When we add those multiplicative inverses to our set of the integers, we have a field the rational numbers. Show 1 more comment. Upcoming Events. Featured on Meta. Now live: A fully responsive profile. The unofficial elections nomination post. Linked Related Hot Network Questions. Mathematics Stack Exchange works best with JavaScript enabled. Accept all cookies Customize settings. Where you should pay special attention is when things don't work in the way you expect.
If you look at the axioms carefully, you might notice that some familiar properties of multiplication are missing. We will single them out next.
A ring R is commutative if the multiplication is commutative. That is, for all ,. Note: The word "commutative" in the phrase "commutative ring" always refers to multiplication since addition is always assumed to be commutative, by Axiom 4.
A ring R is a ring with identity if there is an identity for multiplication. That is, there is an element such that. Note: The word "identity" in the phrase "ring with identity" always refers to an identity for multiplication since there is always an identity for addition called "0" , by Axiom 2. A commutative ring which has an identity element is called a commutative ring with identity.
In a ring with identity, you usually also assume that. Nothing stated so far requires this, so you have to take it as an axiom. In fact, you can show that if in a ring R, then R consists of 0 alone which means that it's not a very interesting ring!
Each of these is a commutative ring with identity. In fact, all of them except are fields. I'll discuss fields below. By the way, it's conventional to use a capital letter with the vertical or diagonal stroke "doubled" as in or to stand for number systems. It is how you would write them by hand. If you're typing them, you usually use a special font; a common one is called Blackboard Bold. You might wonder why I singled out the commutativity and identity axioms, and didn't just make them part of the definition of a ring.
Actually, many people add the identity axiom to the definition of a ring automatically. In fact, there are situations in mathematics where you deal with rings which aren't commutative, or less often lack an identity element.
We'll see, for instance, that matrix multiplication is usually not commutative. The idea is to write proofs using exactly the properties you need. In that way, the things that you prove can be used in a wider variety of situations.
Suppose I had included commutativity of multiplication in the definition of a ring. Then if I proved something about rings, you would not know whether it applied to noncommutative rings without carefully checking the proof to tell whether commutativity was used or not.
If you really need a ring to be commutative in order to prove something, it is better to state that assumption explicitly, so everyone knows not to assume your result holds for noncommutative rings. The next example or collection of examples of rings may not be familiar to you. These rings are the integers mod n. For these rings, n will denote an integer.
Actually, n can be any integer if I modify the discussion a little, but to keep things simple, I'll take. I won't prove this; I'll just show you how to work with these operations, which is sufficient for a linear algebra course. You'll see a rigorous treatment of in abstract algebra. Then divide by n and take the remainder call it r. Since modular arithmetic may be unfamiliar to you, let's do an extended example.
Suppose , so the ring is. Hence, in. You count around the circle clockwise, but when you get to where "6" would be, you're back to 0. To see how works, start at 0. Count 4 numbers clockwise to get to 4, then from there, count 5 numbers clockwise.
You'll find yourself at 3. You can see that as you do computations, you might in the middle get numbers outside. But when you divide by 6 and take the remainder, you'll always wind up with a number in.
Using our circle picture, if you start at 0 and do 80 steps clockwise around the circle, you'll find yourself at 2. Maybe you don't have the patience to actually do this! When we divide by 6 then "discard" the multiples of 6, that is like the fact that you return to 0 on the circle after 6 steps. We see that in doing arithmetic mod 6, multiples of 6 are equal to 0.
And in general, in doing arithmetic mod n, multiples of n are equal to 0. Negative numbers in are additive inverses. Thus, in , because. To deal with negative numbers in general, add a positive multiple of 6 to get a number in the set. For example,. The reason you can add 18 or any multiple of 6 is that 18 divided by 6 leaves a remainder of 0.
In other words, " " in , so adding 18 is like adding 0. In a similar way, you can always convert a negative number mod n to a positive number in by adding multiples of n. For instance,. Recall that subtraction is defined as adding the additive inverse.
Thus, to do in , use the fact that the additive inverse of 2 that is, -2 is equal to We haven't discussed division yet, but maybe the last example tells you how to do it. Just as subtraction is defined as adding the additive inverse, division should be defined as multiplying by the multiplicative inverse. Let's give the definition. Let R be a ring with identity, and let. The multiplicative inverse of x is an element which satisifies.
If we were dealing with real numbers, then , for instance. But going back to the example, we don't have fractions in. So what is say in? By definition, is the element if there is one in which satisfies.
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It only takes a minute to sign up. Connect and share knowledge within a single location that is structured and easy to search. Is a commutative ring a field? A set equipped with addition and multiplication which is abelian over those two operations and it holds distributivity of multiplication over addition? A key difference between an ordinary commutative ring and a field is that in a field, all non-zero elements must be invertible.
For example:. In a commutative ring not every nonzero element has a multiplicative inverse unlike the requirement in a field that every nonzero element has a multiplicative inverse. Sign up to join this community. The best answers are voted up and rise to the top. Stack Overflow for Teams — Collaborate and share knowledge with a private group. Create a free Team What is Teams?
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