrt2)$ : every nonzero element is invertible inside $\Q(\sqrt2)$ , which is therefore a subfield of $\R$ . Examples: $N(3 + 2\sqrt2) = 9 - 8 = 1$ , so

\frac1{3 + 2\sqrt2} = 3 - 2\sqrt2 ; \qquad N(1 + \sqrt2) = -1, \quad \frac1{1 + \sqrt2} = \sqrt2 - 1 .

5. If $\sqrt6 = p/q$ then $6q^2 = p^2$ ; the exponent of $2$ is $1 + 2v_2(q)$ , odd, on the left, and $2v_2(p)$ , even, on the right: impossible. Now suppose $\sqrt3 = a + b\sqrt2$ with $a, b \in \Q$ . Squaring: $3 = a^2 + 2b^2 + 2ab\sqrt2$ . If $ab \neq 0$ , then $\sqrt2 = (3 - a^2 - 2b^2)/(2ab) \in \Q$ : impossible. If $b = 0$ : $\sqrt3 = a \in \Q$ , contradicting Exercise 6.7 ( $p = 3$ ). If $a = 0$ : $\sqrt3 = b\sqrt2$ , and multiplying by $\sqrt2$ : $\sqrt6 = 2b \in \Q$ : impossible. So $\sqrt3 \notin \Q(\sqrt2)$ .

6. The products of basis vectors are

\sqrt2\,\sqrt3 = \sqrt6,\quad \sqrt2\,\sqrt6 = 2\sqrt3,\quad \sqrt3\,\sqrt6 = 3\sqrt2,\quad (\sqrt2)^2 = 2,\ (\sqrt3)^2 = 3,\ (\sqrt6)^2 = 6,

all in $M$ . A product of two elements of $M$ expands by bilinearity into rational combinations of these: $M$ is stable under multiplication.

7. Let $x + y\sqrt3 = 0$ with $x, y \in K = \Q(\sqrt2)$ . If $y \neq 0$ , then $\sqrt3 = -x/y \in K$ (question 4: $K$ is a field), contradicting question 5. So $y = 0$ , then $x = 0$ : $(1, \sqrt3)$ is free over $K$ . It generates: $K + K\sqrt3 = \{(a + b\sqrt2) + (c + d\sqrt2)\sqrt3\} = \operatorname{Vect}_\Q (1, \sqrt2, \sqrt3, \sqrt6) = M$ . Hence $\dim_K M = 2$ .

8. A relation $a + b\sqrt2 + c\sqrt3 + d\sqrt6 = 0$ regroups as $(a + b\sqrt2) + (c + d\sqrt2)\sqrt3 = 0$ with coefficients in $K$ ; by question 7 both vanish, and by question 2 $a = b = 0$ and $c = d = 0$ . So the family is free and $\dim_\Q M = 4$ .

9. Every $x \in M$ writes $x = \sum_j y_j f_j$ with $y_j \in K$ (basis of $M$ over $K$ ), and each $y_j = \sum_i \lambda_{ij} e_i$ with $\lambda_{ij} \in \Q$ (basis of $K$ over $\Q$ ); substituting, $x = \sum_{i,j} \lambda_{ij}\, e_i f_j$ : the products generate $M$ over $\Q$ .

10. Suppose $\sum_{i,j} \lambda_{ij}\, e_i f_j = 0$ with $\lambda_{ij} \in \Q$ . Regroup: $\sum_j \bigl(\sum_i \lambda_{ij} e_i\bigr) f_j = 0$ , and the inner sums belong to $K$ . Freeness of $(f_j)$ over $K$ gives $\sum_i \lambda_{ij} e_i = 0$ for each $j$ ; freeness of $(e_i)$ over $\Q$ then gives $\lambda_{ij} = 0$ for all $i, j$ .

11. By questions 9 and 10, $(e_i f_j)_{i \leq m,\, j \leq n}$ is a basis of $M$ over $\Q$ with $mn$ elements:

\dim_\Q M = m\,n = \dim_\Q K \cdot \dim_K M .

Check: $\dim_\Q K = 2$ and $\dim_K M = 2$ (questions 2 and 7) give $\dim_\Q M = 4$ , which is question 8.

12. The vectors $u a_1, \dots, u a_d$ belong to $A$ (stability). They are free: if $\sum_i \lambda_i\, u a_i = 0$ , then $u \sum_i \lambda_i a_i = 0$ in $\R$ , and $u \neq 0$ forces $\sum_i \lambda_i a_i = 0$ , hence $\lambda_i = 0$ (the $a_i$ form a basis). So $(u a_1, \dots, u a_d)$ is a free family of $d$ vectors in $A$ , $\dim_\Q A = d$ : a basis (Proposition 19.8). In particular $1 \in A$ decomposes as $1 = \sum_i \mu_i\, u a_i = u\,v$ with $v = \sum_i \mu_i a_i \in A$ : the inverse of $u$ lies in $A$ . This recovers question 4 ( $A = \Q(\sqrt2)$ ) and proves at one stroke that $M$ is a subfield of $\R$ (with question 6).

13. The $d + 1$ vectors $1, x, x^2, \dots, x^{d}$ all lie in $A$ (stability under products); a free family of $A$ has at most $d$ vectors (Proposition 19.8), so they are linked: there are rationals $\lambda_0, \dots, \lambda_d$ , not all zero, with $\sum_k \lambda_k x^k = 0$ . The polynomial $P = \sum_k \lambda_k X^k$ is nonzero, has rational coefficients, degree $\leq d$ , and $P(x) = 0$ .

14. $s^2 = 2 + 2\sqrt6 + 3 = 5 + 2\sqrt6$ : coordinates $(5, 0, 0, 2)$ . Then

s^3 = s\,s^2 = (\sqrt2 + \sqrt3)(5 + 2\sqrt6) = 5\sqrt2 + 2\sqrt{12} + 5\sqrt3 + 2\sqrt{18} = 11\sqrt2 + 9\sqrt3 ,

coordinates $(0, 11, 9, 0)$ (using $\sqrt{12} = 2\sqrt3$ , $\sqrt{18} = 3\sqrt2$ ). Finally $s^4 = (s^2)^2 = 25 + 20\sqrt6 + 24 = 49 + 20\sqrt6$ : coordinates $(49, 0, 0, 20)$ .

15. Suppose $a + bs + cs^2 + ds^3 = 0$ . Reading the four coordinates on $(1, \sqrt2, \sqrt3, \sqrt6)$ :

a + 5c = 0,\qquad b + 11d = 0,\qquad b + 9d = 0,\qquad 2c = 0 .

So $c = 0$ , then $a = 0$ ; subtracting the middle equations, $2d = 0$ , then $b = 0$ : $(1, s, s^2, s^3)$ is free. Four free vectors in $M$ of dimension $4$ : a basis, so $\operatorname{Vect}_\Q(1, s, s^2, s^3) = M$ . From question 14, $s^3 - 9s = 2\sqrt2$ and $11s - s^3 = 2\sqrt3$ :

\sqrt2 = \frac{s^3 - 9s}{2}, \qquad \sqrt3 = \frac{11s - s^3}{2} .

16. $s^4 - 10s^2 + 1 = (49 + 20\sqrt6) - 10(5 + 2\sqrt6) + 1 = 0$ . A monic polynomial of degree $\leq 3$ vanishing at $s$ would produce a nontrivial null combination of $(1, s, s^2, s^3)$ , contradicting question 15: $X^4 - 10X^2 + 1$ has least degree. Its roots: $X^2 = 5 \pm 2\sqrt6 = (\sqrt3 \pm \sqrt2)^2$ , so the four real roots are $\pm(\sqrt3 + \sqrt2)$ and $\pm(\sqrt3 - \sqrt2)$ , i.e. $\pm\sqrt2 \pm \sqrt3$ .

17. From $s^4 - 10s^2 + 1 = 0$ : $s\,(10s - s^3) = 1$ , so

\frac1s = 10s - s^3 = 10(\sqrt2 + \sqrt3) - (11\sqrt2 + 9\sqrt3) = \sqrt3 - \sqrt2 ,

consistent with $(\sqrt3 + \sqrt2)(\sqrt3 - \sqrt2) = 1$ . If $a + b\sqrt2 + c\sqrt3 + d\sqrt6 = r \in \Q$ , then $(a - r) + b\sqrt2 + c\sqrt3 + d\sqrt6 = 0$ , and freeness (question 8) forces $b = c = d = 0$ (and $a = r$ ). For $\sqrt2 + \sqrt3 + \sqrt6$ the coordinates $(0, 1, 1, 1)$ are not of this form: irrational — a statement stronger than Exercise 10.7, obtained without any squaring tricks.

18. A rational root $p/q$ (lowest terms) of $X^3 - 2$ satisfies $p^3 = 2q^3$ , and the criterion of Exercise 8.5 gives $p \mid 2$ , $q \mid 1$ : candidates $\pm1, \pm2$ , whose cubes are $\pm1, \pm8 \neq 2$ . No rational root; in particular $t = 2^{1/3} \notin \Q$ , so $(1, t)$ is free over $\Q$ (as in question 2).

19. Suppose $(1, t, t^2)$ linked: some nonzero $P \in \Q[X]$ with $\deg P \leq 2$ has $P(t) = 0$ ; choose such a $P$ of least degree $d \geq 1$ . By question 18, $d \neq 1$ , so $d = 2$ . Euclidean division (Theorem 8.3): $X^3 - 2 = PQ + R$ with $Q, R \in \Q[X]$ , $\deg R < 2$ . Evaluating at $t$ : $0 = P(t)Q(t) + R(t) = R(t)$ , so $R$ vanishes at $t$ ; minimality of $d$ forces $R = 0$ . Then $X^3 - 2 = PQ$ with $\deg Q = 1$ : the rational root of $Q$ is a rational root of $X^3 - 2$ , contradicting question 18. Hence $(1, t, t^2)$ is free and $\dim_\Q A = 3$ . Stability: $t^3 = 2$ reduces every product of $1, t, t^2$ to a combination of them ( $t\cdot t^2 = 2$ , $t^2\cdot t^2 = 2t$ ); $A$ contains $1$ : by question 12, $A$ is a subfield of $\R$ .

20. Suppose $t \in M$ . Then $t^2 \in M$ (stability), so $A \subseteq M$ , and $M$ is a vector space over the field $A$ : the axioms are those of $\R$ -arithmetic, restricted. It is finite-dimensional over $A$ (a finite $\Q$ -generating family generates a fortiori over $A \supseteq \Q$ ). The tower law for $\Q \subseteq A \subseteq M$ gives

4 = \dim_\Q M = \dim_\Q A \cdot \dim_A M = 3\,\dim_A M ,

impossible: $3$ does not divide $4$ . So $2^{1/3} \notin M$ : no rational combination of $1, \sqrt2, \sqrt3, \sqrt6$ equals $2^{1/3}$ .

21. $\Q(\sqrt2) \subseteq M$ , so $t \notin \Q(\sqrt2)$ . If $t + \sqrt2 = r \in \Q$ , then $t = r - \sqrt2 \in \Q(\sqrt2)$ : contradiction — $2^{1/3} + \sqrt2$ is irrational. If $t = a + b\sqrt3$ , then $t \in \operatorname{Vect}_\Q(1, \sqrt3) \subseteq M$ : contradiction again.

22. $B$ is a vector space over the field $A$ (restriction of scalars), finite-dimensional since $\dim_\Q B$ is finite. The tower law for $\Q \subseteq A \subseteq B$ gives $\dim_\Q B = \dim_\Q A \cdot \dim_A B$ : the left factor divides. Subfields of $M$ therefore have $\Q$ -dimension $1$ , $2$ or $4$ : dimension $1$ is $\Q$ itself, dimension $4$ is $M$ .

23. Let $F \subseteq M$ be a subfield with $\dim_\Q F = 2$ , and $w \in F \setminus \Q$ : $(1, w)$ is free, hence a basis of $F$ . By question 13 ( $d = 2$ ), $w^2 = \alpha + \beta w$ for rationals $\alpha, \beta$ . Setting $v = w - \beta/2 \in F \setminus \Q$ :

v^2 = w^2 - \beta w + \frac{\beta^2}4 = \alpha + \frac{\beta^2}4 \;\in\; \Q,

and $F = \operatorname{Vect}(1, v)$ . Now write $v = a + b\sqrt2 + c\sqrt3 + d\sqrt6$ and expand:

v^2 = (a^2 + 2b^2 + 3c^2 + 6d^2) + (2ab + 6cd)\sqrt2 + (2ac + 4bd)\sqrt3 + (2ad + 2bc)\sqrt6 .

The three irrational coordinates vanish: $ab + 3cd = ac + 2bd = ad + bc = 0$ . If $a \neq 0$ : $b = -3cd/a$ , and the second condition becomes $c(a^2 - 6d^2) = 0$ ; since $a^2 = 6d^2$ with $d \neq 0$ would make $\sqrt6 = \abs{a/d}$ rational, either $c = 0$ or $d = 0$ , and in both cases the remaining conditions force $b = c = d = 0$ , i.e. $v \in \Q$ : excluded. So $a = 0$ , and the conditions read $3cd = 2bd = bc = 0$ : at most one of $b, c, d$ is nonzero. Hence $v$ is a rational multiple of $\sqrt2$ , $\sqrt3$ or $\sqrt6$ , and

F \in \bigl\{\Q(\sqrt2),\ \Q(\sqrt3),\ \Q(\sqrt6)\bigr\} ,

each of which is indeed a $2$ -dimensional subfield (stability as in question 6, inverses by question 12): exactly three quadratic subfields.

24. $(1 + \sqrt2 + \sqrt3)(1 + \sqrt2 - \sqrt3) = (1 + \sqrt2)^2 - 3 = 2\sqrt2$ , so

\frac1{1 + \sqrt2 + \sqrt3} = \frac{1 + \sqrt2 - \sqrt3}{2\sqrt2} = \frac{\sqrt2 + 2 - \sqrt6}{4} = \frac12 + \frac14\sqrt2 + 0\cdot\sqrt3 - \frac14\sqrt6 .

(Check: $(1 + \sqrt2 + \sqrt3)(2 + \sqrt2 - \sqrt6) = 4$ after expansion.)

25. (i) $\Q$ -dimension attaches an integer to each subfield of $\R$ , and the tower law makes these integers multiply along inclusions: dimension behaves like an arithmetic invariant, and divisibility constraints become impossibility proofs. (ii) Finite dimension forces every element to satisfy a nonzero rational polynomial equation of degree at most the dimension: finiteness means algebraicity. (iii) The two-out-of-three rule turned “multiplication by $u$ maps a basis to a free family of maximal size” into surjectivity, producing $u^{-1}$ with no formula: inverses came from counting. (iv) The tower law forbids a $3$ -dimensional field inside a $4$ -dimensional one, which is why $2^{1/3}$ cannot be reached from $\sqrt2$ and $\sqrt3$ . The theorem of Part III is Dedekind’s tower law, the opening move of Galois theory, developed in the Year 3 volume.

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