Congruence

2020 James B. Wilson

Colorado State University

 

Homomorphisms

Quotients

Algebra is the study of equations, e.g. 

\[x^2+1=0\]

Abstract algebra is when we see every symbol in an equation as a variable, including +, and =.

 

We are engaged in replacing "=".

Theme

Objective

  • We found congruence is what can replace "=".
  • Homomorphisms give congruences.
  • We now define Quotients to produce homomorphisms.

Problem

Want an equivalence but don't have a function.

Consider the following relation of positive integers

\[(a,b)\equiv (c,d) \Longleftrightarrow ad=bc\]

Recognize this is \(a/b=c/d\) but written without fractions to focus on the problem.

Yes: simply "reduce the fraction", e.g. \((2,4)\mapsto (1,2)\); \((21,27)\mapsto (7,9)\).

Is this the equivalence given by a function, ie.:

\[x\equiv y \pmod{f}\Longleftrightarrow f(x)=f(y)?\]

Consider \[\mathbb{Z}[\sqrt{-5}]=\{a+b\sqrt{-5}\mid a,b\in \mathbb{Z}\}\]

 

Can you "reduce" \(\frac{9}{3}\)?

Easy: \[\frac{9}{3}= \frac{3\cdot 3}{3\cdot 1}=\frac{3}{1}.\]

Not so easy but equally true:

\[\frac{9}{3}=\frac{(4+5)}{3}=\frac{2^2-\sqrt{-5}^2}{3}=\frac{(2+\sqrt{-5})(2-\sqrt{-5})}{3}\]

which does not reduce.

Fact: \(\mathbb{Z}[\sqrt{-5}]\) does not have unique factorization.

Consequence

Some "equals" do not seem to be  based on functions, e.g. fractions over \(\mathbb{Z}[\sqrt{-5}]\) do not have unique reduced forms.

Equivalence

Function

Partition

Recall Disjoint Unions

Given a function \(P:I\to type\)

\[\bigsqcup P:=\bigsqcup_{i:I} P_i.\]

More specifically you have intro rules to give you \[\iota_i:P_i\to \bigsqcup P\] and elimination rule to give you \[\sqcup f:\bigsqcup P\to R\] given \((i:I)\mapsto (f_i:P_i\to R)\) and such that

\[\begin{array}{ccl} P_i &\overset{\iota_i}{\longrightarrow} & \bigsqcup P \\ & {_{f_i}\searrow} & \downarrow \sqcup f\\ & & R\end{array}\]

Partition

A partition of a type (or set) \(A\) is an invertible pair of functions

\[f:A\to \bigsqcup P\qquad g:\bigsqcup P\to A.\]

together with a evidence for \(f\circ g=1\) and \(g\circ f=1\), (see bottom slide for details.)

 

The classes are \(g_i:P_i\to A\) where \[\begin{array}{ccl} X & \overset{\iota_i}{\longrightarrow} & \bigsqcup P\\ & g_i \searrow & \downarrow g\\ & & A\end{array}\]

\(f\)

\(g\)

\(A\)

\(\bigsqcup_{i:I}P_i\)

\(A=\bigsqcup_{X:P} X\)

click box

Informally we "identify" these types hiding \((f,g)\) in notation

Actually a new breed of Type Theory called Homotopy Type Theory (HoTT) can formally identify these.

Treat \(a:X_i\) as \(a:A\) where \(a=g(x_i)\) for some \(x_i:P_i\).

So we can write \(A=\bigcup_{X:P}X\), though "=" here means to use \(f,g\) under the covers to translate.

After "identifying" get an alternative Perspective

A partition of a set \(A\) is a set \(\mathcal{P}\) of subsets where

  • \(A=\bigcup_{X\in P} X\)
  • For \(X,Y\in \mathcal{P}\), If \(X\cap Y\neq \emptyset\) then \(X=Y\).

Claim. Every partition \((f,g):A\leftrightarrow \bigsqcup_{i:I} P_i\) of \(A\) determines a function \(\varphi:A\to I\) where

\[a\equiv a'\pmod{\varphi}\Leftrightarrow f(a),f(a'):P_{\varphi(a)}.\]

 

Proof.  For \(i:I\) set \(\gamma_i:P_i\to I\) as the constant function

\[\gamma_i:(x:P_i)\mapsto (i:I).\]

By the elimination rule for \(\sqcup P\), there is a function \(\gamma:\bigsqcup P\to I\) where \[\gamma(\iota_i(x)) = \gamma_i(x)=i.\]

Define \(\varphi:A\to P\) as \(x\mapsto \gamma(f(x))\) (where \(f:A\to \bigsqcup P\) is given above.)

 

Assignment: show the property of \(a\equiv a'\pmod{\varphi}\)\(\Box\)

Consequence

  • Every partition \(A\leftrightarrow \bigsqcup_{i:I}P_i\) of produces a function \(\varphi:A\to I\).

  • Every function \(\varphi:A\to I\) produces an equivalence.

  • So every partition produces an equivalence.

Congruence

Homomorphism

Quotient

Now add algebra!

The Idea

Fix a set (or type) \(A\) with an operation, say \(*:A^2\to A\).

 

A quotient of \(\langle A,*\rangle\) is a partition such that the operation \(*\) applies on the class of the partition.  We rename the classes cosets!

Technical Definition

Fix a set (or type) \(A\) with an operation, say \(*:A^2\to A\).

 

A quotient of \(\langle A,*\rangle\) is a partition \((f,g):A\leftrightarrow \bigsqcup_{i:I}P_i\), with an operation \(\#:I^2\to I\) where

\[x:P_i, y: P_j\Rightarrow f(g(x)*g(y)): P_{i\# j}.\]

We write this condition as \[P_i*P_j \subset P_{i\# j}.\]

 

In a quotient \(A\leftrightarrow \bigsqcup P\), the \(P_i\) are called cosets.

Shorthand Definition

Fix a set (or type) \(A\) with an operation, say \(*:A^2\to A\), identified with a quotient \(\bigsqcup_{X:P} X\) then for \(X,Y\in P\) 

\[x:X, y:Y\Rightarrow x*y:  X*Y.\]

If it exist it is unique

 

Recall \((f,g):A\leftrightarrow \bigsqcup_{i:I}P_i\) produces a map \(\varphi:A\to I\) where

\[a:A \Rightarrow f(a):P_{\varphi(a)}\]

Hence, \(P_i*P_j \subset P_{i\#j}\) means \[i\#j = \varphi(f(g(x)*g(y))).\]

Example.  \(\langle \mathbb{Z},+\rangle\) as a quotient of \(\langle \mathbb{N}^2,+\rangle\).

 

Define

\[m-n:=\{(s,t)\in\mathbb{N}^2\mid m+t=n+s\}\]

\[\mathbb{Z}:=\{m-n\mid m,n\in\mathbb{N}^2\}.\]

 

Here \(m-n\) is a coset, like fractions \(m/n\), even though it is tempting to think it is substraction now, wait to do so.

Define

\[m-n:=\{(s,t)\in\mathbb{N}^2\mid m+t=n+s\}\]

\[\mathbb{Z}:=\{m-n\mid m,n\in\mathbb{N}^2\}.\]

 

Claim 1. \(\mathbb{Z}\) is a partition of \(\mathbb{N}^2\).

Proof. Assignment.

Define \(m-n:=\{(s,t)\in\mathbb{N}^2\mid m+t=n+s\}\), 

\(\mathbb{Z}:=\{m-n\mid m,n\in\mathbb{N}^2\}.\)

 

Claim 1. \(\mathbb{Z}\) is a quotient of \(\langle\mathbb{N}^2,+\rangle\) where

\[(m,n)+(s,t):=(m+s,n+t)\]

is the addition in \(\mathbb{N}^2\).

Proof. Fix \((s,t)\in (m-n)\) and \((s',t')\in (m'-n')\).

\[(s,t)+(s',t')=(s+s',t+t')\]

and \[\begin{aligned}(m+m')+(t+t') & = (m+t)+(m'+t')\\ & = (n+s)+(n'+s')\\ & = (n+n')+(s+s')\end{aligned}\]

Hence, \((m-n)+(m'-n'):=(m+m')-(n+n')\) contains \((s+s',t+t')\) as required.  So \(\mathbb{Z}\) is a quotient. \(\Box\)

In General

  • A quotient must respect not only binary operators but all operators of an algebra.
  • We abbreviate this writing \[[P_{i_1},\ldots,P_{i_n}]\subset P_{\langle i_1,\ldots,i_n\rangle}\]
  • The product on the indices \(I\) becomes a new smaller algebra with compatible operations.
  • The \(P_i\) are called cosets. 

Conclusion

  • A quotient is an \(\sigma\)-algebraic structure \(A\) with a partition \(Q\) that respect all operators.
  • Quotients are \(\sigma\)-algebraic structures themselves.
  • There is always a homomorphism \(A\to Q\) from an algebra to its quotients.
  • Quotients are third source for new congruences.

Quotients

By James Wilson

Quotients

Definitions and properties of quotients

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