The "modules over Gaussian integers" law

Stream: Equational

[Bruno Le Floch]

The Tarski law 543 characterizes subtraction in abelian groups, aka $\mathbb{Z}$-modules. The law 27213516127903475784245028046357, of order 24 and with four variables EDIT: see below for an equivalent law of order 6, characterizes the operation $x\diamond y = x+\sqrt{-1}y$ on modules over the ring $\mathbb{Z}[\sqrt{-1}]$ of Gaussian integers.

Proof outline.

Specializing to w=z the law expresses that the auxiliary operation x-y = x◇(y◇(y◇y)) obeys law 1774 x = (y-z)-((y-x)-z) hence is abelian subtraction. Hence, left and right multiplications, and cubing C:x↦x◇(x◇x), are all bijective. The variable w appears exactly twice in the law, as w◇(z◇w), so this combination is independent of w. Various specializations involving properties of subtraction allow us to show that x↦0◇x is a group morphism that squares to minus the identity, which concludes.

Details

The law 27213516127903475784245028046357 reads x = (y◇(w◇(z◇w)))◇C((y◇Cx)◇Cz) with Cy=y◇(y◇y). It has exactly two occurrences of w. As mentioned above, when specialized to w=z it reduces to law 27213516127895692600464340196909, which states that x-y = x◇Cy obeys law 1774. There is a distinguished element 0 and all the usual properties of subtraction hold. In particular,

• for each $x$ the map $y \mapsto x-y = L_x(Cy)$ is bijective, so C is injective and left multiplication is surjective;

• for each $y$ the map $x \mapsto x-y = R_{Cy}x$ is bijective, namely right multiplication by a cube is bijective.

The equation takes the form $x = (y \diamond (w \diamond (z \diamond w))) \diamond C(\dots)$ so this last bijectivity property implies that y◇(w◇(z◇w)) is independent of w. It is equal to its value at w=z, namely y◇(w◇(z◇w))=y-z. Now take w=Cv and use z◇Cv=z-v to write y-z = y◇(Cv◇(z-v)). On the left-hand side z↦y-z is bijective, and on the right-hand side we have the composition of the bijective z↦z-v and of two left multiplications. These two left multiplications are known to be surjective, and must compose to a bijection, so they are both bijective. We thus learn that left multiplication is bijective. Next, since we know $L_x\circ C$ is bijective, we learn that $C$ is bijective. Every element is a cube, so right-multiplication by anything is bijective. The magma equipped with ◇ is a quasigroup.

Next, we focus on 0. We have y = y-0 = y◇C0 so C0 is a right unit, and it is thus idempotent, which implies that it cubes to itself, CC0=C0. Since C is a bijective map, we get C0=0. The previous observation becomes that 0 is a right unit. Inserting zero in a different way in y◇(w◇(z◇w))=y-z we find 0◇(0◇z) = 0◇(0◇(z◇0)) = 0-z = -z.

By left-cancellativity, y◇(w◇(z◇w)) = y-z = y◇Cz implies w◇(z◇w)=Cz for all w. In particular Cz = 0◇(z◇0) = 0◇z, so CCz=0◇(0◇z)=-z. We then easily have x◇y = x◇C(-Cy) = x-(-Cy) = x+Cy. There only remains to show that C is a group morphism, which is equivalent to C(x-y)=Cx-Cy.

Write w◇(z◇w)=Cz in terms of addition and C to get w+C(z+Cw) = Cz, so C(z+Cw) = Cz - w. Setting w=Cv we get C(z-v) = Cz - Cv as desired.

[Bruno Le Floch]

There is a similar single-operation description of $\mathbb{Z}[1/n]$ modules for many values of $n$ (but not primes I think) by a single operation of the form $x \diamond y = a_n x + b_n y$ with well-chosen $(a_n, b_n)$. I didn't work out the single law that this operation has to obey. The coefficients $(a_n,b_n)$ have to be such that the sub-magma of $\mathbb{Z}[1/n]$ generated by $1$ under the operation $\diamond$ is all of $\mathbb{Z}[1/n]$.

Examples include $n=k(k+1)$ with $a_n=(k+1)/k$ and $b_n=-1/(k(k+1))$, but also $(a,b)=(3/2,-1/3)$, $(a,b)=(8/5,-1/10)$ and, I think, $(a,b)=(27/25,-1/15)$. Of course, $\mathbb{Z}[1/n]$ only depends on the square-free part of $n$. EDIT: the product of primes dividing $n$.

[Kevin Buzzard]

Normally the "square-free part of n" refers to "the unique squarefree t such that n can be written as m^2t" but here you mean "the product of the primes dividing n".

[Bruno Le Floch]

Thanks, indeed I messed up the terminology.

[Kevin Buzzard]

Are you casually asserting in your message that (for example) every element of $\Z[1/6]$ is in the smallest set containing 1 and satisfying that if $x,y$ are in the set then so is $3x/2-y/3$? Why is this true? I think I'd be in better shape if I could prove that this set contained -1 (then at least it would be closed under negation) and was closed under addition (then it would contain 0 and we'd be in much better shape), or am I going about this in the wrong way?

[Bruno Le Floch]

Yes. According to some mathematica code, for x◇y=3x/2-y/3 one has

• (((x◇x)◇((x◇x)◇(x◇x)))◇(x◇x))◇x = 2x
• (x◇(((((x◇x)◇(x◇((x◇x)◇x)))◇x)◇x)◇x))◇x = x/3
• ((x◇(x◇x))◇((x◇(x◇x))◇x))◇x = 3x/2

so iterating these a couple of times we can get any $2^k 3^l x$. Then (2x/9)◇x=0, so we now have 0 available and we get 0◇3y=-y.

But I realized there is a much simpler choice of operation for all $\mathbb{Z}[1/n]$: just take $x\diamond y = x-y/n$. Then $(((x\diamond x)\diamond x)\diamond \dots)\diamond x = x-kx/n$ for any $k$, and in particular we can get $-x/n$ and $-px$ for any integer $p$ by suitable choices of $k$. Nesting these gives access to any element of $\mathbb{Z}[1/n] x$.

More generally the operation $x+\alpha y$ seems to work for $\mathbb{Z}[1/\alpha]$ for any root $\alpha$ of a monic polynomial with non-negative coefficients, but it is not clear how general those rings are. And using non-trivial coefficients for both $x$ and $y$ may allow more rings.

[Kevin Buzzard]

Ha ha no wonder I couldn't reconstruct the arguments :-) Nice!

[Bruno Le Floch]

Actually, the much simpler order-6 law 86082 x = y ◇ (z ◇ ((y ◇ z) ◇ (w ◇ (x ◇ w)))) also characterizes $\mathbb{Z}[i]$-modules (with x◇y=x+iy)! Other order 6 laws also work.

Full list

Laws that are equivalent: 86082, 86160, 125935, 127386, 127464, 127560, 128214, 184602, 185406, 185442, 186339, 251586, 251959, 252414, 252439, 444561, 446262, 446340.
Laws that are equivalent at least in the finite case: 86046, 86064, 444507, 446226, 446244.
Laws for which I know nothing: 127759, 183687.

[Bruno Le Floch]

Here is a proof that law 86082 characterizes $\mathbb{Z}[i]$-modules: 86082.pdf

Proof in LaTeX format for reference

\documentclass{article}
\usepackage{amsmath,amsfonts,amsthm,mathtools}
\newcommand{\ZZ}{\mathbb{Z}}
\newcommand{\op}{\diamond}
\newcommand{\id}{\mathrm{id}}
\newtheorem{lemma}{Lemma}
\begin{document}
The order-6 law 86082 $x = y \op (z \op ((y \op z) \op (w \op (x \op w))))$ characterizes the operation $x \op y = x+iy$ in $\ZZ[i]$-modules (with $i^2=-1$).  Other order-6 laws also work.

Let $i_x(y) = L_x \circ R_x (y) = x \op (y \op x)$.
The law can be written as $L_y \circ L_z \circ L_{y \op z} \circ i_w = \id$.
We first prove that $i_v=i_w$ for all $v,w$.  We calculate
\begin{equation}\label{technical}
  \begin{aligned}
    (x \op y) \op i_z(u)
    & \overset{86082}{=} L_x \circ L_y \circ L_{x \op y} \circ i_{i_z(u)} \bigl((x \op y) \op i_z(u)\bigr)
    \\
    & = L_x \circ L_y \Bigl( L_{x \op y} \circ L_{i_z(u)} \circ L_{(x \op y) \op i_z(u)} \circ i_z(u) \Bigr)
      \overset{86082}{=} x \op (y \op u)
  \end{aligned}
\end{equation}
Using this identity $L_{x \op y}\circ i_z = L_x \circ L_y$ in the equation yields $L_y \circ L_z \circ L_y \circ L_z = \id$.
We learn in particular that left-multiplication is bijective.

Taking $u = x \op y$ in \eqref{technical} and further left-multiplying by $z$ yields
\begin{equation}\label{right-unit}
  z = (L_z\circ L_{x\op y})^2(z) = z \op (x \op (y \op (x \op y))) .
\end{equation}
Invertibility of $L_z$ means that
\begin{equation}\label{xyxy}
  x \op (y \op (x \op y)) = x \op i_y(x) \text{ is independent of $x$ and $y$.}
\end{equation}
Invertibility of $L_x$ allows us to deduce $i_y(x)$ is $y$-independent (but depends on~$x$).  We henceforth drop the subscript on the function~$i$.

Next, we define the operation $x - y = x \op i(y)$.  We shall show that the magma equipped with $-$ is an abelian group with subtraction and that $i$ is a group morphism squaring to negation.

The observation~\eqref{xyxy} states that $x-x = x \op i(x) = x \op (x \op (x \op x))$ is independent of~$x$.  We denote $0 = x - x$.  Observe that~\eqref{right-unit} becomes $z = z \op 0$ so that zero is a right-unit.  In particular,
\begin{equation}
i(x) = i_0(x) = 0 \op (x \op 0) = 0 \op x
\end{equation}
and $i(0)=0$.
Observe that $i(i(x)) = 0 \op i(x) = 0 - x$.  We call this operation negation and denote it simply by $-x$.

Taking $z=0$ in the identity $L_y \circ L_z \circ L_y \circ L_z = \id$ yields $y - (y - x) = x$.  In particular for $y=0$ we see that $-(-x)=x$, namely $i(i(i(i(x)))) = x$.

It is then natural to introduce addition as
\begin{equation}
  x+y = x - (-y) = x \op i(i(i(y))) = x \op (0 \op (0 \op (0 \op y))) .
\end{equation}
One has
\begin{equation}
  x+0 = x , \qquad 0+x = 0 \op i(i(i(x))) = i(i(i(i(x)))) = x .
\end{equation}
In terms of addition, $x - y = x - (-(-y)) = x + (-y)$, and the magma operation is expressed as $x \op y = x \op i(i(i(i(y)))) = x + i(y)$.

The technical identity~\eqref{technical} evaluated at $x=0$ tells us that $i$ is a magma morphism,
\begin{equation}
  i(y) \op i(u) = i(y \op u) ,
\end{equation}
and thus $i(x+y)=i(x)+i(y)$.
Its square, negation, also is, so that $-(x + y) = (-x) + (-y)$.


We now return to the main equation with $(y,z)$ replaced by $(-y,i(z))$, write it in terms of addition, and use that $i$ is a morphism, to get $x = -y + (-z + ((y + z) + x))$.  Using that $t - (t - u) = u$, and how subtraction and addition are related through the negation involution, we get
\begin{equation}
  z + (y + x) = (y + z) + x .
\end{equation}
For $x = 0$ this yields commutativity $z+y=y+z$.  Then, by swapping operands around, we get associativity $(x+y)+z = x+(y+z)$.  Together with what we know about negation and $0$, we deduce that $(M,+)$ is an abelian group.  It is equipped with a group automorphism $i$ with $i(i(x)) = -x$, which completes the description of $M$ as a $\mathbb{Z}[i]$-module.

\end{document}

[Bruno Le Floch]

Here are order-6 characterizations of the operation $x \diamond y = x + \omega y$ in $\mathbb{Z}[\omega]$ modules, where $\omega=e^{2\pi i/3}$.

Lists of law numbers

• Characterizations: [85914, 86034, 130757, 130783, 131577, 131603, 144232, 194058, 264007, 264067, 268121, 268194, 279823, 280567, 300583, 337710, 383143, 461983].
• Work in finite magmas, unknown generally: [143049, 384114, 465969].
• Other candidates: [64470, 85063, 101569, 102829, 134897, 136009, 144378, 159529, 160421, 188509, 188563, 188673, 188793, 192959, 279873, 280541, 300369, 383089, 384365, 461957, 462803, 463072, 465889, 466797, 467298, 478363, 478590].

[Bruno Le Floch]

Models of law 546 are affine spaces over $\mathbb{Z}[i]$-modules. Law 546 $x = y \diamond (z \diamond (x \diamond (z \diamond y)))$ is a close neighbor of Tarski's axiom in which we simply swap the last two variables. All of its models are linear and of the form $x \diamond y = - x + i y + \text{const}$ in a $\mathbb{Z}[i]$-module. The magma operation is enough to reconstruct the module structure up to a choice of reference element $u$: for instance $x-y = (x \diamond u) \diamond (u \diamond y)$.

[Bruno Le Floch]

Several other low-order laws only have linear models! (Here I won't list duals.)

• Law 556 characterizes the operation $x\diamond y=i(x+y)+\text{const}$, again in an affine version of $\mathbb{Z}[i]$-module, with the group subtraction being given by $x - y = (((x\diamond u)\diamond u)\diamond y)\diamond u$ for some fixed $u$.

• Law 898, a neighbor of the law 895 characterizing abelian groups of exponent 2, also only has linear models. Specifically, the operation $x + y = ((u \diamond x)\diamond (y\diamond u))\diamond u$ obeys 895. In particular, finite magma sizes are powers of 2. There is some more structure involving a seventh root of unity (but in characteristic 2).

• Law 2 is the trivial law

• Law 4 is the left projection
• Law 41 is the constant law
• Law 543 is Tarski's axiom for abelian group subtraction
• Law 895 is addition in (abelian) groups of exponent 2

I think that is it, but I haven't ruled out completely [467, 677, 704, 1076, 1083, 1110, 1117, 1279, 1286, 1313, 1516], in the sense that I didn't work out a nonlinear model for these.

EDIT: Actually, for 1117, finite ⇒ left-cancellative ⇒ quasigroup ⇒ linear, but it has many infinite models that are not left-cancellative. And 556 implies 1117 and left-cancellative, hence implies linearity (without finiteness assumption).