Finite: ℤ/m and the places

Hypercomplex algebras introduces i_k, j_k and ε_k as formal generators with a square attached. But you do not have to adjoin them — ℤ/m already contains them, and which ones it contains is decided one prime at a time. This page follows that all the way down.

A square root of 1 is a root in every channel

The Chinese Remainder Theorem splits the ring: for $m = \prod p_i^{a_i}$,

$$\mathbb{Z}/m \;≅\; \prod_i \mathbb{Z}/p_i^{a_i}$$

and a ring isomorphism carries $x^2 = 1$ to the same equation in each factor independently. So a square root of 1 mod m is a choice of root in each channel — and an odd prime power $p^a$ has exactly two, $± 1$, because $(\mathbb{Z}/p^a)^×$ is cyclic and so has a unique element of order 2.

Counting is then immediate. For odd m,

$$\#\{x : x^2 ≡ 1 \pmod m\} \;=\; 2^{ω(m)}$$

with $ω(m)$ the number of distinct primes dividing m. These $2^{ω(m)}$ residues are exactly the $± 1$ sign vectors over the CRT channels — a spectral basis, one coordinate per place. Two of them, $1$ and $m-1$, are the trivial pair; every other one is a genuine split unit, a square root of 1 that is not $± 1$.

#Sign vectors, as residues

15 = 3·5 gives 2² = 4; 1155 = 3·5·7·11 gives 2⁴ = 16 — one residue per ± choice across four places.

In$\mathrm{PowerModList}(1, \frac{1}{2}, 15)$Out$\bigl\lbrack1, 4, 11, 14\bigr\rbrack$In$\mathrm{PowerModList}(1, \frac{1}{2}, 1\,155)$Out$\bigl\lbrack1, 34, 76, 274, 386, 419, 461, 496, 659, 694, 736, 769, 881, 1\,079, 1\,121, 1\,154\bigr\rbrack$

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<notatio-cell value="PowerModList(1, 1 / 2, 15)" data-allow-mismatch="children" /> <notatio-cell value="PowerModList(1, 1 / 2, 1155)" data-allow-mismatch="children" />

The rows of that second list are the sign vectors: 1 is $(+,+,+,+)$, 1154 is $(-,-,-,-)$, and the fourteen in between are the mixed ones. Reading a residue as a vector of signs is the whole content of the CRT here.

The 2-adic channel is the exception

The clean law is stated for odd m because 2 misbehaves, and it misbehaves in a way worth seeing: $\mathbb{Z}/2$ has one root of 1, $\mathbb{Z}/4$ has two, and $\mathbb{Z}/2^a$ for $a ≥ 3$ has four — namely $± 1$ and $± 1 + 2^{a-1}$. The unit group stops being cyclic at 8, so it acquires a second independent element of order 2.

#Powers of two break the count

In$\mathrm{PowerModList}(1, \frac{1}{2}, 2)$Out$\bigl\lbrack1\bigr\rbrack$In$\mathrm{PowerModList}(1, \frac{1}{2}, 4)$Out$\bigl\lbrack1, 3\bigr\rbrack$In$\mathrm{PowerModList}(1, \frac{1}{2}, 8)$Out$\bigl\lbrack1, 3, 5, 7\bigr\rbrack$In$\mathrm{PowerModList}(1, \frac{1}{2}, 16)$Out$\bigl\lbrack1, 7, 9, 15\bigr\rbrack$

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<notatio-cell value="PowerModList(1, 1 / 2, 2)" data-allow-mismatch="children" /> <notatio-cell value="PowerModList(1, 1 / 2, 4)" data-allow-mismatch="children" /> <notatio-cell value="PowerModList(1, 1 / 2, 8)" data-allow-mismatch="children" /> <notatio-cell value="PowerModList(1, 1 / 2, 16)" data-allow-mismatch="children" />

Imaginary units are rarer, and local

Now ask the same question for $x^2 = -1$. Again it decomposes channel by channel, but now most channels have nothing to offer: $\mathbb{Z}/p$ has a square root of $-1$ iff$p ≡ 1 \pmod 4$. So

$$x^2 ≡ -1 \pmod m \text{ is solvable} \iff \text{every odd prime factor of } m ≡ 1 \pmod 4, \text{ and } 4 \nmid m.$$

Half the primes qualify, by Dirichlet. This is the asymmetry that makes split units the generic case and imaginary units special — and it is the same asymmetry that makes $\mathbb{R}[j]/(j^2-1) ≅ \mathbb{R} × \mathbb{R}$ split while $\mathbb{R}[i]/(i^2+1) ≅ \mathbb{C}$ does not.

#Where √−1 lives

65 = 5·13, both ≡ 1 (mod 4), so two channels each contributing two roots. 15 = 3·5 has none: 3 ≡ 3 (mod 4) kills it.

In$\mathrm{PowerModList}(-1, \frac{1}{2}, 5)$Out$\bigl\lbrack2, 3\bigr\rbrack$In$\mathrm{PowerModList}(-1, \frac{1}{2}, 65)$Out$\bigl\lbrack8, 18, 47, 57\bigr\rbrack$In$\mathrm{PowerModList}(-1, \frac{1}{2}, 15)$Out$\bigl\lbrack\bigr\rbrack$

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<notatio-cell value="PowerModList(-1, 1 / 2, 5)" data-allow-mismatch="children" /> <notatio-cell value="PowerModList(-1, 1 / 2, 65)" data-allow-mismatch="children" /> <notatio-cell value="PowerModList(-1, 1 / 2, 15)" data-allow-mismatch="children" />

Hensel lifting up a prime power

Finding a root mod $p$ is a search; getting it mod $p^a$ is Hensel's lemma. If $r^2 ≡ -1 \pmod{p^k}$ then the correction

$$r' = r - \frac{r^2+1}{2r} \pmod{p^{k+1}}$$

is a root one power higher, and $2r$ is invertible because $p$ is odd and $r \not≡ 0$. Each step is Newton's method on $f(x) = x^2+1$ over the p-adics, and it converges for exactly the reason Newton's method does — which is why a root mod $p$ is enough to give a root mod $p^a$ for every $a$, and in the limit a root in $\mathbb{Z}_p$.

Both of these are one head: Wolfram's PowerModList(a, 1/r, m), the $r$-th roots of $a$ mod $m$. The split units are its $(1, 1/2)$ case and the imaginary units its $(-1, 1/2)$ case, so there is no bespoke head to learn — and the same call reaches past squares.

#One head, any root

Three cube roots of 8 mod 13, because 3 divides 12.

In$\mathrm{PowerModList}(8, \frac{1}{3}, 13)$Out$\bigl\lbrack2, 5, 6\bigr\rbrack$

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<notatio-cell value="PowerModList(8, 1 / 3, 13)" data-allow-mismatch="children" />

#2² ≡ −1 (mod 5), lifted

7² = 49 = 50 − 1, so 7 is the lift of 2 to ℤ/25; 182 is the lift to ℤ/625.

In$\mathrm{PowerModList}(-1, \frac{1}{2}, 5)$Out$\bigl\lbrack2, 3\bigr\rbrack$In$\mathrm{PowerModList}(-1, \frac{1}{2}, 25)$Out$\bigl\lbrack7, 18\bigr\rbrack$In$\mathrm{PowerModList}(-1, \frac{1}{2}, 625)$Out$\bigl\lbrack182, 443\bigr\rbrack$

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<notatio-cell value="PowerModList(-1, 1 / 2, 5)" data-allow-mismatch="children" /> <notatio-cell value="PowerModList(-1, 1 / 2, 25)" data-allow-mismatch="children" /> <notatio-cell value="PowerModList(-1, 1 / 2, 625)" data-allow-mismatch="children" />

The identities transport

A split unit is not just a curiosity of ℤ/m: it is a homomorphism. If $x^2 = 1$ in ℤ/m then

$$\mathbb{R}[j]/(j^2-1) \longrightarrow \mathbb{Z}/m, \qquad j ↦ x$$

respects addition and multiplication, so every identity the symbolic algebra proves becomes a fact about ℤ/m. The zero divisor and the idempotent both survive the trip: $(1+j)(1-j) = 0$ becomes $(1+4)(1-4) = -15 ≡ 0$, and $\tfrac{1+j}{2}$ — an idempotent because $j^2 = 1$ — lands on $5 \cdot 8 = 40 ≡ 10$, with $10^2 = 100 ≡ 10$.

#The same statement, twice

Left: the symbolic identity. Right: its image under j ↦ 4 in ℤ/15.

In$(1+j_1)(1-j_1)$Out$0$In$(1+4)(1-4)\bmod15$Out$0$In$(\frac{1+j_1}{2})^2$Out$\frac{j_1}{2}+\frac{1}{2}$In$10×10\bmod15$Out$10$

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<notatio-cell value="(1 + j_1) * (1 - j_1)" data-allow-mismatch="children" /> <notatio-cell value="(1 + 4) * (1 - 4) % 15" data-allow-mismatch="children" /> <notatio-cell value="((1 + j_1) / 2) ^ 2" data-allow-mismatch="children" /> <notatio-cell value="10 * 10 % 15" data-allow-mismatch="children" />

This is why the idempotents matter. The split units, the CRT sign vectors and the orthogonal idempotents $\tfrac{1 ± j}{2}$ are three descriptions of one thing: a decomposition of the ring into independent channels.

Split, inert, degenerate — the square as a local invariant

Step back and the three columns of the grid stop looking like three arbitrary choices. Over a field, the quadratic algebra $K[x]/(x^2-d)$ has discriminant $4d$, and exactly three things can happen:

$d$	algebra	unit
a square in $K$	$K × K$ — splits	j
a non-square	a quadratic field	i
zero	$K[\varepsilon]$ — degenerate, the two roots collide	\epsilon

So j, i and ε are not three unrelated inventions: they are the three outcomes of one construction, selected by whether $d$ is a square, a non-square, or zero. And "is $d$ a square" is a local question — it has a different answer at each place.

That is what the mod-m results above are really saying. Take $d = -1$:

• at the real place, $-1$ is not a square, so $\mathbb{R}[x]/(x^2+1) = \mathbb{C}$ — inert;
• at $p ≡ 1 \pmod 4$, $-1$ is a square in $\mathbb{Z}_p$, so the algebra splits;
• at $p ≡ 3 \pmod 4$ it is not, and the algebra is the unramified quadratic extension of $\mathbb{Q}_p$;
• at $p = 2$ it is the awkward case, as it was in the counting above.

Whereas $d = +1$ splits at every place, which is why split units are everywhere and $\mathbb{R}[j]/(j^2-1) ≅ \mathbb{R}^2$ already at the real place. A hypercomplex unit's character is not a property of the unit — it is a property of the unit at a place.

Where this is heading: profinite and adelic

The rings $\mathbb{Z}/m$ are not separate objects. They form a system under divisibility — $\mathbb{Z}/m → \mathbb{Z}/n$ whenever $n \mid m$ — and the inverse limit of that system is the profinite integers

$$\widehat{\mathbb{Z}} \;=\; \varprojlim_m \mathbb{Z}/m \;≅\; \prod_p \mathbb{Z}_p,$$

so the CRT splitting at each finite level is the shadow of one product over all primes. Under that identification the split units assemble too: the elements of order dividing 2 in $\widehat{\mathbb{Z}}^×$ are a sign per prime, $\prod_p \{± 1\}$ — a Cantor-set of split units, of which the $2^{ω(m)}$ residues above are the finite truncations. Hensel lifting is the statement that a root mod $p$ already determines its coherent lift into $\mathbb{Z}_p$, so "which units exist" stabilises per place.

Adjoining the archimedean place gives the adeles, $\mathbb{A}_{\mathbb{Q}} = \mathbb{R} × {\prod_p}' \mathbb{Q}_p$, and the split/inert/degenerate trichotomy above is then a statement about all places at once: a quadratic algebra over $\mathbb{Q}$ has a local character at each place, and those characters are not independent — quadratic reciprocity is precisely a constraint tying them together, so the local behaviours have to multiply out consistently. Which means the ± sign vectors that index the split units of ℤ/m are a fragment of something global, not just bookkeeping per modulus.

That is the direction, stated honestly as a direction: this page demonstrates the finite levels and the per-place criterion, and nothing here computes an adelic object. The pieces that would make it concrete — a Norm that names its ambient level, coefficients that live in ℤ/m rather than only mapping into it, and a per-place signature rather than one global square — are all still to build.