 We have seen when we were studying the isomorphism theorems for groups that kernels of group homomorphisms were actually equivalent to the notion of a normal subgroup. And the other direction is also true that normal subgroups are the only types of subgroups of a group which can then be, you know, modded out when you take a quotient group of some kind. That is, normal subgroups are the only things you can quotient out to get a well-defined multiplication. Essentially, normal subgroups and kernels are equivalent notions inside of group theory. What is the ring theoretic analog of this? We've now introduced the idea of a ring homomorphism and we saw previously for groups that kernels were always normal subgroups. So what do kernels of ring homomorphisms, what do they form? It is sort of natural to think that from the isomorphism theorems that the kernel is what you want to mod out to get a quotient ring. And that's exactly what you want to do for ring theory, right? You quotient out a kernel of a homomorphism. But what if I don't want to explain it using homomorphism? What if I want to do it entirely intrinsic, internal to the ring itself, without reference to some external ring that we're mapping towards? How do you do that? What is a normal subring, so to speak? Well, the short answer is it's an ideal, but let me give you, before we define what an ideal is, let us explain what conditions must be on the set i so that our mod i makes a well-defined quotient ring, right? Which we'll of course define in a future lecture here. Now, just so you're aware, r, if we forget the multiplication, is an abelian group. Therefore, every subring of r will automatically be an abelian subgroup. And since r is an abelian group itself with respect to addition, every subgroup of r is a normal subgroup. So if since r, with respect to addition, is abelian, if you mod out by anything, you'll get an abelian group. So if we want to make a quotient ring, right, we want this to be well-defined, it's only multiplication that we're really concerned about. We need that multiplication. We want multiplication here to be well-defined. If we can get that, we're as good as gold. But we don't know yet. Sorry about the wrap around there. So let's think about multiplication for a moment. But also, as we mod out by i, as we have this additive structure, we can talk about the additive cosets. Because the symbol r mod i is the collection of all additive cosets of this subring i that you see here on the screen. And so a typical element of r mod i would be something like r plus i. So I want to consider the problem of multiplication. If I take two arbitrary cosets, additive cosets, mind you, in the set r mod i, it would look something like r plus i and s plus i. What would happen with their product? Now, for the moment being, I'm going to consider the Frobenius product. That is, I'm going to take the element-wise product of sets. So recall, we did this with groups, of course. If you have two subsets, c and d of your group, not necessarily subgroups, but there's just two subsets, we can define the product of the subsets, often referred to as the Frobenius product. This is going to be the collection of all possible products, c and d, where c belongs to c and d belongs to d. Like so. This is not just true in groups. If we have any binary operation, if we have any set with a binary operation, we can take subsets of that set and then take the set-wise product, if it's multiplication or sum, if it's addition. We can combine the sets using the binary operation that's in play here. So that's what one means when you talk about r plus s, or r plus i, excuse me. You can treat r like a single ten, and so we're just taking all of the possible sums between something with i and something from r, which of course is just r in that situation. So that makes sense. We can also talk about the product here. Now, when it comes to rings, since the ring multiplication does not form a group, we don't get all the nice cute little properties we had when we talked about Frobenius products with group subsets. So when that's a little bit more carefully, quality is not always available to us. Now, because products of sets are defined element-wise, and elements have, of course, additive associativity, additive commutativity, multiplicative associativity, distributed property, those properties will be inherited on the set products as well, because again, it just works element-wise. So if you take r plus i times s plus i as sets, we have the FOIL method, because the distributed property is going to work here, in which case we'll get r times s, we'll get r times i, we'll get i times s, and we'll get i times i. Now, we have to be careful here. We don't know that our ring multiplication is commutative, so the set r times i is not necessarily the same thing as i times r, and same thing with s i versus i s, right? We're okay when it comes to addition, right? So r plus i is the same thing as i plus r. So with regard to addition, right, we have a normal subgroup that the left and right cosets are the same, but when it comes to multiplication, we can't necessarily commute things around. But with looking at this right here, what we can't say is the following. Since we know that i is a subgroup, the product of any two things from i will belong to i. So we can actually say that i... Let me write that again. I'll just write that as i times i, which we can actually define that to be i squared. This is a subset of i, right? Because if we take two things in i, its product will be an i. So i squared is a subset of i. Now, as we're talking about multiplication, which only has a semi-group structure in general, we can't actually guarantee equality in this situation. For groups, if you take a subgroup h and you square it, you'll get back all of h. That has to do with the inversion axiom, but we don't have that for general semi-groups. So therefore we only get a best in inequality here, that i times i does belong inside of i, but there actually do exist counter examples where i squared is strictly less than i. You can take a look in the lecture notes in the footnote if you want to see a counter example of such a thing. I'm not going to provide it in this video though. So we do get that this belongs to that. Now let's consider the case where s equals zero. If s equals zero, that means s plus i equals just i. Now i is the additive identity. So if I'm making a ring using r mod i, that means i has to be the zero element. We have to identify this with zero, so to speak. So if I take r plus i and I times it by i, if this is a ring, this should equal i. And therefore we need this to belong to i if this is going to be well-defined. All right, so backing up a little bit. If this then became just r plus i times i, well since s is zero, that will just kill that thing off. Zero times anything will just be zero and adding zero doesn't make much of a difference. So this does simplify to be looking like this. r plus i times i will just be r i plus i which that should be part of i. If in fact this is a well-defined multiplication, that's what we need right there. And in the other direction, if you set r to be zero, then again you take off these parts and you need that i s plus i is less than i right there. So this is sort of an interesting thing and this is going to then lead to the definition that we're going for. If r i itself was a subset of i, then i plus i would then equal i and we would be good. And if i s was less than i, that is if i times s actually was inside of i, then we also get what we're looking for here. And so that's what leads to the definition of an ideal. That's why we call this symbol i. If r is a ring and i is a subring of r, we say that i is an ideal. If the subsets r i and i s are, if these are contained inside i itself, we call this an ideal. Particularly this is called a two-sided ideal in general, because again with non-commuted rings, we do have to distinguish between these sometimes. If you only assume the first part, r i is contained in i, you call this a left ideal. If you only assume the second one, the second property that i s is contained in i, we call that a right ideal. And so having both of them is called a two-sided ideal or when there's no confusion whatsoever, it's just called an ideal for short. We will see in just a second that modding out an ideal is what you need for that well-defined operation, particularly we're going to do that via ring homomorphisms here. Because as we saw on the previous slide, if these two subsets belong to i, we're going to get a well-defined ring multiplication on the additive cosets. And this definition of an ideal is exactly what gives us the kernel of a homomorphism. In particular, the main result for this video here is that if phi is a ring homomorphism between two rings r and s, then the kernel is an ideal. So we already know, let's look at the proof of this thing, we already know that if we look at just the additive structure, just the additive structure, then the kernel of phi will be an additive subgroup of r. This will be proved with group theory. Commutivity was never a concern in that situation, but we're going to get that kernel of phi as an additive subgroup of r. All right? To make it a subring, we then have to prove that it's closed under multiplication, but being an ideal has a stronger, has a stronger multiplicative closure principle. So what I want you to see is the following. If r i is less than i as a subset, right? What that means is for all x inside of i, we have that r x, which of course belongs to r i. Since this is belonging inside of i here, this says that this belongs to i. And so if you take anything in x and times it by something in the ring, you're going to be inside the ideal. So let me rewrite that one more time. What we get here is that for all x inside of the ideal, and for all r inside of the ring, we have that r x belongs to i. So to be a subring, we just need to have, if you have two elements of the subring, the products inside the subring, but with an ideal, if you take any element of the ideal times any element of the ring, and that element can be external to the ideal, then their product is inside the ideal. The ideal sort of gobbles everyone up. Times it by anyone, you're inside of the ideal. And that's important because that's what zero does, right? Zero times anything is equal to zero, right? So the ideal in a quotient ring has to mimic that property. So it has the property that anything times an element of the ideal is still in the ideal. The ideal has this absorbing multiplicative property. Now I'm not saying that r x is equal to x, but r x is some other element of the ideal. We also get this in the other direction as well, right? If you take the right condition, we're going to get that x times s belongs to the ideal. As long as x belongs to the ideal and s is any element of the ring, that's what we're given with these ideals, all right? So to prove, so we already have that the kernel is an additive subgroup of r, and as things are abelian, it's necessarily a normal subgroup, right? We need to prove just that the ideal closure property happens because if the ideal multiplicative property is true, then general ring closure also is true because if I can take r times an arbitrary element of the ideal, then, and as r could be any element of the ring, then if they're both belonging to the ideal, then we still get closure there. So we just have to prove this stronger condition here. So what we're going to do is we're going to take r to be an arbitrary element of the ring r, and x is going to be an arbitrary element of the kernel. So in particular, I know that phi of x equals zero. I don't know what phi of r is, but I don't need to because if I take phi of rx, because it's a ring homomorphism, we can factor this as phi of r times phi of x since x belongs to the kernel, phi of x equals zero, like we observed above, and anything times zero is equal to zero. And so phi of rx is zero, meaning that rx belongs to the kernel, and therefore the set r times kernel is a subset of the kernel, which admittedly all you have to do is show this, same thing there, by a similar argument, if you take multiplication on the right, so the kernel times s is going to be a subset of the kernel. And this shows us then that the kernel is always an ideal. So this idea of closure, right, that for all r in the ring and for all x in the ideal, we have that rx comma xr belong to the ideal. This ideal closure is exactly what kernels do for ring homomorphisms, and therefore this is exactly the object we need if we want to quotient out to form a so-called quotient ring. And so ideals are, in fact, the ring theoretic analog of a normal subgroup. And so that brings us to the end of lecture 39 for Math 42.20. Thanks for watching, but I also want to remind us that this is actually the end of lecture 13 for Math 42.30, abstract algebra 2. So if you want to learn more about quotient rings, please check out the next video. If you liked what you saw or learned something, give the video a like, give all these videos a like. I appreciate that one. And by all means, subscribe to the channel if you want to see some more of mathematical videos like this in the future. Thanks, everyone. I'll see you next time.