The Reduced Norm of a Cyclic Algebra

Q: Why is the order of a in F^×/N(K^×) always a divisor of s?

Because fixes F pointwise, so for a ∈ F^× the field norm is N_K/F(a) = a,(a)…^s-1(a) = a^s. Hence a^s is always a norm, and the quotient group has exponent dividing s. Wedderburn's hypothesis is that the order is as large as it can possibly be.

Q: Where does the proof of (14.9) actually use that the ideal is proper?

In the bound 1 ≤ d ≤ s-1. If d were 0 the ideal would be all of B; if d were s the ideal would be (z-a) itself and the conclusion a^s ∈ N(K^×) would be true and useless. The whole force of the argument is that a genuine intermediate left ideal produces a genuine intermediate power of a that is a norm.

Q: Is the reduced norm the same as the determinant in a matrix representation?

Yes, once you use a splitting field. Under D ⊗_F K ≅ M_s(K), the reduced norm of is the determinant of the corresponding matrix, and the value happens to lie in F even though the matrix has entries in K. That is exactly why it is well defined independently of the splitting field chosen.

Q: How does this specialise to quaternion algebras?

Take K = F( a) and cyclic parameter b, so that (K/F,,b) is the quaternion algebra with i^2=a, j^2=b. Then M() = pmatrix b_0 & b_1 b,(b_1) & (b_0)pmatrix and n() = N(b_0) - b,N(b_1). Writing b_0 = w + x a and b_1 = y + z a gives N(b_0) = w^2-ax^2 and N(b_1) = y^2-az^2, so n() = w^2-ax^2-by^2+abz^2 — the quaternion norm form.

Q: Does a nonzero reduced norm really give a two-sided inverse?

Yes. M() is the matrix of right multiplication by on the finite-dimensional K-space D, so n() ≠ 0 makes that map bijective and produces with = 1. In a finite-dimensional algebra a one-sided inverse is two-sided, so is a unit.

Executive Summary

A cyclic algebra $D = (K / F, σ, a)$ of degree $s$ is a left $K$ -vector space of dimension $s$ with basis $1, x, \dots, x^{s - 1}$ . Right multiplication by $α \in D$ is a $K$ -linear map of that space; its determinant is the reduced norm $n (α)$ , which lands in $F$ and is multiplicative.

The reduced norm does two jobs. It converts invertibility into a determinant condition — $α$ is a unit exactly when $n (α) \neq 0$ — and, in a polynomial version over the skew polynomial ring $K [t; σ]$ , it proves Wedderburn's criterion: if the class of $a$ in $F^{\times} / N_{K / F} (K^{\times})$ has order exactly $s$ , then $D$ is a division algebra. The same matrix also exhibits $K$ as a splitting field, $D \otimes_{F} K ≅ M_{s} (K)$ .

$n : D \to F$ Reduced norm

$n (α) \neq 0$ Unit criterion

order $s$ Wedderburn's hypothesis on

a

1914Wedderburn

Overview

For prime degree, deciding whether $(K / F, σ, a)$ is a division algebra is easy: it is one exactly when $a$ is not a norm. For composite degree the criterion fails, because Wedderburn–Artin allows intermediate matrix sizes. Wedderburn's 1914 answer replaces "not a norm" by a statement about the order of $a$ in the quotient group $F^{\times} / N_{K / F} (K^{\times})$ .

a, a^{2}, \dots, a^{s - 1} \notin N_{K / F} (K^{\times}) ⟹ (K / F, σ, a) is a division algebra .

(14.9)

Note that $a^{s} = N_{K / F} (a)$ automatically, so the order of the class of $a$ always divides $s$ ; the hypothesis is that it is as large as possible.

The modern proof is cohomological. Lam gives instead the classical argument, following Dickson with simplifications due to Tignol, and its engine is an elementary but non-obvious device: a determinant defined on the skew polynomial ring $B = K [t; σ]$ , regarded as a free module of rank $s$ over the central polynomial ring $K [z]$ with $z = t^{s}$ .

This page is the technical heart of the classical treatment. The statements it supports — the splitting criterion, the structure of $D$ , the prime-degree corollary — are on Cyclic Algebras; the degree- $2$ specialisation, where the reduced norm becomes a quadratic form, is on Generalised Quaternion Algebras.

Learning Objectives

Set up $B = K [t; σ]$ as a free left $K [z]$ -module of rank $s$ , $z = t^{s}$ .
Write the matrix of right multiplication and define the polynomial norm $n$ .
Prove $n (β) \in F [z]$ and that its constant term is $N_{K / F}$ of the constant term of $β_{0}$ .
Prove Wedderburn's theorem $(14.9)$ by applying $n$ to a factorisation of $z - a$ .
Specialise to $D$ and prove $α \in U (D) \Leftrightarrow n (α) \neq 0$ .
Deduce $D \otimes_{F} K ≅ M_{s} (K)$ and read off the explicit matrices for $x$ and for $b \in K$ .
Compute the degree- $3$ reduced norm form and apply it to Dickson's example over $ℚ$ .

Definitions

Construction(14.9a)The polynomial norm on

K [t; σ]

Let $K / F$ be cyclic Galois of degree $s$ with $Gal (K / F) = ⟨ σ ⟩$ , and let $B = K [t; σ]$ with $t b = σ (b) t$ . Put $z = t^{s}$ . Then $z \in Z (B)$ , the subring $K [z]$ is a commutative polynomial ring, and $B$ is a free left $K [z]$ -module with basis $1, t, \dots, t^{s - 1}$ .

Right multiplication $R_{β} : γ \mapsto γ β$ is a left $K [z]$ -module endomorphism of $B$ , because left multiplication by $K [z]$ and right multiplication by $β$ commute. Define $n (β) = \det R_{β} \in K [z]$ .

Since $\det$ is multiplicative and $R_{β β'} = R_{β'} \circ R_{β}$ , we get $n (β β') = n (β) n (β')$ .

Writing $β = β_{0} + β_{1} t + \dots + β_{s - 1} t^{s - 1}$ with $β_{i} \in K [z]$ , and extending $σ$ to $B$ by $σ (t) = t$ so that $t β = σ (β) t$ , the matrix of $R_{β}$ in the basis $1, t, \dots, t^{s - 1}$ is

(\begin{matrix} β_{0} & β_{1} & β_{2} & \dots & β_{s - 1} \\ z σ (β_{s - 1}) & σ (β_{0}) & σ (β_{1}) & \dots & σ (β_{s - 2}) \\ z σ^{2} (β_{s - 2}) & z σ^{2} (β_{s - 1}) & σ^{2} (β_{0}) & \dots & σ^{2} (β_{s - 3}) \\ ⋮ & ⋮ & ⋮ & ⋱ & ⋮ \\ z σ^{s - 1} (β_{1}) & z σ^{s - 1} (β_{2}) & z σ^{s - 1} (β_{3}) & \dots & σ^{s - 1} (β_{0}) \end{matrix})

(*)

Row $i$ records $t^{i} β = \sum_{j} σ^{i} (β_{j}) t^{i + j}$ , with $t^{i + j}$ replaced by $z t^{i + j - s}$ whenever $i + j \geq s$ .

$z$: $t^{s}$ , a central element of $B = K [t; σ]$ because $σ^{s} = id$ .
$n (β)$: The determinant of $(*)$ ; an element of $K [z]$ , and in fact of $F [z]$ .
$N$: $N_{K / F}$ , the field norm $c \mapsto c σ (c) \dots σ^{s - 1} (c)$ .
$\deg_{t}$ , $\deg_{z}$: Degree in $t$ as an element of $B$ ; degree in $z$ as a polynomial in $K [z]$ .
$M (α)$: For $α \in D$ , the matrix of right multiplication by $α$ on $D$ in the left $K$ -basis $1, x, \dots, x^{s - 1}$ .
Reduced norm $n (α)$: $\det M (α) \in F$ . It is the specialisation of the polynomial norm at $z = a$ .

Core Concepts

Why the determinant lands in $F [z]$

Two symmetries pin it down. Applying $σ$ entrywise to $(*)$ produces the matrix of $R_{σ (β)}$ , so $n (σ (β)) = σ (n (β))$ . On the other hand $t β = σ (β) t$ gives $n (t) n (β) = n (σ (β)) n (t)$ , and $n (t) = (- 1)^{s - 1} z \neq 0$ in the domain $K [z]$ , so $n (β) = n (σ (β))$ . Combining, $σ (n (β)) = n (β)$ : the coefficients are $σ$ -fixed, hence in $F$ .

Why the constant term is a field norm

Set $z = 0$ in $(*)$ . The factor $z$ occurs exactly in the entries strictly below the diagonal, so all of those vanish and what remains is upper triangular, with diagonal entries $β_{0}, σ (β_{0}), \dots, σ^{s - 1} (β_{0})$ evaluated at $z = 0$ . If $b_{0}$ denotes the constant term of $β_{0}$ , the determinant at $z = 0$ is therefore $b_{0} σ (b_{0}) \dots σ^{s - 1} (b_{0}) = N (b_{0})$ .

Specialising $z \mapsto a$

The surjection $B ↠ D = B / (t^{s} - a)$ sends $t \mapsto x$ and $z \mapsto a$ . Under it, $B$ as a free $K [z]$ -module of rank $s$ becomes $D$ as a $K$ -vector space of dimension $s$ , and $(*)$ becomes the matrix $M (α)$ with $z$ replaced by $a$ and $β_{i}$ by $b_{i}$ . This compatibility is the whole content of the commutative diagram Lam draws after $(14.10)$ .

F [z] \subseteq K [z] \subseteq B = K [t; σ]

z \mapsto a, t \mapsto x

F \subseteq K \subseteq D = ⨁_{i} K x^{i}

Key Results

Lemma(14.10)Properties of the polynomial norm

Let $B = K [t; σ]$ , $z = t^{s}$ , and $β = β_{0} + β_{1} t + \dots + β_{s - 1} t^{s - 1}$ with $β_{i} \in K [z]$ .

$n (β) \in F [z]$ , and the constant term of $n (β)$ equals $N_{K / F} (b_{0})$ , where $b_{0}$ is the constant term of $β_{0}$ . In particular $n |_{K} = N_{K / F}$ .
If moreover every $β_{i}$ lies in $K$ (so $β$ is a polynomial in $t$ of degree $d = \deg_{t} β \leq s - 1$ ), then $\deg_{z} n (β) = d$ , and if $β$ is monic in $t$ the leading coefficient of $n (β)$ is $(- 1)^{d (s - 1)}$ .

Proof

(1). From $t β = σ (β) t$ and multiplicativity, $n (t) n (β) = n (σ (β)) n (t)$ . Direct inspection of $(*)$ for $β = t$ gives $n (t) = (- 1)^{s - 1} z$ , a nonzero element of the integral domain $K [z]$ , so $n (β) = n (σ (β))$ . Applying $σ$ to every entry of $(*)$ turns it into the matrix for $σ (β)$ — note $σ (z) = z$ — so $n (σ (β)) = σ (n (β))$ . Hence $n (β)$ is fixed by $σ$ , i.e. $n (β) \in F [z]$ . Setting $z = 0$ in $(*)$ leaves a triangular matrix with diagonal $β_{0} (0), σ (β_{0}) (0), \dots, σ^{s - 1} (β_{0}) (0)$ , whose determinant is $N (b_{0})$ .

(2). With all $β_{i} \in K$ and $β$ monic of degree $d$ , set $β_{d} = 1$ and $β_{i} = 0$ for $i > d$ in $(*)$ . Each entry is either a constant or $z$ times a constant, and the $z$ 's occur exactly in the strictly lower-left region. Expanding the determinant, the terms of top degree in $z$ come from the permutation contributions using $d$ of the $z$ -entries, all of which involve the entries carrying $β_{d} = 1$ ; this yields $\deg_{z} n (β) = d$ with leading coefficient the sign of the corresponding permutation, namely $(- 1)^{d (s - 1)}$ .

Theorem(14.9)Wedderburn's division criterion

Let $K / F$ be a cyclic Galois extension of degree $s$ with $Gal (K / F) = ⟨ σ ⟩$ , and let $a \in F^{\times}$ . Suppose the image of $a$ in the abelian group $F^{\times} / N_{K / F} (K^{\times})$ has order exactly $s$ — equivalently, $a^{d} \notin N_{K / F} (K^{\times})$ for every $d$ with $1 \leq d \leq s - 1$ . Then $D = (K / F, σ, a)$ is a division $F$ -algebra.

Proof

Suppose not. Write $D = B / (z - a)$ with $B = K [t; σ]$ and $z = t^{s}$ ; note $t^{s} - a = z - a$ is central in $B$ . Since $D$ is simple artinian and not a division ring, it has a nonzero proper left ideal, whose preimage is a left ideal $𝔅$ with $(z - a) ⊊ 𝔅 ⊊ B$ .

$B$ is a principal left ideal domain, so $𝔅 = B β$ with $β$ monic; scaling, write $β = t^{d} + b_{d - 1} t^{d - 1} + \dots + b_{0}$ with $b_{i} \in K$ . Counting $F$ -dimensions, $\dim_{F} B / B β = d \cdot \dim_{F} K = d s$ , and $0 < \dim_{F} B / 𝔅 < \dim_{F} D = s^{2}$ forces $1 \leq d \leq s - 1$ .

Since $z - a \in B β$ , there is $β' \in B$ with $z - a = β' β$ . Apply the polynomial norm. Right multiplication by the central element $z - a$ is multiplication by a scalar on a free module of rank $s$ , so $n (z - a) = (z - a)^{s}$ , and therefore

n (β') n (β) = (z - a)^{s} in F [z] .

By $(14.10)$ , $n (β), n (β') \in F [z]$ , and $n (β)$ has degree $d$ with leading coefficient $(- 1)^{d (s - 1)}$ . Since $F [z]$ is a unique factorisation domain and $z - a$ is irreducible, $n (β)$ is a constant times a power of $z - a$ ; matching degrees and leading coefficients,

n (β) = (- 1)^{d (s - 1)} (z - a)^{d} .

Compare constant terms. On the left, $(14.10) (1)$ gives $N (b_{0})$ ; on the right, $(- 1)^{d (s - 1)} (- a)^{d} = (- 1)^{d s} a^{d}$ . Hence $N (b_{0}) = (- 1)^{d s} a^{d}$ , and since $N (- 1) = (- 1)^{s}$ ,

a^{d} = (- 1)^{d s} N (b_{0}) = N ((- 1)^{d} b_{0}) \in N_{K / F} (K^{\times}) .

This contradicts the hypothesis, because $1 \leq d \leq s - 1$ . Therefore $D$ is a division algebra.

Definition(14.11)The reduced norm on

D

For $α = b_{0} + b_{1} x + \dots + b_{s - 1} x^{s - 1} \in D = (K / F, σ, a)$ with $b_{i} \in K$ , let $M (α)$ be the matrix of the left $K$ -linear map $R_{α} : γ \mapsto γ α$ in the basis $1, x, \dots, x^{s - 1}$ — that is, $(*)$ with $z$ replaced by $a$ and $β_{i}$ by $b_{i}$ :

M (α) = (\begin{matrix} b_{0} & b_{1} & \dots & b_{s - 1} \\ a σ (b_{s - 1}) & σ (b_{0}) & \dots & σ (b_{s - 2}) \\ ⋮ & ⋮ & ⋱ & ⋮ \\ a σ^{s - 1} (b_{1}) & a σ^{s - 1} (b_{2}) & \dots & σ^{s - 1} (b_{0}) \end{matrix}), n (α) = \det M (α) .

The same $σ$ -invariance argument as in $(14.10) (1)$ gives $n (α) \in F$ . The map $n$ is the reduced norm of $D$ ; it restricts to $N_{K / F}$ on $K$ and satisfies $n (x) = (- 1)^{s - 1} a$ .

Proposition(14.11a)The reduced norm detects units

With $D = (K / F, σ, a)$ as above, $n : D \to F$ is multiplicative, and for $α \in D$

α \in U (D) ⟺ n (α) \neq 0 .

Consequently $D$ is a division algebra if and only if the reduced norm vanishes only at $0$ .

Proof

Multiplicativity is $\det (R_{α α'}) = \det (R_{α'} R_{α})$ . For the criterion: $M (α)$ is by construction the matrix of the $K$ -linear endomorphism $R_{α}$ of the $s$ -dimensional $K$ -space $D$ , so $n (α) \neq 0$ iff $R_{α}$ is bijective. If $R_{α}$ is bijective there is $γ$ with $γ α = 1$ ; then $R_{γ} R_{α} = R_{α γ}$ is also bijective — because $R_{α}$ is and $\dim_{K} D < \infty$ forces $R_{γ}$ to be bijective too — so $α γ$ is a unit and hence $α$ has a right inverse as well; thus $α \in U (D)$ . Conversely if $α \in U (D)$ then $R_{α}$ has inverse $R_{α^{- 1}}$ , so $n (α) n (α^{- 1}) = n (1) = 1$ and $n (α) \neq 0$ .

Theorem(14.12)–(14.13)

K

is a splitting field

The map $M : D \to End (_{K} D) ≅ M_{s} (K)$ is an injective $F$ -algebra homomorphism, described on generators by

M (x) = (\begin{matrix} 0 & 1 & 0 & \dots & 0 \\ 0 & 0 & 1 & \dots & 0 \\ ⋮ & ⋱ & ⋮ \\ 0 & 0 & 0 & \dots & 1 \\ a & 0 & 0 & \dots & 0 \end{matrix}), M (b) = (\begin{matrix} b \\ σ (b) \\ ⋱ \\ σ^{s - 1} (b) \end{matrix}) (b \in K) .

Since $M (D)$ commutes elementwise with the scalar matrices $K \subseteq M_{s} (K)$ , it induces a $K$ -algebra map $M \otimes 1 : D \otimes_{F} K \to M_{s} (K)$ . Both sides have $K$ -dimension $s^{2}$ , and $D \otimes_{F} K$ is a simple $K$ -algebra because $D$ is central simple over $F$ ; hence

M \otimes 1 : D \otimes_{F} K \tilde{⟶} M_{s} (K) .

So $K$ — a maximal subfield of $D$ — is a splitting field for $D$ , and $n$ is the restriction to $D$ of the determinant on $M_{s} (K)$ .

RemarkWedderburn's condition is sufficient, not necessary

For composite $s$ the converse of $(14.9)$ fails. Following Brauer and Tignol, Lam constructs a cyclic division algebra $D = (K / F, σ, - 1)$ of degree $s = 4$ , in which the class of $- 1$ has order at most $2$ in $F^{\times} / N (K^{\times})$ : take $K = ℚ (y, z)$ with $σ (y) = z$ , $σ (z) = - y$ of order $4$ , and $F = K^{⟨ σ ⟩}$ . The argument shows first that the centraliser $C_{D} (x^{2})$ is a quaternion division algebra, then that $D$ itself has no zero divisors by a grading argument. Over an algebraic number field, by contrast, the converse of $(14.9)$ does hold — but that is a theorem of class field theory.

Proof Techniques and Method

How these proofs work, and which move to reuse.

Assume failureIf

D

is not a division algebra, it has a proper nonzero left ideal.

Pull back to $B$

B = K [t; σ]

is a principal left ideal domain, so the ideal is

B β

with

β

monic of degree

d

1 \leq d \leq s - 1

Factor the central element

z - a = β' β

, an equation between elements of a noncommutative ring.

Apply the determinant

n

converts the noncommutative factorisation into

n (β') n (β) = (z - a)^{s}

in the commutative ring

F [z]

Read constant termsUnique factorisation pins

n (β) = \pm (z - a)^{d}

; its constant term is a field norm, giving

a^{d} \in N (K^{\times})

— the contradiction.

Reusable move

Commutativise by determinant

A multiplicative map from a noncommutative ring to a commutative one lets unique factorisation do the work. The trick is to find a module structure over a central subring; here it is $K [z]$ with $z = t^{s}$ .

Reusable move

Degrees and constant terms carry the arithmetic

Degree in $z$ recovers the ideal-theoretic degree $d$ ; the constant term recovers a field norm. Two coefficients of one polynomial deliver the whole theorem.

Reusable move

Right multiplication is left linear

Whenever a ring is a module over a subring on one side, multiplication on the other side is module-linear. That is why $R_{α}$ has a matrix over $K$ at all, and why $\det$ is available.

Reusable move

Split by base change to a maximal subfield

The matrix representation $M$ already lives over $K$ ; tensoring up and comparing dimensions upgrades an embedding to an isomorphism. This is the standard route to splitting fields.

Worked Example

The reduced norm in degree three

Let $s = 3$ and $α = b_{0} + b_{1} x + b_{2} x^{2} \in (K / F, σ, a)$ . Then

M (α) = (\begin{matrix} b_{0} & b_{1} & b_{2} \\ a σ (b_{2}) & σ (b_{0}) & σ (b_{1}) \\ a σ^{2} (b_{1}) & a σ^{2} (b_{2}) & σ^{2} (b_{0}) \end{matrix})

(E.1)

and expanding the determinant gives the classical cubic norm form

n (α) = N (b_{0}) + a N (b_{1}) + a^{2} N (b_{2}) - a (b_{0} σ (b_{1}) σ^{2} (b_{2}) + b_{1} σ (b_{2}) σ^{2} (b_{0}) + b_{2} σ (b_{0}) σ^{2} (b_{1})),

(E.2)

Setting $b_{1} = b_{2} = 0$ recovers $N_{K / F} (b_{0})$ ; setting $b_{0} = b_{2} = 0$ gives $a N (b_{1})$ , consistent with $n (x) = (- 1)^{2} a = a$ .

Dickson's nine-dimensional division algebra over $ℚ$

Let $ζ = e^{2 π i / 7}$ and $E = ℚ (ζ)$ , so $\dim_{ℚ} E = 6$ and $Gal (E / ℚ) = ⟨ τ ⟩$ with $τ (ζ) = ζ^{3}$ . Let $K$ be the unique cubic subfield of $E$ ; then $K / ℚ$ is cyclic of degree $3$ . Setting $v = ζ + ζ^{- 1} = 2 \cos (2 π / 7)$ one finds $v^{3} + v^{2} - 2 v - 1 = 0$ , and since $f (X) = X^{3} + X^{2} - 2 X - 1$ is irreducible over $ℚ$ , $K = ℚ (v) = E \cap ℝ$ . With $σ = τ^{2} |_{K}$ the conjugates are

v = 2 \cos \frac{2 π}{7}, σ (v) = v^{2} - 2 = 2 \cos \frac{4 π}{7}, σ^{2} (v) = 1 - v - v^{2} = 2 \cos \frac{6 π}{7} .

(E.3)

Their sum is $- 1$ and their product is $1$ , matching the coefficients of $f$ .

For $α = p + q v + r v^{2}$ with $p, q, r \in ℚ$ , left multiplication by $α$ on $K$ has matrix (in the basis $1, v, v^{2}$ , using $v^{3} = - v^{2} + 2 v + 1$ and $v^{4} = 3 v^{2} - v - 1$ )

(\begin{matrix} p & r & q - r \\ q & p + 2 r & 2 q - r \\ r & q - r & p - q + 3 r \end{matrix}), N (α) = p^{3} + q^{3} + r^{3} - p^{2} q - 2 p q^{2} + 5 p^{2} r + 6 p r^{2} - q^{2} r - 2 q r^{2} - p q r .

(E.4)

Claim. If $n$ is an even integer lying in $N (K^{\times})$ then $8 ∣ n$ . Write $n = N ((p + q v + r v^{2}) / m)$ with $p, q, r, m \in ℤ$ and $m > 0$ minimal, so $m^{3} n = N (p + q v + r v^{2})$ . Reducing $(E . 4)$ modulo $2$ and using $u \equiv u^{2} \equiv u^{3}$ there,

m^{3} n \equiv p + q + r + p q + p r + q r + p q r \equiv 1 + (p + 1) (q + 1) (r + 1) (\mod 2) .

(E.5)

As $n$ is even, the right side is $\equiv 0$ , forcing $p, q, r$ all even; minimality of $m$ then makes $m$ odd. Writing $p = 2 p_{0}$ , $q = 2 q_{0}$ , $r = 2 r_{0}$ and using homogeneity of degree $3$ , $m^{3} n = 8 N (p_{0} + q_{0} v + r_{0} v^{2}) \in 8 ℤ$ , and $m$ odd gives $8 ∣ n$ .

K = ℚ (v), D = K \oplus K x \oplus K x^{2}, v^{3} + v^{2} - 2 v - 1 = 0, x^{3} = 2, x v = (v^{2} - 2) x .

(14.14)

Dickson's explicit presentation. The reduced norm of $b_{0} + b_{1} x + b_{2} x^{2}$ is $(E . 2)$ with $a = 2$ .

Comparison and Classification

Three norms attached to a cyclic algebra of degree $s$
Norm	Domain and codomain	Degree as a form	Relation
Field norm $N_{K / F}$	$K^{\times} \to F^{\times}$	$s$	the restriction of $n$ to $K$
Reduced norm $n$	$D \to F$	$s$	$\det M (α)$ ; $n (x) = (- 1)^{s - 1} a$
Regular norm $\det_{F}$	$D \to F$	$s^{2}$	determinant of the $F$ -linear regular representation; equals $n^{s}$
Polynomial norm on $B$	$K [t; σ] \to F [z]$	graded by $\deg_{t}$	specialises to $n$ at $z = a$

Which criterion decides the division property
	$s$ prime	$s$ composite, general $F$	$F$ a number field
$a \notin N_{K / F} (K^{\times})$	yes	no	no
class of $a$ has order $s$	yes	partial	yes
reduced norm anisotropic	yes	yes	yes
no proper left ideal in $B$ above $z - a$	yes	yes	yes

Which criterion decides the division property

In the second column, the order condition is sufficient but not necessary, which is what the entry marked partial records.

Relationship Map

The reduced norm is the bridge between the algebra and the arithmetic of its centre.

α \in D

M (α) \in M_{s} (K)

n (α) = \det M (α) \in F

unit iff nonzero

Downwards

Degree $2$

$n$ becomes the quaternion norm form $w^{2} - a x^{2} - b y^{2} + a b z^{2}$ , and the unit criterion becomes anisotropy of a quadratic form.

Sideways

Crossed products

The same determinant construction works for a crossed product over any Galois group, with $End$ of a free module of rank $| G |$ over the fixed field.

Upwards

Reduced norm on any central simple algebra

For general central simple $A$ of degree $n$ , choose a splitting field $L$ ; $\det$ on $A \otimes L ≅ M_{n} (L)$ descends to $A$ , and the cyclic case is the computable instance.

The relation $\det_{F} = n^{s}$ between the regular and the reduced norm explains the word reduced: the ordinary determinant of an element acting on the $s^{2}$ -dimensional space $D$ is an $s$ -th power, and $n$ is that root.

Applications and Industry Use

Applications here means where this structure is used — inside mathematics and in the engineering and computing disciplines that consume it.

Wireless communications

Coding gain of space–time codes

For a code drawn from a cyclic division algebra, the minimum determinant of codeword differences is a minimum of reduced norms. Non-vanishing determinant designs are engineered by bounding $n$ away from zero on an order of the algebra.

Number theory

Reduced norms and class groups

The reduced norm map on the ideles of a central simple algebra is the input to the Hasse–Schilling–Maass norm theorem and to the computation of class numbers of orders.

Algebraic groups

${SL}_{1} (D)$

The reduced-norm-one elements form an algebraic group, an inner form of ${SL}_{s}$ . Its rational points are studied through exactly the matrix $M (α)$ on this page.

Symbolic computation

Splitting algorithms

Explicit splitting of a cyclic algebra over a number field is implemented by writing $M \otimes 1$ down and solving a norm equation; libraries return the isomorphism $D \otimes_{F} K ≅ M_{s} (K)$ in this form.

Cryptography

Norm-form hardness

Deciding whether a value is represented by a norm form of degree $s$ in $s^{2}$ variables is the algebraic core of several proposals; the reduced norm is the canonical such form.

Invariant theory

Generic division algebras

The reduced norm is the fundamental invariant polynomial of a central simple algebra and controls the study of generic matrices and Amitsur's non-crossed products.

The space–time coding application is the one where the determinant is literally the engineering figure of merit: pairwise error probability at high signal-to-noise ratio is governed by the determinant of the difference of two transmitted matrices, and choosing codewords inside a division algebra guarantees that determinant is never zero.

Standards and Notation

Standards here covers notation, symbol and markup standards, and reference implementations, rather than material or design codes.

This page

n (α)

for the reduced norm, following Lam's §14

Common notation

{Nrd}_{D / F}

, and

{Trd}_{D / F}

for the reduced trace

Regular versus reduced

N_{D / F} = {Nrd}^{s}

and

{tr}_{D / F} = s \cdot Trd

Reduced characteristic polynomialDegree

s

, with

Nrd

as constant term up to sign

MagmaReducedNorm, ReducedCharacteristicPolynomial, IsDivisionRing

Sage / Parialgnorm and algtomatrix in Pari/GP for algebras given by a cyclic presentation

MarkupPresentation MathML per ISO/IEC 40314; matrix layout per mtable

Computational Notes

Computational notes cover algorithms, cost and library behaviour rather than manufacturing process.

**Building $M (α)$ ** costs $O (s^{2})$ applications of powers of $σ$ to elements of $K$ ; if $K$ is given by a basis over $F$ , each is a fixed $\dim_{F} K$ -square matrix, so the total is $O (s^{4})$ operations in $F$ .
**Evaluating $n (α)$ ** is one $s \times s$ determinant over $K$ , so $O (s^{3})$ operations in $K$ , that is $O (s^{5})$ in $F$ with naive arithmetic. For $s \leq 4$ the expanded forms are small enough to hardcode, and $(E . 2)$ is the degree- $3$ case.
Deciding the division property by Wedderburn's criterion needs the order of $a$ in $F^{\times} / N_{K / F} (K^{\times})$ — that is $s - 1$ norm-equation tests. Over a number field each is decidable; over a general field none need be.
**Testing anisotropy of $n$ directly** is not a finite computation for infinite $F$ : it asks whether a degree- $s$ form in $s^{2}$ variables has a nontrivial zero, which is undecidable in general.
**Exact arithmetic in $D$ ** is best done through $M$ : multiply matrices over $K$ rather than reducing words in $x$ , since the reduction $x^{s} \mapsto a$ is already built into the matrix.

Failure Modes and Common Mistakes

Do not omit the bound $1 \leq d \leq s - 1$ ; without it the conclusion $a^{d} \in N (K^{\times})$ is vacuous, since $a^{s} = N (a)$ always.
Do not assume $n (β)$ lies in $F [z]$ without the $σ$ -invariance argument — a priori the determinant only lies in $K [z]$ .
Do not read $(*)$ as a matrix over $K$ : its entries lie in $K [z]$ , and the specialisation $z \mapsto a$ is a separate step.
Do not conclude from $D \otimes_{F} K ≅ M_{s} (K)$ that $D$ splits over $F$ — a division algebra always splits over its maximal subfields.

Quick Reference

Setting

K / F

cyclic of degree

s

Gal = ⟨ σ ⟩

a \in F^{\times}

D = (K / F, σ, a)

Polynomial norm

n : K [t; σ] \to F [z]

z = t^{s}

n (β) = \det

(*)

Key values

n (t) = (- 1)^{s - 1} z

n (b) = N_{K / F} (b)

n (z - a) = (z - a)^{s}

Reduced norm

n (α) = \det M (α) \in F

, multiplicative

Unit test

α \in U (D) \Leftrightarrow n (α) \neq 0

Wedderburnclass of

a

of order

s

F^{\times} / N (K^{\times}) \Rightarrow D

division

Splitting

D \otimes_{F} K ≅ M_{s} (K)

Reduced vs regular

\det_{F} (α) = n (α)^{s}

Statements and their hypotheses
Statement	Hypotheses	Reference
$n (β) \in F [z]$ ; constant term $N (b_{0})$	$β \in K [t; σ]$ written over $K [z]$	(14.10)(1)
$\deg_{z} n (β) = \deg_{t} β$ ; leading coeff $(- 1)^{d (s - 1)}$	all coefficients of $β$ in $K$ , $β$ monic	(14.10)(2)
$D$ is a division algebra	class of $a$ has order $s$ in $F^{\times} / N (K^{\times})$	(14.9)
$n (α) \in F$ , multiplicative, detects units	$D = (K / F, σ, a)$	(14.11)
$M (x)$ cyclic with corner $a$ ; $M (b)$ diagonal	same	(14.12)
$D \otimes_{F} K ≅ M_{s} (K)$	$D$ central simple over $F$ , $K$ maximal subfield	(14.13)

Frequently Asked Questions

Why is the order of

a

F^{\times} / N (K^{\times})

always a divisor of

s

Because $σ$ fixes $F$ pointwise, so for $a \in F^{\times}$ the field norm is $N_{K / F} (a) = a σ (a) \dots σ^{s - 1} (a) = a^{s}$ . Hence $a^{s}$ is always a norm, and the quotient group has exponent dividing $s$ . Wedderburn's hypothesis is that the order is as large as it can possibly be.

Where does the proof of

(14.9)

actually use that the ideal is proper?

In the bound $1 \leq d \leq s - 1$ . If $d$ were $0$ the ideal would be all of $B$ ; if $d$ were $s$ the ideal would be $(z - a)$ itself and the conclusion $a^{s} \in N (K^{\times})$ would be true and useless. The whole force of the argument is that a genuine intermediate left ideal produces a genuine intermediate power of $a$ that is a norm.

Is the reduced norm the same as the determinant in a matrix representation?

Yes, once you use a splitting field. Under $D \otimes_{F} K ≅ M_{s} (K)$ , the reduced norm of $α$ is the determinant of the corresponding matrix, and the value happens to lie in $F$ even though the matrix has entries in $K$ . That is exactly why it is well defined independently of the splitting field chosen.

How does this specialise to quaternion algebras?

Take $K = F (\sqrt{a})$ and cyclic parameter $b$ , so that $(K / F, σ, b)$ is the quaternion algebra with $i^{2} = a$ , $j^{2} = b$ . Then $M (α) = (\begin{matrix} b_{0} & b_{1} \\ b σ (b_{1}) & σ (b_{0}) \end{matrix})$ and $n (α) = N (b_{0}) - b N (b_{1})$ . Writing $b_{0} = w + x \sqrt{a}$ and $b_{1} = y + z \sqrt{a}$ gives $N (b_{0}) = w^{2} - a x^{2}$ and $N (b_{1}) = y^{2} - a z^{2}$ , so $n (α) = w^{2} - a x^{2} - b y^{2} + a b z^{2}$ — the quaternion norm form.

Does a nonzero reduced norm really give a two-sided inverse?

Yes. $M (α)$ is the matrix of right multiplication by $α$ on the finite-dimensional $K$ -space $D$ , so $n (α) \neq 0$ makes that map bijective and produces $γ$ with $γ α = 1$ . In a finite-dimensional algebra a one-sided inverse is two-sided, so $α$ is a unit.

Why does Lam give a non-cohomological proof?

Because §14 is meant to be readable before any Brauer group theory is available. The cohomological proof computes the order of the class of the algebra in $Br (F)$ via $2$ -cocycles; the determinant proof needs only a principal ideal domain, unique factorisation in $F [z]$ , and two coefficients of one polynomial.

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §14, (14.9)–(14.14), pp. 236–239.
J. H. M. Wedderburn, “A type of primitive algebra”, Transactions of the American Mathematical Society 15 (1914), 162–166.
L. E. Dickson, Algebren und ihre Zahlentheorie, Orell Füssli, Zürich, 1927; Appendix 1 contains the classical proof Lam follows.
N. Jacobson, Basic Algebra II, 2nd edition, W. H. Freeman, 1989, Chapter 4 (reduced norms and traces of central simple algebras).
R. S. Pierce, Associative Algebras, Graduate Texts in Mathematics 88, Springer-Verlag, 1982, Chapter 16.
P. Gille and T. Szamuely, Central Simple Algebras and Galois Cohomology, Cambridge University Press, 2006, Chapters 2–4.

AI Suggested Questions

Give the cohomological proof of Wedderburn's theorem and compare the two arguments step by step.
Prove that the determinant of the regular representation of a central simple algebra of degree $s$ equals the $s$ -th power of the reduced norm.
Write out the reduced norm form of a cyclic algebra of degree $4$ and analyse its singular locus.
For which fields $F$ is Wedderburn's order condition also necessary, and what is the proof over a number field?
Describe Brauer's and Tignol's degree- $4$ example in full and verify that $- 1$ has order $2$ modulo norms there.
How is the minimum reduced norm of a lattice in a cyclic division algebra bounded below, and why does that matter for space–time codes?
Explain how the reduced norm defines the algebraic group ${SL}_{1} (D)$ and what its rational points look like.

Reduced Norms

Executive Summary

Overview

Learning Objectives

Definitions

Core Concepts

Why the determinant lands in $F [z]$

Why the constant term is a field norm

Specialising $z \mapsto a$

Key Results

Proof Techniques and Method

Commutativise by determinant

Degrees and constant terms carry the arithmetic

Right multiplication is left linear

Split by base change to a maximal subfield

Worked Example

The reduced norm in degree three

Dickson's nine-dimensional division algebra over $ℚ$

Comparison and Classification

Relationship Map

Degree $2$

Crossed products

Reduced norm on any central simple algebra

Applications and Industry Use

Coding gain of space–time codes

Reduced norms and class groups

${SL}_{1} (D)$

Splitting algorithms

Norm-form hardness

Generic division algebras

Standards and Notation

Computational Notes

Failure Modes and Common Mistakes

Quick Reference

Frequently Asked Questions

References

AI Suggested Questions

Continue learning

01Executive Summary

02Overview

03Learning Objectives

04Definitions

05Core Concepts

Why the determinant lands in F[z]

Why the constant term is a field norm

Specialising z↦a

06Key Results

07Proof Techniques and Method

Commutativise by determinant

Degrees and constant terms carry the arithmetic

Right multiplication is left linear

Split by base change to a maximal subfield

08Worked Example

The reduced norm in degree three

Dickson's nine-dimensional division algebra over ℚ

09Comparison and Classification

10Relationship Map

Degree 2

Crossed products

Reduced norm on any central simple algebra

11Applications and Industry Use

Coding gain of space–time codes

Reduced norms and class groups

SL1(D)

Splitting algorithms

Norm-form hardness

Generic division algebras

12Standards and Notation

13Computational Notes

14Failure Modes and Common Mistakes

15Quick Reference

16Frequently Asked Questions

17Related KEVOS Topics

18References

19AI Suggested Questions

Continue learning

Executive Summary

Overview

Learning Objectives

Definitions

Core Concepts

Why the determinant lands in $F [z]$

Specialising $z \mapsto a$

Key Results

Proof Techniques and Method

Worked Example

Dickson's nine-dimensional division algebra over $ℚ$

Comparison and Classification

Relationship Map

Degree $2$

Applications and Industry Use

${SL}_{1} (D)$

Standards and Notation

Computational Notes

Failure Modes and Common Mistakes

Quick Reference

Frequently Asked Questions

Related KEVOS Topics

References

AI Suggested Questions