 So the afternoon, so last time we have been started to talk about rings. So we introduced rings and talked about subrings and units and so on. So it was all rather elementary. So now I will just proceed. So in particular, we want to start by introducing division rings and fields. So the first is a definition of a ring with the unit ring. So ring R with one is called a division ring. Well, if every non-zero element is invertible. So if R star is equal to R without zero. So R star, I recall, was the multiplicative group of units. So the elements which are invertible for the multiplication. So every non-zero element has an inverse. So we will in a moment not really deal with division rings. So we are more interested in the commutative division rings which are called fields. Commutative division ring is a field. So the last third or so of this course will be devoted to fields in particular to field extensions. So that means you have one field sitting, a smaller field sitting inside a bigger one. So I will write explicitly once again the actions for a field. So that we have them once together. So that means a set R together with two binary operations. So which are addition from R times R to R. And multiplication from R times R to R is called a field. Well, so two binary operations and two distinguished elements. So zero, which is different from one. The two elements zero and one which are different from each other is called a field. And first the addition, so R together with a plus is a commutative ring with neutral element zero. A compensative group, this was part of the definition of a ring. So second one is that if I take R without zero with a multiplication, this is a commutative group with neutral element one. And secondly, these two structures have to be somehow compatible. This is by the distributive law, so which just says that if I take A times B plus C, this is equal to AB plus AC for all ABC and R. So this is the definition of the field. As I said later, we will study this a little bit more. For now, so you know a few fields. No, we have Q, so the rational numbers, real numbers, and the complex numbers are fields. I want to also show that there's some finite fields. And for this I show that every finite integral domain is a field. So composition, every finite integral domain is a field. So recall that an integral domain is a commutative ring with one which has no zero devices. So there are no two non-zero elements whose product is zero. And we want to show that such a thing is a field. So that in such a thing, every non-zero element is invertible for the multiplication. So what we have to show, as it is already a commutative ring, we have to show that it is that the units are without zero. So in other words, we have to take a non-zero element in I to find its inverse. So we have to find its inverse element. Well, so we look at this distributive law. So the distributive law says, which we have in any ring, says that A, so now for this given A times E plus C is equal to AB plus AC. So in other words, that means if we take the map of multiple line by A from R to R, which sends an element B to A times B, this map is a group homomorphism for the addition on R from R plus to R plus. Because it says precisely that if I apply this map to B plus C, it's the same as if I first apply it to B and then to C and take the sum in R. Okay, so it's a group homomorphism and we know that R has no zero divisors. So in particular, there is no C in R without zero such that A times C is equal to zero. Obviously, A times zero is equal to zero. In other words, if I take this group homomorphism, then its kernel consists only of zero. If you have a group homomorphism whose kernel is just zero, it means it's injected. Now, we use the property that this is a finite integral domain. If you have a finite set and you have a map from this finite set to itself, which is injective, then it must also be surjective because you have as many elements on both sides, sometimes called the pigeonhole principle. So as R is finite, if the map is injective from R to itself, it must be bijective. In particular, if I take the element one in R, this must lie in the image. So let an element in R with A times B is equal to one. Well, so this exists because the map is bijective, but it gets surjective, so it means B is equal to A to the minus one. So we have found an inverse for our element for our non-zero element, and so it means that every non-zero element is a unit. Okay, so this gives us, we already know some finite integral domains because we had seen that if we take Zp, where p is a prime number, we know that this is an integral domain, and so, and thus it is a field, and it has with p elements. So for every prime number p, we have this field with p elements, and later we will see that for every prime power p to the n, there is a field with p to the n elements, and we will soon introduce the notion of isomorphism and homomorphism of rings and fields, and then we will find that for every prime power p to the n, there is a unique field with p to the n elements up to isomorphism. But as we will see later. Okay, so now we want to come to ring homomorphisms. So it's the usual thing whenever you introduce sets with a structure, you're interested in homomorphisms between them, which means maps between such sets which are compatible with the structure. Now for a group, the structure was the multiplication, and so the homomorphism must be a map between the groups which is compatible with the multiplication. And for a field, for a ring, you have two operations, addition and multiplication, so it must be compatible with both. So we talk about ring homomorphisms. So definition A and B be rings. So a map from A to B is called ring homomorphism. If it's compatible with addition and multiplication, so it sends the sum to the sum and partial product. So if for all A and B in A, we have phi of A plus B is equal to phi of A plus phi of B. Here's the sum in A, here's the sum in B. And phi of A times B is equal to phi of A times phi of B. That seems the obvious definition to make. And then we have the same words that we use for homomorphisms of groups. So the image of phi is phi of A. So that's the image as a map. And the kernel of phi is the kernel of phi, which is defined to be the inverse image of the zero element of B. So if you have a ring homomorphism, it's in particular a homomorphism of commutative groups if you just take the addition on both sides. And then the kernel is the same as the kernel was for this group homomorphism between the additive groups. So and, say, a bijective ring homomorphism, as usual, is a ring isomorphism. And if you have an isomorphism from a ring to itself, it's called automorphism. Phi from A to A is an automorphism. OK, I mean, this is a bit small, but anyway, it's the opposite. So now, as I just said, we have just defined the kernel to be the kernel of the homomorphism of the additive groups. So a ring homomorphism from phi from A to B is in particular a group homomorphism between the additive groups from A with the addition to B with the addition. And by definition, the kernel is the same as the kernel of this group homomorphism. Thus, phi is injective if and only if the kernel of phi is equal to 0, because that we know for the group homomorphisms of the additive group. And then, as usual, I don't even write it down. If you have a ring isomorphism, then it's straightforward to see that the inverse map is also a ring isomorphism. And the composition of ring homomorphism is a ring homomorphism. And last time, I introduced the notion of a suffering. So we also have that if phi from A to B is a ring homomorphism, then its image is a suffering of B. This is essentially directly from definition. It just means it's a subgroup of the additive group and the product of any two elements also lies in it. That was the definition of a suffering. And this is straightforward to check. If you want, you can check it as an exercise. So if you remember, when we were talking about groups, we had that the image of a group homomorphism is a subgroup, and the kernel of a group homomorphism is also a subgroup, but it's something better, namely a normal subgroup. And something similar is true here. If you'd have the kernel of a ring homomorphism, it's also a subgroup, but it's something better, namely an ideal. And so we want to introduce that now. Definition. So a subset in R is called an ideal if following holds. First, it's a subgroup for the additive group. So the sum of any two elements in I lies in I, and also the negative for any element which won't just the difference of any two elements in I lies in I. And the second statement is something more than for a subgroup, namely for all, say, x in I, and all A at x times A is in I, and A times x is in I. So to be a subring of R, we would have to know this for all x and A in I, we need xA is in I. So this is a weaker condition. So in particular, an ideal is a subring, but it's something that, so for the moment, so typical, so example, so same. And if the set 0 is an ideal, and if I take the whole of R, this is an ideal in R, this is basically obvious from the definitions. Let's now see what I just said before that I didn't write it. I just said it that the kernel of a ring homomorphism is an ideal, so let me state that lemma that phi from A to B, a ring homomorphism, maybe I should say that so here, as I'm not assuming that the ring is commutative at this stage, these two conditions are two different conditions, and I require them both. So that's for me an ideal. So in some books, you call that two-sided ideals, because multiplying on both sides, you still are in I. And then you also have left ideals and right ideals. But we will not be concerned with left ideals and right ideals, we are only concerned with two-sided ideals like this. Actually soon, we will restrict our attention to commutative rings where it makes no difference anyway. So let phi from A to B be a ring homomorphism, then the kernel of phi is an ideal in A. More generally, if I is an ideal in B, then its inverse image is an ideal in A. So first, you should be clear that this first statement follows from the second, because I've just told you, and it's anyway obvious, that 0 is an ideal in B. So I just applied to this, I get that the kernel, which the inverse image of 0 is an ideal in A. So now, I have only to show the second state. So let's take our ideal in B, and we just have to check the definition. So as i is a subgroup of the additive group of B, its inverse image is a subgroup of the additive group of A, by what we have learned when we did groups. So we only have to show this second condition. So let x be an element in phi to the minus 1. I and let A be an element in A. So we have to show that Ax and xA are in I. Then we have shown that it's an ideal. So if we take phi, this kind of, if we take phi of Ax, as it's a ring home of phi of A times phi of Ax, this is phi of Ax is phi maps x to an element in I. That's what's written here. So this is an element in I. And I multiplied with any element. So phi of A is some element in A. But as this is an ideal, it follows that this lies in I. And similarly, phi of xA element I. So it's basically obvious. And so this was this statement. So the inverse image of an ideal. Yeah, yeah. Sorry, sorry, sorry. That's what I obviously meant. I should also have written that. So here it's OK. And so Ax in phi to the minus 1 of I and xA in phi to the minus 1 of I. OK, thank you. So I want to say slightly more precise description to compare the ideals. If you have a subjective ring homomorphism, I want to compare the ideals. Say in A and B, if the ring homomorphism goes from A to B. It's again rather a simple thing, but it's useful to know sometimes. Lemma, or let phi from A to B be a subjective ring homomorphism, which sends an ideal in B to its inverse image is a projection from the ideals in B. The ideals in A which contain the kernel of phi. So we can describe all the ideals in B in terms of some of the ideals in A, namely those which contain the kernel. This is actually quite simple. We can write down the inverse map, namely just taking the image of an ideal under phi. So we define the inverse map. So this is the other way around. So I write again the idea from the ideals in A containing kernel phi to the ideals in A in B. Well, we want to use phi to do it. So the most obvious way we just take the image, so we take an ideal J in A which contains the kernel of phi is mapped to its image under phi. So I claim that this is the inverse map to this. So we know that this map sends an ideal in B to an ideal in A. And obviously, an ideal in A containing the kernel. Because this ideal contains the element 0, and its inverse element contains the inverse image of 0, which is the kernel. So the map does this. And now we want to write down the inverse map. So here's written down the inverse map. And now we want to see their inverse to each other. The first thing we might want to see is, however, that it sends that it actually does what is claimed here. That it sends an ideal in A, which contains the kernel to the ideal in ideal in B. Actually, it sends OK. So let's J be an ideal in A. I don't even need to have that contains the kernel of phi. Then we want to see that its image is an ideal in B using that the map is subjected. So to see phi of J is an ideal in B. Well, that's quite simple. Obviously, we know that phi of J is a subgroup of B with addition. Because phi is a group homomorphism for the active group, and J is a subgroup of A. So that's fine. So we have to only see that whenever I multiply with any element of B, I stay there. So let E in B and Y element in B of J. Then I can write, as the map is subjective, I can write B equal to phi of A, where A is in A, and Y equal to phi of X, where X is an element in J. Well, and then obviously, if I take, say, EY, this is equal to phi of A, phi of X, which is the same as phi of AX. And so it is in phi of J. And similarly, Y is an element in phi of J. So we see that if we have a surjective ring homomorphism, the image of an ideal is an ideal. So in particular, it holds under our X assumption that this ideal was supposed to contain the kernel of it. So now let J be an ideal in A, which is what we want to do. First, let I be an ideal. So we want now to show that this map is the inverse map. So we have to show that both compositions of this map with this map in both directions are the identity. So now we take I, an ideal in A. Do I want this? Yeah. No, I'd be an ideal in B. Then if we take phi of phi of to the minus 1 of I, this is obviously equal to I. In fact, that's true for any subset of P, which is just the property of being, this means all, this is the set of all elements which map to elements in I. And then I apply phi to it so I get I. This is by definition without any assumptions on what I is, obviously. And conversely, let J be an ideal in A containing the kernel of phi. And then we have to then it's, again, the following is clear. If we take the phi to the minus 1 of phi of J, this will contain J. Again, this is true regardless of what J is. If I have any subset of A, then phi to the minus 1 of phi of it contains the set. Because again, just by definition of phi to the minus 1, these are all the elements in A which map to elements in the image of J, in other words. And so certainly any element of J lies inside the set. Because that's, again, just the definition. So we have to see the other inclusion. So which one do I want to? So let's set an element in phi to the minus 1 of phi of J. We have to show, so we want to show the other inclusion, then we are done. So we have to show that Z actually lies in J. Obviously, if it's the fact that it lies in the inverse image of phi of J, means that if I take phi of Z, this is an element in phi of J. That means there exists an element X in J such that phi of Z is equal to phi of X. That's what it means to lie in the image of J. So if I take phi of X minus of Z minus X, so phi of Z minus phi of X, we know that this is 0 because they're equal. So this is phi of Z minus X. So it means that Z minus X is an element in the kernel of phi, which we know is contained in J because that was our assumption. So thus, we can write Z, which was our element, which we wanted to show is in J. You can write this as X plus Z minus X. And this is an element in J. This is an element in J. So this is an element in J. So our element Z is in J. So we find that phi to the minus phi of J is equal to J, and which means that this map is inverse to this. And so they are bijective to each other. OK, so this is again a quite elementary thing. So today, I mostly do very simple things, but also history. I just start again slowly, and then we come to more. I'm more interested in things later, but not really today. So another thing that one has is we had defined the subgroup generated by some element. So we can also look at the ideal generated by some elements in a ring, which is in some sense easier. So the definition, let R be a ring, and say A1, 2, An, some elements in R. The ideal generated by them is just, so it's denoted like we had denoted the subgroup generated by it, but always be clear from the context which one we mean. In this case, it's just a set of all linear combinations of these elements with coefficients in R. So this is the set of all A1, R1, plus An, Rn, R. Yeah, so I was not. So I want this to be a commutative ring, because otherwise this will not work. And we have our elements A1, 2, An, in R. So with this, we look at all these linear combinations where the Ri are elements in R. So we just look at all these combinations. And so it's easy to see. It's very trivial. This is an ideal in R. Of the simplest cases, obviously, that this ideal is generated by only one element. So for A in R, the set A, the ideal generated by this one element, which I could also just write as A times R, which is a set of all AR with R in R, it will be called the principal ideal generated by A. So for instance, in Z, we have N, which is just equal to NZ, which are all elements divisible by Z. We will see later that in Z, all ideals are principal ideals. So every ideal can be written as an ideal generated by just one element. And the ring with this property will be called the principal ideal ring. But as I said, we'll see that later. So just one. So for security, say let R be an integral domain. So we find that the principal ideal generated by an element, by two different elements, will be the same if they only differ by multiplication by unit. So that A and B, the elements of R, then the ideal generated by A is equal to the ideal generated by B. If and only if there is a unit, a invertible element for the multiplication, U in R is equal to U times A. So it's easy to decide when two principal ideals are equal. That's again basically obvious. So if A is equal to B, then it means in particular that B lies in the ideal generated by A. So it's a multiple of A. Then B is equal to U A for some U in R. The only thing we have to see is that this U is a unit. Well, we also have that A lies in the ideal generated by B. So we also have that A is equal to W times B for some B for some W in R. So if we put this in here, we have that B is equal to W, so to U times A. And A is equal to W times B. But as we are in an integral domain, we can cancel the factor B. So it follows that 1 is equal to U times W. And so this means that W is the inverse of U and U is a unit. So this shows this direction. This direction we have to, is in some sense, obvious if they are related in this way, we want to say that they generate the same element. Conversely, if B is equal to U A with U a unit, well, first B lies in the ideal generated by A, because it is A multiplied by something. So the ideal generated by B consists of all elements of R which you obtain by multiplying B by something. But these are then also, that means I can also multiply the, so this is all W times B, but it's all the same as all W U times B. So it follows that the ideal generated by B is contained in the ideal generated by A. And now, obviously, if B is equal to U A, then A is equal to U to the minus 1 B. And so we can change the role of A and B. So also, equal to U to the minus 1 B plus A is contained. So first, A is an element in B, and thus, A is contained in the ideal generated by B. OK, this was another simple remark. So then, we want to see that the ideals, so in a field, they're essentially no ideals. We know that in any ring, the set consisting of the element 0 and the whole of the ring are ideals. And in a field, the claim is these are the only ones. And this also will lead to the fact that any non-zero homomorphism which starts from a field has to be injective. So let's state this in Mark. So let I be an ideal in a ring. So if this ideal contains a unit, then it is the whole of R. Second statement, which is an easy consequence, is that the only ideals in a field, or k, are 0 and k. And the third, which is a consequence of that, is that, so let k be a field, and assume we have a homomorphism from k to any ring, a homomorphism to a ring. Then either phi is injective or phi is the 0 map. So by this, I mean that phi of x is equal to 0 for all x and k. Or phi is injective. So this is all very simple, actually. So let's do the first one. So assume we have such an element which lies in i and is a unit. So let A be an element of i, which is also a unit. So i is an ideal. So the inverse element of A lies in R. So if I take A to the minus 1 times A, this will lie in i. Because the product of an element in i with an element in R is equal to 1. We find that the element 1 lies in i. And so thus, if I take any element x in R, then I can write x equal to x times 1. And so this is now a product of an element in R with an element in i. And so this is an element in i. So this is the first one. Second one is a direct consequence. We know, I mean, we use that the units in a field are precisely all the elements in the field which are not 0. This is the definition of a field. And so if I have an ideal which does not only consist of the element 0, it contains a unit. And therefore, it is equal to the rule of k. So this follows directly from 1. And the third one is a direct consequence of 2. Because the kernel of phi is an ideal in k. And thus, either the kernel of phi is equal to 0, which means phi is injective, or the kernel of phi is equal to the whole of k. It's the only other ideal, which means phi is 0. So when we were discussing normal subgroups, one thing that one could do with normal subgroups was that you could take the quotient group by the normal subgroup. So the set of cosets by this normal subgroup was naturally a group. And so as I said, the ideals are somehow analogous to the normal subgroups in a group. So we also want to take the quotient of a ring by an ideal and see that that is also a ring. So let me do this. Let R in ring and I in ideal. So we know in part of the definition of a ring is that the additive group of the ring is a commutative group. So R comma plus is commutative. So I, which is a subgroup of R comma plus, is a normal subgroup. So we do have the quotient group for the addition. So thus we have the quotient group, R mod I, which consists of all the cosets with respect to I. So this set of all X plus I with X and R, where X plus I is just the other word for the equivalence class. So also the equivalence class of X for the equivalence relation as we had before for the groups, that X is equivalent to Y if and only if the difference is an element in I. Then the equivalence classes will be precisely these. And the set of equivalence classes is this quotient set. And this is a group. So this is, as I said, is a group with X plus I plus Y plus I in defined to be X plus Y plus I. So in future, however, to simplify rotation, I will always write this just as equivalence class like that. And now we want to show that this thing is also a ring. So we're going to call it the theorem. That's the Godfrey composition. So we have the same situation for R as a ring. And I and R is an ideal. So R mod I with the operations plus Y, this is what we have just written, equal to X plus Y. And X times Y is a ring. So the zero element will be R of 0. And the natural projection, we had the natural projection before, just a map which sends every element to its equivalence class. So pi from R to R mod I, which sends any element X in R to its equivalence class or its coset, is a ring homomorphism. So this is very similar to what we did for the groups. It's also not more difficult. We just have to. So essentially we also have already seen it. So we have seen that so that we get an additive group if we just take this sum. And we have to then therefore show that the product is well defined and makes this thing into a ring. And that this is a ring homomorphism. So this is a ring homomorphism. And obviously it is a surjective ring homomorphism. That the map is surjective is obvious because this map is obviously surjective. OK. Not so high. So by what we have just seen, we have that R mod I together with this addition that we had defined. So we were right to do this as before with X plus Y equal to X plus Y is a commutative ring, commutative group. And this map pi is a group homomorphism. So from R plus to R i plus is a group homomorphism. This we have already seen because this we already know from what we did about groups. OK. Maybe I should also say here that this thing has, this homomorphism has obviously a kernel. So we can, given any ideal, we can find a group homomorphism by taking this quotient so that the kernel is that ideal i. And obviously this is a group homomorphism with kernel i because the inverse image of 0 is the equivalence class of 0, which is i. So we want to show the product is very defined. So that means if we have two different representatives for this class of X and the class of Y, we still get the same thing for the product. So if the class of X is equal to the class of X prime and the class of Y is equal to the class of Y prime, then this means in other words that X minus X prime lies in i and Y minus Y prime lies in i. So then if I have XY minus X prime Y prime, I can kind of write this in a different way to show that this difference also lies in i. In the I write this as X minus X prime times Y minus, I hope it's correct. Anyway, you can check this, X prime times Y prime minus Y. So anyway, the things that I want is this one, this one, and these two are the same. So that's OK. And you see that this thing is an element in i and this thing is an element in i. So it follows that this whole product lies in i and this whole product lies in i. So the whole thing lies in i. That means the product of X, the class of product of XY is equal to that for X prime, Y prime. So the product is very defined. Only depends on the equivalence class. And then to see that, so the product is very defined. So to show now that this gives us a ring, we have to show associativity and distributivity. Associativity, the associative law and the distributive law. Well, this is completely trivial because it follows directly from that in R. So if I just do it for the associative law, so I want to say I take A times B times C. My definition, this is A times B, C. And this is the same as A, B, C. Now you can already see that we don't remember the brackets anymore, so it's obviously anyway A, B. OK, so it's just moving around these dotted brackets, so it's kind of clear. The associative law and the distributive law is as simple. And at pi as a ring homomorphism is also obvious. So if I take elements, so for x, y in R, I have pi of x, y. My definition is the class of x times y, which is class of x instead of y, which is pi of x times pi of y. So this is all. So anyway, so we find this quotient ring. And so if you have an idea, you can divide by the idea. You get a new ring, which works precisely in the obvious way. OK, so we can, if we want to, in particular, so we have this ring homomorphism from R to R mod i. The kernel is i. So we can, therefore, describe the ideals that contain i, a rolling projection between the ideals of R mod i and the ideals in R containing R. As we've seen, the projection is given by taking the inverse image of the natural projection. OK, now finally, how much time? Finally, I want to show also the other thing we showed for groups so that we have the homomorphism theorem. And we can basically, if we have a homomorphism between rings, and then we can divide by any ideal, which is contained in the kernel. So again, it's not really so much of a theorem. But anyway, it's a universal property of the quotient. So let phi from a to b be a ring homomorphism. And do we take an ideal in A, which is contained in the kernel of phi? Then we can, one says then one can factor the map over the quotient by i or through the quotient by i. So then there's a unique homomorphism phi bar from a mod i to b. Such that you either write this as a commutative diagram, such that this diagram A phi b, you have the natural projection to a mod i. And here we have phi bar commutes. This just means if you go this way, same as if you go first down here and then there. So in other words, phi bar is equal to phi bar composed of i. And we can see furthermore, the kernel of phi bar is equal to the kernel of phi modulo i. So the quotient by this ideal. And the image of a mod i, the image under phi bar is the same as the image of phi. Well, this takes much longer to state this than actually to prove it. If you look at it, we have imposed this condition that phi bar should satisfy this property that I have composed it with phi, with phi, then I get phi. But this will determine the phi bar first. So the uniqueness of the homomorphism will be obvious. Namely, so if that phi equal to phi bar composed with pi, this fact means that if I take phi bar of any element x, so of any plus x of any element in a mod i, this has to be, by this formula, equal to phi of x. So if phi bar exists, it is unique, because it's written down the formula for it. And obviously, we'll use this definition. We only have to define to prove it's well defined. We have to show with this definition it's well defined. Well, and that's kind of obvious. So if the class of x is equal to the class y, then, as we have seen now a few times, x minus y is an element in i, which is contained in the kernel of phi. So it follows that phi of x minus y is equal to 0, which is equal to phi of x minus phi of y. So that's it follows. So it means that this phi bar is well defined. The image of the class depends, does not depend on the representative. You always get the same result. And it's immediately to check that this is a ring homomorphism. We've done similar things already a couple of times today. You just check that if you take the sum, it maps to the sum. If you take the product, it maps to the product. This follows immediately from the fact that phi is a ring homomorphism. And then it is clear we have to find phi of an equivalence class to be phi of the representative. So therefore, the image of phi bar is the same as the image of phi. Because every element in the image of phi is obtained. This way, we just take any representation. We take the class for any class. So if we have something which is in the image of phi, we just take the equivalence class of the thing that maps to it, and this will map the phi bar. So really by definition. Basically, also by definition, we have that the class of x is in the kernel phi bar, if and only if phi bar of is equal to 0. This is equal to phi of x. So that means if and only if phi of x is equal to 0 for A for 1 or equivalently for every representative of x, so this means if and only if. So if and only if x is in the kernel of phi, which is if and only if the class of x is in the kernel of phi, or to go up. Because we have just the equivalence class that just taken the class, or to go up. So maybe I can also finish this section by finishing with the homomorphism theorem, which is a straightforward equation. It says, like with the groups, that if we have a subjective ring homomorphism with kernel i, then this gives us an isomorphism of the quotient by i with what you are mapping. So let phi from A to B be a subjective ring homomorphism with kernel i. Then the map phi bar from A mod i to B is an isomorphism. So in particular, A mod i is isomorphic to B. And that's basically clear from the universe property. So by the universal property, we have that. So this phi bar is a, so this map was a subjective. So phi bar is a subjective ring homomorphism with kernel. What's the kernel? So the kernel of phi modulo i. So the 0 is i mod i. So the kernel is just a 0 element of A mod i. So that means it is also injective. So maybe I stop. So I have, as you see today, we did not really do anything interesting. Maybe I went a little bit fast because of it. I don't know. On the other hand, there was no argument which was not very easy. So just everything we did was kind of analogous to what we did for the case of groups. And the proofs are very simple, similar. So hope next time we will do a little bit more. We will talk about prime ideals and maximal ideals, which is creeper and principle ideal domains. So anyway, we'll see each other on Friday.