## A strange zero divisor

Let be a three element set. Equip its power set with two operations and defined as: and , for . Then is a ring. The zero-divisors in this ring are the singleton subsets and the two element subsets.

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

The zero element in is obviously the empty set . In addition, any singleton subset has an empty intersection with another singleton subset: (we’ve assumed implicitly that ).

Consider a two element subset, e.g. . We have . Similarly, and . So the two element subsets are also zero-divisors.

The entire set itself, namely 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!

## Difference of squares in a ring

Let , where is a ring. Today’s snippet focuses on the expression , which may not always factor as .

#### if and only if the ring is commutative

Suppose that . Expand the right side using the distributive property of the ring:

Thus, the ring is commutative under multiplication.

Conversely, suppose that for all we have . Then:

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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 such that for some positive integer . 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 , where is a commutative ring with identity. If is a unit and , then is a unit.

We have to produce the inverse of .

Let’s get a motivation from real numbers, where the inverse of is . Write as:

Let be the inverse of . We’ll utilize the fact that the ring is commutative to prove that the inverse of is . Indeed:

Similarly, . Therefore is a unit.

#### An example from

In , we have , so is nilpotent. Also , so is (trivially) a unit.

Let and . Since the inverse of is , we take . All terms as in the above proof.

The inverse of is . In this case, it is . But in , is the “same” as . So, the inverse of is also .

This concludes today’s snippet. Thanks for visiting!

## Group of units in a ring

Today’s snippet focuses on the set , consisting of all the units of . Here, is a commutative ring with identity .

#### is a group under multiplication

Since , it follows that , and so the set is non-empty. We now show that is a group.

• identity under multiplication belongs to .
• associativity of multiplication is inherited from the ring structure.
• closure under multiplication: Let . Then both and are units. Let their inverses be and , respectively. Now, is a unit because we can produce its inverse, namely :

In other words, .

• Finally, if , then so is . Each element thus has a multiplicative inverse.

#### Three examples from the rings:

1. . In particular, and , so each element of is its own inverse.
2. . Here, . Every non-zero element of is a unit. Being already a commutative ring with identity, is a field.
3. . Here also, each element of is its own inverse, since and .

This concludes today’s snippet. Thanks for visiting!

## Examples of rings — endomorphism rings

Today’s snippet focuses on the set — the set of linear maps . It is also denoted by . Here, is a vector space over a field . In reality, the full power of the vector space structure of is not needed; any abelian group will do.

#### The set is a ring

For , define (point-wise addition) and (composition). Then endowed with these two operations is a ring.

(The set is non-empty; it contains at least the identity linear transformation that sends each element of to itself. Thus it already possesses a multiplicative identity.)

We first show that is an abelian group. Readily follows from the vector space structure of , and basic facts about linear maps.

• closure under addition — adding two linear maps gives a linear map; so if , then .
• it contains a zero element; this corresponds to the linear transformation that sends every to .
• addition is associative — this is inherited from the vector space structure of . Indeed, if and , then

• addition is commutative — again inherited from the vector space
• for any , there is an additive inverse such that , the zero linear transformation. This will be defined as follows: for any , .

Next, the set satisfies:

• composition is associative — always, even if the maps involved are not linear maps
• composition distributes over addition — let . For any we have:

Similarly, composition also distributes over addition from the right.

Thus we have a ring structure on .

Today’s snippet ends here. Thanks for stopping by!

## 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 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 .

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

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

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

#### Inverse

Let be a ring with identity. If an element has a left inverse and a right inverse , then .

Let the (multiplicative) identity be .

Since is a left inverse of , we have: .

Since is a right inverse of , we have: .

Thus .

Today’s snippets end here. Thanks for visiting!