 I wanted to present some examples, some important examples of ring homomorphisms. One of the most classic ones is gonna be a homomorphism from the ring of integers to the ring of integers mod n, right? And this, the homomorphism is actually similar to what we did when we considered cyclic groups previously in this lecture series. I can take any integer m and just map it to that same integer, but then reduce it mod n. This is in fact gonna be a ring homomorphism. This is pretty easy to see here, that if I have two numbers, m plus m prime, if you take phi of that, this just becomes m plus m prime mod n, like so. Which of course, this is the same thing as m mod n plus m prime mod n, like so. In which case then you get phi of m plus phi of n m prime, excuse me. And this is a property we proved about modular arithmetic at the very beginning of this lecture series. A similar statement can also be used to show that multiplication is preserved here. So phi of m times m prime, right? This thing becomes m m prime mod n. But then again, the properties with modular arithmetic that preserves these things, this is the same thing as m mod n times m prime mod n, which then gives you phi of m and phi of m prime, like so. So again, these are all properties of modular arithmetic we've proven previously. So I'll just erase them from the screen. This is sort of the canonical ring homomorphism from the integers to z mod n. And n is somewhat irrelevant in this situation. We can map z to any to any modules we want to. In fact, we can generalize this example right here that if s is any ring with unity, right? If s is any ring with unity, then we can actually construct a ring homomorphism from the integers to s via the following rule here that the integer m will map to m times the unity of s. So what this of course means is we're taking the sum of i s and we do this m times, that's what it means. So this really is just one plus one plus one and we do this m times. That's how we should interpret that little symbol there. Why is this a ring homomorphism? Well, it's basically the same argument that we did before. And again, I do want you to see this as a generalization of this previous observation that you're just mapping m to m times one mod n. That's all you're doing there because m, when you're working in z n just means one plus one plus one plus one m times you just reduce it mod n. So that's what we're doing right here as well. So let's consider y is phi with this rule a ring homomorphism we consider addition, right? So you take phi of m plus n you're gonna get m plus n i s for which the distributor property applies in the situation for which we get m one plus n one but that's just phi of m plus phi of n. So this is an additive homomorphism. What about multiplication? Well, in terms of multiplication here you're gonna get that this is m n times one which of course one has the nice property is that it's item potent. One times one is equal to one. And then working through this this becomes m one plus n one. Now again, I should even be more careful because after all, what are we doing here? This means one plus one plus one plus all the way down. There's a bunch of ones here and so we're getting m n many ones. Here we have m many ones and here we have n many ones. If you foil that thing out using the distributor property you end up with all these ones. I mean, you get a bunch of one times one so the item potent statement is necessary. But what I'm trying to point out here is that this multiplication with integers is not necessarily the ring multiplication. It's more like a scalar multiplication is how we should be interpreting it here because when we talk about three times x in a ring we don't mean you're times in three by x using the multiplication of the ring. What we mean is you have x plus x plus x you add it together three times. This makes sense for any ring that is with any ring you can scale by an integer. If it's a negative integer of course you interpret that to be the additive inverse of course. But nonetheless, this makes sense for any arbitrary ring we have this idea of scalar multiplication by an integer and this preserves, right? And because we can do this scalar multiplication for any ring basically we have this universal ring homomorphism from the integers into an arbitrary ring with unity. So basically as long as the ring has unity in some respect it has something that looks like the integers for which if this map is one to one this would suggest that S has a sub ring isomorphic to the integers. Now, of course it could also be that this ring maybe it doesn't have a sub ring isomorphic to integers but it might, it will have a sub ring isomorphic to ZN where it could be the case that one plus one plus one all the way down to one. It could, this could be that there's some combination of adding ones together that gives you zero. And if N is the smallest number of ones that add up to be zero we call this the characteristic of the ring. So the characteristic of S will be N in that situation. It's the smallest number of ones that add up to be zero. If there's no number of ones that have to be zero we actually call it a characteristic zero ring. And that's mostly because the integers themselves are isomorphic to Z zero because Z zero well Z in general a Z mod ZN, excuse me NZ that's what I meant to say. But if you have a zero there then you don't use kill off the identity which doesn't do anything that's just the identity, right? So we can identify Z with Z sub zero. And so the characteristic is then this modulus of this Z mod N, Z mod zero being zero right there. And so in general the kernel of this map from Z to S here is going to be some number times Z and that number N is the so-called characteristic of the ring, it's the smallest number that makes adding ones together go to zero. But of course if one goes to zero and you add up one that happens for every element of the ring. So let's say I have some element of the ring X and you do it N times, N times this then becomes N times X which is the same thing as N one times your identity call it R right here, multiplicative identity for which if you add together one N times that becomes zero and anything times zero is equal to zero. So if the identity, if the unity I should say adds up to zero and when you add together N times then every element will have that property as well. Hence the characteristics of very important invariant of a ring. And so as you map the integers into a ring this will happen in such a way that oh just map the one the integer one to the unity of that ring. That's why we call unity is one because in some regards behaves just like the integers and the kernel then would have to do with the characteristic of the ring. Let's look at another example. This time let R be a ring and let X be any set under the sun. X doesn't have to have any algebraic structure. It does not have to be a ring. It does not have to be a group. It does not have to be a semi group. It doesn't have to be anything. It's just a set. Then we're gonna build a ring. Now that ring is not gonna be on X. We're gonna take R to the X which remember R to the X is the set of all functions of the form F maps X to R. So the superscript here is the domain and then the base using this exponential notation is the co-domain. So R to the X is the set of all functions that map from X to R. Now we can give R to the X a ring structure. We do that by doing component wise addition and multiplication so we can add together functions. So if we have two functions F and G that belong to R to the X we define addition of the functions F plus G by the rule that if you evaluate at X we define this to be F of X plus G of X. We can also do the same thing with multiplication. How do you define F times G? Well to define a function you just have to tell me what does it do to each element? So if you allow X to be an arbitrary element we define F times G evaluated at X to be F of X times G of X. Now notice F of X and G of X both belong to R. They're in the ring R. So since R is a ring we can add together F of X and G of X and so then we define the sum of the functions using that observation. And again F of X and G of X belong to R so their product belongs to R because it's a ring. So F times G is defined to be the function that always maps X to the product of F of X and G of X. So if the co-domain is a ring we can then define a ring structure on the functions. That has a set of functions I should say. And it'll inherit all the nice ring properties. Addition, multiplication will be associative. Addition will be commutative. We will have a zero element, right? For R of X here the zero element will be the zero function which is the constant function that sends every element to zero. If R has a unity, R to the X will also have a unity which will be the constant function that F of X is always equal to one. You get something like that and the distributed property works because basically here's piggybacking on the properties that R has. If R is a commutative ring then R to the X will be a commutative ring as well. So we can build a ring structure using just functions and component-wise operations. Now with this ring in mind, R to the X, it's now a ring we can actually define a ring homomorphism from R to the X to R where the R to the X, these are these R-valued functions and therefore we can actually evaluate the functions and this makes a ring homomorphism. We actually call this the evaluation map. So for a fixed element X, we define a homomorphism phi sub X where we send a function F to its evaluation F of X. Like so, this will define a ring homomorphism. We can prove that it's closed under, excuse me, that it preserves addition and multiplication, right? Let's see that really quickly. If I have two functions, phi sub X, you have F plus G right here, this maps it to F plus G of X, but by definition F plus G of X is equal to F of X plus G of X, like so, which F of X is then phi sub X of F and then G of X is phi sub X of G, right? So we actually defined addition of functions exactly to be what it must be for this function, this evaluation function to be additively homomorphic. And the same is also true for multiplication here. If I take phi sub X of F times G, this will map to phi times G evaluated at X, but by definition, what does F times G means? It means this is the function that maps X to F of X times G of X for which individually F of X is phi sub X of F and then individually G of X is gonna be phi sub X of G, like so. And so again, the operations of addition and multiplication on R to the X are exactly the operations that they need to be to make the evaluation map be homomorphic. And this is something we do in algebra a lot. We can build an algebra. In this case, we can build a ring so that what we really care about is maybe not the ring itself but we care about the homomorphism. We built a ring exactly so that evaluation is now a homomorphism in this setting. A very important special case of this is if we take the polynomial ring R to the X and map it to R, we can also do this by evaluation. And so we can evaluate each of the X's with a specific value, call it T or whatever. This is by analog gonna be the case as well. All right, one last example that's really important when we talk about is the endomorphism ring. This is gonna be very similar to the R to the X we saw a moment ago, but not exactly. So there's a slight change here. So let's explain that. So we're gonna take R to be a ring and we're gonna take functions from R to R. So it's kind of like what we did a moment ago with R to the X, but now the set X is R itself. An endomorphism on R is a ring homomorphism of the form R to R. So this is a ring homomorphism from R back into itself. All right, so unlike the previous example we had R to the X where X didn't necessarily have any algebraic structure and then the functions we were considering had no algebraic structure whatsoever. Things are a little bit different now. Now we're talking about R to the R where the domain has an algebraic structure and now we want the maps from in this situation from R to R, we want this to be homomorphisms, ring homomorphisms. That's a very important difference here. There is algebraic structure on these functions now and because of that we're gonna change what multiplication means in the situation. Now instead of writing R to the R because if we wrote that, this would actually imply we were talking about the function ring we had done on the previous slide and so we use a different notation. We're gonna use end of R, so the endomorphism ring on R. This will actually be a ring with unity. The unity is guaranteed this time with R to the X. It had a unity only if R had a unity but even if R doesn't have unity the endomorphism ring will have unity and that's because we change with multiplication means. Addition is gonna be exactly what it was a moment ago. We define addition as just the usual function addition. So F plus G evaluated at X is still gonna be F of X plus G of X. That part's the same. So if you just look at the additive structure in the endomorphism ring R and R to the R, the function ring has exactly the same structure and with regard to multiplication though, multiplication is gonna be defined to be function composition. So if you take F times G, this is equal to F composed with G, all right? And so therefore if we take F times G evaluated at X, this actually means F of G of X in the usual function composition sense. Because these are endomorphisms, the domain and the co-domain are exactly the same. So if I take any two endomorphisms I can always compose them together. This will give us a non-commutative ring. R to the X is commutative if R is commutative. Even if R is commutative here, then it doesn't matter. The endomorphism ring will be non-commutative because function composition, they just don't commute. It does have an identity. The identity element is gonna be the identity function in the usual sense. And in fact, when you look at this definition, we don't even need R to be a ring. We can actually get away with R being an Abelian group. If R is just an Abelian group, then the same definition would apply. We can actually build a ring using an Abelian group. Although oftentimes we start off with a ring and that's why I defined it the way I did. And so these are some very important examples of, these are some very important examples of rings that we actually can construct using homomorphisms. The last thing of course I wanna mention is if you look at the group of units, given any ring, right? If we look at those rings which have units that are multiplicly invertible, the group of units here is going to be the automorphism group of R, which remember an automorphism is a isomorphism from R back into itself. For which of course we define ring homomorphisms. What's a ring isomorphism? It's gonna be a bijective ring homomorphism. And so those endomorphisms which are invertible exactly are the bijective endomorphisms. These are the automorphisms. This would be the group of units of the ring. And so the study of the endomorphism ring is a very, very important topic in ring theory which we might see sometime in the future.