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.

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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*}

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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:


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.

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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:

  1. 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.
  2. 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.
  3. 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.

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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.

Thus we have a ring structure on \left(\textbf{End}_{K}(V),+,\circ\right).

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Uniqueness of identities and inverses

Today is a unique day in our jurisdiction, and we’ve also chosen to celebrate it in a unique way.

Identity element

If a (commutative) ring R has a multiplicative identity, then the multiplicative identity is unique.

The emphasis on multiplicative is to distinguish it from the additive identity of the underlying abelian group (R,+).

As is customary, suppose there are two multiplicative identity elements, namely e and e'.

Then we have e=ee'=e', by using the fact that both e and e' are identities in turn.

That’s it! The simplest proof in the whole of abstract algebra.


Let R be a ring with identity. If an element x\in R has a left inverse l and a right inverse r, then l=r.

Let the (multiplicative) identity be 1.

Since l is a left inverse of x, we have: lx=1.

Since r is a right inverse of x, we have: xr=1.

Thus l=l1=l(xr)=(lx)r=1r=r.

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