Uncategorized – Commutative

August 20, 2020August 20, 2020

A strange zero divisor

Let $S=\{a,b,c\}$ be a three element set. Equip its power set $\mathcal{P}(S)$ with two operations $+$ and $.$ defined as: $A+B=(A-B)\cup(B-A)$ and $A.B=A\cap B$ , for $A,B\in\mathcal{P}(S)$ . Then $\left(\mathcal{P}(S),+,.\right)$ is a ring. The zero-divisors in this ring are the singleton subsets and the two element subsets.

Recall that an element $a$ in a ring is a zero-divisor if $a\neq 0$ and there exists another $b\neq 0$ for which $a.b=0$ (technically, a left zero-divisor).

The zero element in $\left(\mathcal{P}(S),+,.\right)$ is obviously the empty set $\varnothing$ . In addition, any singleton subset has an empty intersection with another singleton subset: $\{a\}\cap\{b\}=\varnothing$ (we’ve assumed implicitly that $a\neq b$ ).

Consider a two element subset, e.g. $\{a,b\}$ . We have $\{a,b\}\cap\{c\}=\varnothing$ . Similarly, $\{b,c\}\cap\{a\}=\varnothing$ and $\{a,c\}\cap\{b\}=\varnothing$ . So the two element subsets are also zero-divisors.

The entire set itself, namely $\{a,b,c\}$ is not a zero-divisor.

What is strange about these zero divisors as the title suggests? Nothing.

This concludes today’s snippet. Thanks for visiting!

July 17, 2020July 18, 2020

Difference of squares in a ring

Let $a,b\in R$ , where $R$ is a ring. Today’s snippet focuses on the expression $a^2-b^2$ , which may not always factor as $(a-b)(a+b)$ .

$a^2-b^2=(a-b)(a+b)$ if and only if the ring is commutative

Suppose that $a^2-b^2=(a-b)(a+b)$ . Expand the right side using the distributive property of the ring:

$\begin{equation*} \begin{split} (a-b)(a+b)&=a^2+ab-ba-b^2\\ a^2-b^2&=a^2+ab-ba-b^2\\ \therefore 0&=ab-ba\\ ab&=ba \end{split} \end{equation*}$

Thus, the ring is commutative under multiplication.

Conversely, suppose that for all $a,b\in R$ we have $ab=ba$ . Then:

$\begin{equation*} \begin{split} a^2-b^2&=a^2+ab-ab-b^2\\ &=a^2+ab-ba-b^2\\ &=a(a+b)-b(a+b)\\ &=(a-b)(a+b) \end{split} \end{equation*}$

This concludes today’s snippet. Thanks for visiting!

July 10, 2020July 18, 2020

When a nilpotent element plus a unit is a unit

In a ring, recall that a nilpotent element is an element $a$ such that $a^n=0$ for some positive integer $n$ . Today’s snippet focuses on adding a nilpotent element and a unit to produce a unit.

Unit from a nilpotent element and a unit

Let $a,b\in R$ , where $R$ is a commutative ring with identity. If $a$ is a unit and $b^2=0$ , then $a+b$ is a unit.

We have to produce the inverse of $a+b$ .

Let’s get a motivation from real numbers, where the inverse of $a+b$ is $\frac{1}{a+b}$ . Write as:

$\left[a\left(1+\frac{b}{a}\right)\right]^{-1}=a^{-1}\left(1+\frac{b}{a}\right)^{-1}=a^{-1}\left(1+a^{-1}b\right)^{-1}$

Let $u$ be the inverse of $a$ . We’ll utilize the fact that the ring $R$ is commutative to prove that the inverse of $a+b$ is $u-u^2b$ . Indeed:

$\begin{equation*} \begin{split} (a+b)(u-u^2b)&=au-a(u^2b)+bu-b(u^2b)\\ &=au-(au)(ub)+bu-bu(ub)\\ &=au-ub+bu-(ub)(bu)\\ &=au-ub+ub-u(0)u\\ &=1 \end{split} \end{equation*}$

Similarly, $(u-u^2b)(a+b)=1$ . Therefore $a+b$ is a unit.

An example from $\mathbb{Z}_{4}=\{0,1,2,3\}$

In $\mathbb{Z}_{4}$ , we have $2.2=0$ , so $2$ is nilpotent. Also $1.1=1$ , so $1$ is (trivially) a unit.

Let $a=1$ and $b=2$ . Since the inverse of $1$ is $1$ , we take $u=1$ . All terms as in the above proof.

The inverse of $a+b$ is $u-u^2b$ . In this case, it is $1-1^2.2=1-2=-1$ . But in $\mathbb{Z}_{4}$ , $-1$ is the “same” as $3$ . So, the inverse of $a+b=1+2=3$ is also $3$ .

This concludes today’s snippet. Thanks for visiting!

July 7, 2020July 9, 2020

Group of units in a ring

Today’s snippet focuses on the set $U(R)$ , consisting of all the units of $R$ . Here, $R$ is a commutative ring with identity $1$ .

$U(R)$ is a group under multiplication

Since $1.1=1$ , it follows that $1\in U(R)$ , and so the set $U(R)$ is non-empty. We now show that $\left(U(R),.\right)$ is a group.

identity under multiplication belongs to $U(R)$ .
associativity of multiplication is inherited from the ring structure.
closure under multiplication: Let $x,y\in U(R)$ . Then both $x$ and $y$ are units. Let their inverses be $x^{-1}$ and $y^{-1}$ , respectively. Now, $xy$ is a unit because we can produce its inverse, namely $y^{-1}x^{-1}$ :
$\begin{equation*} \begin{split} xy(y^{-1}x^{-1})&=x(yy^{-1})x^{-1}\\ &=x(1)x^{-1}\\ &=1\\ y^{-1}x^{-1}(xy)&=y^{-1}(x^{-1}x)y\\ &=y^{-1}(1)y\\ &=1 \end{split} \end{equation*}$

In other words, $x,y\in U(R)\implies xy\in U(R)$ .
Finally, if $x\in U(R)$ , then so is $x^{-1}$ . Each element thus has a multiplicative inverse.

Three examples from the $\mathbb{Z}_{n}$ rings:

$U(\mathbb{Z}_{4})=\{1,3\}$ . In particular, $1.1=1$ and $3.3=1$ , so each element of $U(\mathbb{Z}_{4})$ is its own inverse.
$U(\mathbb{Z}_{5})=\{1,2,3,4\}$ . Here, $1.1=1,2.3=3.2=1,4.4=1$ . Every non-zero element of $\mathbb{Z}_{5}$ is a unit. Being already a commutative ring with identity, $(\mathbb{Z}_{5},+,.)$ is a field.
$U(\mathbb{Z}_{6})=\{1,5\}$ . Here also, each element of $U(\mathbb{Z}_{6})$ is its own inverse, since $1.1=1$ and $5.5=1$ .

This concludes today’s snippet. Thanks for visiting!

July 3, 2020July 3, 2020

Examples of rings — endomorphism rings

Today’s snippet focuses on the set $\textbf{End}_{K}(V)$ — the set of linear maps $V\rightarrow V$ . It is also denoted by $\textbf{Hom}_{K}(V,V)$ . Here, $V$ is a vector space over a field $K$ . In reality, the full power of the vector space structure of $V$ is not needed; any abelian group will do.

The set $\textbf{End}_{K}(V)$ is a ring

For $S,T\in\textbf{End}_{K}(V)$ , define $(S+T)(v)=S(v)+T(v)$ (point-wise addition) and $(ST)(v)=(S\circ T)(v)=S(T(v))$ (composition). Then $\textbf{End}_{K}(V)$ endowed with these two operations is a ring.

(The set $\textbf{End}_{K}(V)$ is non-empty; it contains at least the identity linear transformation $1:V\rightarrow V$ that sends each element of $V$ to itself. Thus it already possesses a multiplicative identity.)

We first show that $\left(\textbf{End}_{K}(V),+\right)$ is an abelian group. Readily follows from the vector space structure of $V$ , and basic facts about linear maps.

closure under addition — adding two linear maps gives a linear map; so if $S,T\in\textbf{End}_{K}(V)$ , then $S+T\in\textbf{End}_{K}(V)$ .
it contains a zero element; this corresponds to the linear transformation $0_V:V\rightarrow V$ that sends every $v\in V$ to $0\in V$ .
addition is associative — this is inherited from the vector space structure of $V$ . Indeed, if $S,T,U\in \textbf{End}_{K}(V)$ and $v\in V$ , then
$\begin{equation*}\begin{split}(S+(T+U))(v)&=S(v)+(T+U)(v)\\ &=S(v)+T(v)+U(v)\\ &=(S(v)+T(v))+U(v)\\ &=(S+T)(v)+U(v)\\ &=((S+T)+U)(v)\\ \therefore S+(T+U)&=(S+T)+U \end{split} \end{equation*}$
addition is commutative — again inherited from the vector space $V$
for any $S\in\textbf{End}_{K}(V)$ , there is an additive inverse $-S\in\textbf{End}_{K}(V)$ such that $S+(-S)=0_{V}$ , the zero linear transformation. This $-S$ will be defined as follows: for any $v\in V$ , $(-S)(v)=-S(v)$ .

Next, the set $\left(\textbf{End}_{K}(V),\circ\right)$ satisfies:

composition is associative — always, even if the maps involved are not linear maps
composition distributes over addition — let $S,T,U\in\textbf{End}_{K}(V)$ . For any $v\in V$ we have:
$\begin{equation*}\begin{split}(S\circ(T+U))(v)&=S((T+U)(v))\\ &=S(T(v)+U(v))\\ &=S(T(v))+S(U(v))\\ &=(S\circ T)(v)+(S\circ U)(v)\\ \therefore S\circ (T+U)&=S\circ T+S\circ U\end{split}\end{equation}$

Similarly, composition also distributes over addition from the right.