 Welcome back to our lecture series Math 42-20, Abstract Algebra 1 for students at Southern Utah University. As usual, I'll be your professor today, Dr. Andrew Misteldine. This video is the start of lecture 39, which is actually the last lecture for Math 42-20, and we're going to talk today about ring homomorphisms and ideals. I also want to point out that this lecture is also lecture 13 from Abstract Algebra 2, that is Math 42-30. Let's get to our titular topic though, the idea of a ring homomorphism. If you've been following along with this lecture series, then we've already talked about the idea of a group homomorphism, which remember if we have two groups, say G maps into H and we have some map fee, of the homomorphic property for that situation, would then take that, if we have two elements of G say X and Y, their product when put inside of the function phi, this will be equivalent to phi of X times phi of Y. All right, that is to say for a group homomorphism, the homomorphism preserves the group operation. A product before the function is equal to the product after the function. So that is a product in the domain G is connected to a product in the co-domain H. Now, when it comes to rings, it's always important to remember that there is a group in play here, right? Because a ring has two operations, both addition and multiplication. So let me relabel my symbols here. So to make us think of them as rings, we have R and S right here, we'll so-called fee the map. Now, R and S, if we ignore their multiplication is a group, in fact, the Abelian group with respect to addition. So really we should be thinking of it as, okay, you have R with respect to addition, maps to S with respect to addition, right? We want this to still be a group homomorphism. So if I take any two elements of R, call them little R and little S, right? If I take fee of R plus S, I want this to be a group homomorphism so that fee of R plus S will be the same thing as fee of R plus fee of S. So we want the homomorphic property with respect to the addition of the rings. Therefore, a ring homomorphism is going to be a group homomorphism with respect to the Abelian group structure of addition on the rings. That is, if we forget the multiplication, we want it to be a group homomorphism on addition. But rings have more than just their additive group structure. They also have multiplication. The second binary operation we want also to be reserved when we talk about a ring homomorphism. So just like with groups, if we take a product of two elements in R, again called little R and little S, we want that fee of R times S is the same thing as fee of R times fee of S. So if we restrict just to the multiplication, right? If you have R with multiplication and S with multiplication, we want that the ring homomorphism, it's gonna be just a homomorphism of the semi group structure. Because remember with a ring, the multiplication doesn't necessarily have all of the same axioms that the addition does. Basically for us, when we talk about rings, unless we add any other adjectives, we're only requiring that ring multiplication be associative. We don't necessarily need identity, otherwise we call it ring with unity. It doesn't have to be commutative, otherwise we call it a commutative ring. And all those other things, right? We just want, with just the multiplication structure, it's just a semi group. So really we're just asking, oh, for a ring homomorphism, we need it to be a group homomorphism on its additive structure. And it is to be a semi group homomorphism for its multiplicative structure. But of course with rings, the multiplication and the addition interact with each other because of the distributive property. So we can say a lot. But all that we require for the definition of ring homomorphism is that the map preserves the ring addition and the ring multiplication, okay? Now, like we did with group homomorphisms, there's a notion of a kernel. Now, with group homomorphisms, the kernel was a set of all elements that maps to the group identity, okay? But with two operations in mind, how do you define that? Well, like I said a moment ago, when we talked about the ring multiplication, we don't necessarily have any requirements on anything other than associativity. The ring might not even have an identity. That is a possibility. But the groups, the additive structure is always an appealing group structure. It does have an identity. In particular, this is what we call the zero element of the ring. So we define the kernel of a ring homomorphism just to be the kernel of the underlying abelian group structure on addition. That is to say the kernel of a ring homomorphism, we'll call it kernel of the, like usual, is gonna be all elements of the domain that map to zero. That is the things that map to the additive, the additive identity in that case. So ring homomorphisms preserve the additive structure, it preserves the multiplicative structure, and we defined the kernel to be everything that maps to the additive identity, aka zero. And so let's look at some properties of ring homomorphisms. The proofs of these I'm gonna leave as an exercise to the viewer right now. But their proofs are fundamentally the same as we saw with groups. So I would reference you back to that video if you get stuck on some of these, but these would be good properties to check out yourself here. So first of all, the kernel of a ring homomorphism is always a subring of R, the domain, right? Because these are things in R that map to zero. And so it's not too hard to go through the details of this thing, right? So if you take phi of R plus S, since it's homomorphic, you get phi of R plus phi of S. Like so, as R and S both belong to the kernel, you'll end up with zero plus zero, which is equal to zero, not so bad. Why is it closed under multiplication? Well, if you have phi of R times S there, well, this becomes phi of R times phi of S for which if those are both zero, you end up with zero times zero, which is equal to zero, like so. Now, as zero is an absorbing element, anything times zero is equal to zero. It turns out we don't even need both of them to be inside of the kernel. If only one of them was in the kernel, the product would still be zero, which in fact seems to suggest the kernel has a stronger multiplicative closure principle than just being a subring. We'll talk about this in a later video for this lecture. It turns out that this is actually what we call an ideal. An ideal is then the ring analog of a normal subgroup. Remember that normal subgroups for a group, it's a group but it's also closed under conjugation, which is a stronger multiplicative closure than just a standard subgroup. Kernels always for group homomorphism, kernels are always normal subgroups and vice versa. The two notions are actually logically equivalent. We will see that for ring homomorphisms, the kernel is equivalent to what we call an ideal which has a stronger multiplicative structure than just a subring. But for the moment be, we won't worry so much about that. You can also argue that the image of a ring homomorphism is a subring of the co-domain. So image of phi is a subring of S, right? And I'll leave it to you to show that. The image of the identity, I should say of the zero element of R will map to the zero element of S that always happens with the ring homomorphism. Because after all, if we forget the multiplication, then a ring homomorphism is just an additive group homomorphism and the identity has to map to the identity, all right? In which case zero then maps to zero in that situation. A curious observations, the next one here, the image of the unity, the one element of the domain will map to the unity of the image, right? Which this may or may not be the same thing as the unity of S. It might be the S doesn't have any unity, but the image will, right? And the argument is fairly simple, right? If I take phi of one R and you times it by something in the image, right? So this would look like something phi of R. Well, by the homomorphic property, this becomes phi of one R times R. And as one R is the unity of R, this becomes just phi of R. So phi of one R acted like the identity of just the image. The problem is if you take some element of S, some element of S that doesn't belong to the image, the homomorphic property doesn't come into play here. You might not necessarily get a full blown unity. Well, why is that the case? Well, for groups, it didn't matter. They always mapped onto the identity, but for unity here, for the ring multiplication, you might not necessarily get the unity of the ring. It could actually be something different, right? And the reason for that is that with regards to the homomorphism on multiplication, it's only a semi-group homomorphism. For addition, it is in fact a group homomorphism. And with semi-groups, we only are requiring things about associativity. You can't necessarily guarantee that the identity maps to the identity. And why is that, of course? Well, basically what you'll see is that with a group homomorphism, the proof that the identity maps to the identity requires we use something about inverses. Inverses don't necessarily exist in a general semi-group. In particular, when it comes to rings, we don't have any assumptions about the inverses of elements, right? Unless we start talking about things like integral domains or fields or whatever. But for an arbitrary ring, we don't have that. And so that's what prevents this potentially from connecting to the unity of the ring itself. So I mean, I should mention that if R is a ring with unity, right? Then the image of phi itself will be a ring with unity, but it doesn't necessarily have to be the same unity. That's, again, the thing I'm really trying to emphasize here. The image will be a ring with unity, but it might not have the same unity as S. So it's not a subring with unity. It's just a subring which happens to have a unity which may or may not coincide. And that's where this thing is gonna map over to. Now, if we really want the unity of the domain to map to the unity of the co-domain, we might have to add in an extra axiom. So we had those previously. We might have to write in the axiom that the unity of R maps to the unity of S, right? If we do that, we often refer to the homomorphism as a unital ring homomorphism because that extra axiom's there. With our discussion, we are assuming ring homomorphisms are non-unital. That is to say, we don't have any guarantee that the unity of R maps to the unity of S. It'll map to something, but of course, we're not claiming that it'll be the unity of S whatsoever. So that's a very important little caveat. We have to mention when it comes to ring homomorphisms. The remaining properties are fairly straightforward and identical really when it comes to their group analogs. If you have a subring of the domain, its image will be a subring of S. Now, be of course cautious that even if this R prime is a unital subring, that is, it's a subring of R and they have the same unities, then I can't claim that phi of R prime is a unital subring of S. That is to say, it'll be a subring of S and if R prime has unity, then phi of R prime will have unity, but it might not be the same unity as in S. And we've explored examples like this in the past. Looking at the next one here, if S prime is a subring of the co-domain, then its pre-image will be a subring of the domain R itself. Now, one again has to be cautious when it comes to unity here, that even if S prime is a ring with unity, phi inverse of S might not have any unity whatsoever. Because it could be that phi doesn't map onto the unity of S prime so that it might not be present in the pre-image. And this is again, getting back to the idea of a unital ring homomorphism, unless we specifically require the ring homomorphism to preserve unity, in which case really a unital ring homomorphism is a homomorphism between rings with unity. Unless we require that, then we don't necessarily guarantee those things in general. That's why these statements are listed as subrings, not necessarily with unity, because again, there's some issues going on there unless we require the third unital axiom there. But things like commutivity is pretty simple. If R is commutative, then the image will be commutative as well.