Name: 環例（抽象代數 (Ring Examples (Abstract Algebra))
Uploaded: 2021-01-14T08:50:35.000Z
Duration: 7 min 18 s
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The key structures in Abstract Algebra are groups, rings, fields, vector spaces,

and modules. You start with groups because the other four structures are

built upon them. All of these concepts are fairly abstract, so it's helpful to

learn lots of concrete examples to help keep you grounded and test out

everything that you learn. Today, let's look at examples of rings.

Before we dive into the examples, let's remind ourselves of the definition of a ring.

Loosely speaking, a ring is a set of elements where you can freely add, subtract, and

multiply with the usual rules of arithmetic. And while addition is

of elements with two operations: addition and multiplication.

It's an abelian group under addition. Multiplication obeys the associative

property, and addition and multiplication are connected by the left and right

distributive laws. In a ring, multiplication may not be commutative. If

And if the ring has a multiplicative identity

one, we call it a ring with identity. Every ring has the additive identity 0,

since it's a group under addition. So when you say ring with identity,

everyone knows that you mean ring with multiplicative identity.

A classic example of a ring are the integers.

The set of integers is denoted with a double stroke Z.

This is from the German word Zahlen, meaning number. Historically,

number meant integer. That's why number theory is a study of the integers.

Nowadays, however, number has a more general meaning, but we stick with the

letter Z for the integers. Another example of a ring is a set of

polynomials with real coefficients, you can freely add, subtract, and multiply any

two polynomials, and you'll get a third polynomial. Each polynomial has an

additive inverse. For any polynomial F, its additive inverse is a polynomial

where you take the opposite of each term. The additive identity is zero.

Don't forget that zero is a polynomial. On the multiplication side, there is an

identity element: the number one. And for real polynomials, multiplication is

commutative. This is a commutative ring with an identity element.

In this last example, we looked at polynomials with real number coefficients.

But you can use any ring for the coefficients: the integers, the complex numbers, or how

about the integers mod n? You can even have polynomials where the coefficients

are matrices. Any ring will do. So if you are handed a ring R, you can use it

to make a new ring: the ring of polynomials with coefficients in R.

Here is its notation. Be careful you use brackets and not parentheses. They mean

different things. Brackets means the set of polynomials;

while parentheses means the set of rational functions, which are fractions of polynomials.

So far every ring that we've seen have been commutative, and

has had an identity element for multiplication. Let's now look at rings

where this is not the case. An example of a ring without an identity element is the

set of even integers. If you add, subtract, or multiply any two even numbers, you'll

get another even number. But the number one is not in this set, because one is

odd. So while this is a commutative ring, it does not have an identity element for multiplication.

Let's now look at a ring that's not commutative. The classic

example here are matrices. For instance, the set of 2 by 2 matrices with whole

number entries are a ring, but it's not commutative. To see why, consider the two

matrices 1 2 3 4 and 1  0   negative 1   1. Here, A times B does not equal B times A.

So this ring is not commutative. We call this a non-commutative ring. As a

consolation prize, however, it does have an identity element: the 2 by 2 identity matrix.

Just for fun, let's see a ring that is not commutative and does not

have an identity: the 2 by 2 matrices with even entries. This definitely does

not have an identity, since the identity matrix has ones along the diagonal and

one is an odd number. This poor ring is having an identity crisis.

Already we've seen a wide variety of rings: commutative, non-commutative,

identity, no identity - but all the examples have had something in common:

they have all been infinite. Are there rings with only a finite number of

elements?  I'm glad you asked, because yes, yes there are.

The integers mod n are a finite ring  with only n elements.

You may recognize this from your study of groups. This is the notation for a quotient group.

Z is a group and n Z is a normal subgroup.

The quotient group is the group of cosets. This notation is carried over to rings.

Here, Z is the ring, and n Z is an ideal. In this context, we call it a quotient ring.

The integers mod n is a finite ring. It is also commutative, and contains an

identity element 1. This gives us a nice list of finite rings: the integers mod 2,

mod 3, mod 4, mod 5, and so on. But something special happens when N is a

prime number. If n is prime, the integers mod n is now a field. Here's where you

need to be careful. Every field is a ring, but not every ring is a field.

If you were to draw a Venn diagram, the set of all fields would lie inside the set the

all rings. And since every ring is a commutative group under addition, the set

of rings lies inside the set of groups. From the examples that we've seen,

we can add more details to this diagram. Inside the set of rings are the

commutative rings and the rings with identity. Every field is commutative and

has an identity, so the set of fields lies in the overlap between commutative rings

and rings with identity. I know what you're thinking you've shown

me lots of examples of rings: infinite rings, finite rings, commutative rings,

non-commutative rings... but you haven't yet shown me a finite non-commutative ring.

This I'm going to leave as a challenge for you. Here's a hint: you've seen a

finite ring: the integers mod n; a non commutative ring: the two by two matrices.

In the comments below, help each other understand how you can combine these to

make a finite non-commutative ring. By doing so, you'll be joining the

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