 It is important to realize that although we require additive inverses in every ring, we are not making any requirements about elements of the ring having multiplicative inverses. That is, reciprocals we don't necessarily require. But there do exist rings for which we could require multiplicative inverses. Remember, the axioms of a ring tell us that with respect to addition, we have an abelian group. It's associative, commutative, identity, inverses for addition. With respect to multiplication, the only thing we require for a ring is that the multiplication be associative, and multiplication distributes over addition from the left and from the right. That's basically it. That's all that we require when it comes to rings. We can add assumptions about commutivity. We can add assumptions about unity. But we don't necessarily require them in general. So let's suppose we do have a commutative ring with unity. So addition forms an abelian group. Multiplication is commutative associative with identity. And we have the distributive laws, of course. We say that in addition to that, if a commutative ring with unity satisfies one additional axiom, this is called a field. And this final axiom is going to be the multiplicative inverse axiom, for which if we take any non-zero number in the ring, so r cannot be zero, then it's guaranteed to have a multiplicative inverse, sometimes called the reciprocal of the element, such that r inverse r is equal to 1, the unity of the ring. Or you get r times r inverse in that situation, like so. Now, we don't allow zero to be inside of, that is, we don't allow zero to be an invertible element. And when we talk about an element being invertible, we're talking about multiplicatively, because every element of a ring has an additive inverse. Okay? We don't allow zero for basically the following reason. Okay? If we allowed zero to have an inverse, we'll basically pick your favorite two elements inside the ring. Right? Here's going to be a proof that x equals y. Well, you start off with zero equals zero, which is certainly a true statement. Then you're going to say that, okay, zero x is equal to zero y, because we've shown previously because of the distributive laws that zero is this dominant element inside of the ring. For which then if you can multiply both sides by zero inverse, zero inverse, here you can re-associate, cancel out the zero, and then you would end up with x equals y. So basically what I'm saying is the only ring for which zero could be invertible would be the zero ring itself, because we've just proved that every element x equals y, every element is equal to each other elements, in particular they're all equal to zero. So only the zero ring could be a ring for which zero is invertible, and that's kind of a cheap shot because the zero ring is the only ring for which zero is also the unity of the ring, which we don't allow that to be a commutative ring with unity. So it's imperative that we don't allow zero to be invertible, but if every other element of the ring is invertible, then we call this a field, okay? So in every ring we always have a well-defined operation of addition, subtraction, and multiplication. Fields are exactly those commutative rings for which division is possible. So let's look at some examples that we know and love very well. So the rings Q, R, and C, with their usual multiplication addition, define examples of infinite fields. On the other hand, the ring of integers is not a field. It is a commutative ring with unity, but the only elements that have multiplicative inverses would be plus or minus one. No other integer has an inverse, which is itself an integer. So Z is very far from being a field. Most of the elements of this infinite set, all about two of them, do not have multiplicative inverses. What about the finite ring Zn, right? So if we take the set zero up to n minus one with addition and modular multiplication, can we make that into a field? Are there inverses in that situation? Now from the group theory perspective, we have taken a look at the elements of Zn which are multiplicatively invertible, okay? So this would consist of what we usually call Zn star. Some people call this U Zn, or probably more commonly U of n. Now I don't particularly like this notation because this makes me think of the unitary group, but it sometimes is used here. The reason why they use the symbol U is that we're talking about right now units, which is something we'll talk a little bit more later on in this lecture 38, but not in this video. So Zn star, this is going to be a set of all integers which are coprime to n, and which we've seen previously by the Euclidean algorithm, those integers which are coprime to n will be exactly those integers which have a multiplicative inverse with respect to n, multiplication by n. And that's exactly because by the Euclidean algorithm, if n and say some number a are coprime, then we can write a linear combination. You'd like some ab plus say like some mn is equal to 1, in which case then we'll see that b is equal to the multiplicative inverse of a mod n. So we've seen that previously using the using Euclidean algorithm. Now, if Zn were a field, if Zn is a field, then this actually suggests that every element is invertible except for 0. This would tell us that Zn star must equal Zn take away 0. That is, everything must be coprime to n except for 0 itself, which that would mean that every number less than n, other than 0, if it's coprime, that means none of them can be divisors of n. That forces n to be a prime number. And so this is then a very important result to mention here that the ring Zn is a field if and only if n is prime. So finite fields, amongst finite fields, I should say Zp is one of the most famous and simplest of all finite fields to describe because that's gonna be these modular finite fields like so. And we play around in this lecture series when we talk about algebraic coding theory, the finite field Z2 was a very important field for us because that's where we did our linear algebra. Linear algebra requires that our scalars be belonging to a field. Now, I wanna mention that there are other finite fields. They're a little bit more interesting to construct that don't necessarily come from modular arithmetic in any manner. So what I'm gonna do is give you the Cayley table for the finite field of order four, right? So let's just call the elements zero, one, two, and three because we need four elements. And so in terms of addition, well, we're gonna make zero be the additive identity. So we can kind of fill in what's gonna go in there, right? And then if we were to talk about multiplication a little bit, see if I can fit that in here. If we're gonna describe the multiplication of this thing, zero, one, two, and three, zero, one, two, and three. Well, since zero is the additive identity, it's gonna be the zero element. So multiplication by zero is given, right? Then we're gonna let one be the multiplicative identity, the unity of the ring. So we're gonna get one, two, and three, two, and three. So that's information that's guaranteed when we construct this field, right? Because we want it to be a ring after all. And so the fill in the rest of it, what I'm gonna do is I'm gonna have one plus one be zero. So basically I wanna work mod two. One plus two is gonna equal three, but one plus three is gonna equal two in this. It's gonna be commutative, so we can copy that down. I want two plus two to be zero, and then two plus three has to equal one, since it's gonna form a Latin square. And then likewise, we're gonna get three plus two, which is gonna equal one, and then I want three plus three to equal zero. So you'll notice that along the diagonals, that every element when you add it to itself, you get zero. It's kind of like when you add, when you get like mod two, right? If you take an element plus an element, you get zero, that happens when you work mod two. Now in terms of multiplication, you're gonna get that two times two is three, and three times two is then one, one, and then the last one right here would then have to be a two. So this gives us the idea of a field of order four. In fact, this is the only field of order four, and so this is sometimes called F4 for short. And now it's not usually given this label right here, because again, we sometimes look at this thing right here. You look at the additive group. It is a group after all. This is none other than just the Klein four group, up to isomorphism. So you could relabel things a little bit. And so in terms of the Klein four group, you can think of it as like it's Z2 cross Z2. And so you have one generator, which comes from the unity of the ring, and then you have this other generator just comes from something else. So what some people do when they talk about this group is sometimes instead of just calling it two, we'll call it X, for which then three you notice is just one plus X. So you get one plus X, one plus X like so. And you can redo all of this table. So you get X, X plus one, X plus one, X, X plus one, X, X, X plus one. And so in terms of addition, you see it's just the Klein four group. You see there's only four possible elements. You get zero, which means zero plus zero X. You get one, which means one plus zero X. You get X, which means zero plus one X. And then you get X plus one, which obviously is one plus X like so. So we can describe, you see there's sort of like that, you see the order pairs right here, right? You could think of zero, zero, one, zero, zero, one, and then one, one. So we can make that identification with the Klein four group. All right. Well, how does the multiplication work? If we're going to play that game, spring up the multiplication a little bit. If we relabel this thing, we would get here an X and X plus one. So this should look like X and X plus one. That's not too surprising since that's the unity. That's an X. So then the thing that kind of gets really interesting is here, you're going to take X squared. So this is what's curious about this room, is this element X squared is equal to X plus one. In some regard, because that's the only element it could be, if this is going to be a field. So you get X plus one times one. You also see that if you take X times X plus one, you get the identity. And this isn't too surprising because after all, if you take X times X plus one, you're going to get X squared plus X. For which X squared, like we saw earlier, is going to be X plus one. For which in this field, if you ever take an element and add it to itself, you get zero. So this gives you back one. So in fact, X and X plus one are inverses of each other. Thus filling this out, this table.