 So last time we have been started to talk about rings. So we introduced rings and talked about sub rings 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 in the first definition, a ring with a unit ring, so ring R with 1 is called a division ring. Well, if every non-zero element is invertible. So if R star is equal to R without 0. 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. The 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 the 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 0 which is different from 1. The two elements 0 and 1 which are different from each other is called a field. The first, the addition, so R together with a plus, is a commutative ring with neutral element 0, a complexative group. This was part of the definition of a ring. So second one is that if I take R without 0 with a multiplication, this is a commutative group with neutral element 1. And secondly these two structures have to be somehow compatible and this is by the distributive law. So which just says that if I take A times E 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. We have Q, so the rational numbers, the real numbers, and the complex numbers are fields. I want to also show that there are some finite fields. And for this I, so a field with finitely many elements. 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 1, which has no zero devices. 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 the units are without zero. So in other words, we have to take a non-zero element in R and have 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, it says that a, so now for this given a times b plus c is equal to a b plus a c. 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. So it's a group homomorphism. And we know that R has no zero divisors. So in particular, there is no, say, 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 is surjective, so it means b is equal to a to the minus one. And 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. 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 of the 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 by objective 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. Okay, 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 zero. 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 homomorphisms 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. So 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 is also the negative for any element. We want just the difference of any two elements in I lies in I. And the second statement is something more than for a subgroup, 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 X, A is in I. So this is a weaker condition. So in particular, an ideal is a subring, but it's something better. So for the moment, so typical, so say, I have set zero is an ideal. And if I take the whole of R, this is an ideal in R. This is basically obvious from the missions. Let's now see what I just said before that I didn't write it. I just said that the kernel of a ring homomorphism is an ideal. So let me state that lemma that 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 V from A to B be a ring homomorphism. Then the kernel of V 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 zero is an ideal in B. So I just applied to this, I get that the kernel, which the inverse image of zero 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 V to the minus 1 of 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 homomorphism, this is phi of A times phi of X. This is phi of X is, you know, 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, 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. 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 also have written that. So here it's okay. And so AX in phi to the minus one of I and XA in phi to the minus one of I. Okay, 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 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 to 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. We need 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 ideal from the ideals in A containing kernel phi to the ideals in A and 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 zero and its inverse element contains the inverse image of zero, 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, okay. So let's J be an ideal in A. I don't even need to have to contain the kernel of phi. Then we want to see that its image is an ideal in B using that the map is subject. 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 LTF 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 phi of, 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 we want to go. 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 V. 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 B. It's just the property of being, this means all, this is the set of all elements which map to elements in I. And then I map phi to it so I get I. This is by definition without any assumptions on what I is. And conversely, let J be an ideal in A containing the kernel of phi. 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, you know, that's, again, just the definition. So we have to see the other inclusion. So which one do I want to answer? 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. Actually, okay, 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 zero 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. Okay, so this is again quite elementary thing. So today I mostly do very simple things but I just start again slowly and then we come to more. It's like more interesting 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 element in a ring which is in some sense easier. So definition that R be a ring and say V1 to Vn 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. So I want this to be a commutative ring because otherwise this will not work and we have our elements A1 to 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. 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, an 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 say that 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 a unit. So that A and B the elements are then the ideal generated by A is equal to the general ideal generated by B. If and only if there is a unit and a vertical 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 but 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 W in R. So if we put this in here then 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 one is equal to U times W and this means that W is the inverse of U and U is a unit. So this shows this direction. This direction we have to, in some sense, be 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 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 minus 1 B and so we can change the role of A and B. Also A is 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. Okay, this was another simple remark. So then we want to see that the ideals, so in a field there are essentially no ideals. We know that in any ring the set consisting of the element zero and the whole of the ring are ideals and in a field the claim is these are the only ones. And this also leads 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 called K are zero 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 zero map. So by this I mean that phi of X is equal to zero 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 in element 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 one times A this will lie in I because the product of an element in I with an element in R is equal to one. We find that the element one lies in I and so thus if I take any element X in R then I can write X equal to X times one 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 zero. This is the definition of a field. And so if I, if I have an ideal which does not only consist of the element zero it contains a unit and therefore it is equal to the whole of K. So this follows directly from one. And the third one is a direct consequence of two because the kernel of I is an ideal in K and thus either the kernel of V is equal to zero which means phi is injective or the kernel of V is equal to the whole of K. It's the only other ideal which means phi is a zero. 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 kind of said 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 ring and I in ideal. So we know 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 other word for the equivalence class 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 we will precisely see 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 defined to be X plus Y plus I. So in future however to simplify notation I will always write this just as equivalence class like that and now I want to show that this thing is also a ring. So when I call it the theorem that's about what we mean for position. So we have the same situation for R the 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 Y is a ring. So the zero element will be as of zero and the natural projection we had the natural projection before just a map which sends every element 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 subjective ring homomorphism that the map is subjective is obvious because this map is obviously subjective. Okay, not so high. So by what we have just seen we have that R mod I together with this addition that we had defined we will write it as 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. Maybe I should also say here that this thing has this homomorphism has obviously as 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 zero is the equivalence class of zero I. So we want to show the product is well 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 X, Y 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 okay 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 X, Y is equal to that for X prime, Y prime so the product is way defined only depends on the equivalence class and then you know to see that so the product is way defined so to show now that this gives us a ring we have to show associativity and distributivity the associative law and the distributive law well this is completely trivial because it follows directly from that in I 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 okay so it's just moving around these brackets so it's kind of clear this associative law and the distributive law is as simple and at pi as a ring homomorphism is also obvious so if I take elements 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 the inclusion 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 okay so we can if we want 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 we have to contain I for all the ring there's a projection between the ideals of R mod I and the ideals in R containing I as we've seen the projection is given by taking the inverse image of the natural projection we finally 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 it's again it's not really so much of a theorem anyway it's a universal property of the quotient so let phi from A to B the ring homomorphism and do we take an ideal in A which is contained in the kernel of phi then we can this 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 either write this as a commutative diagram such that this diagram A phi B we 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 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 if you look at it we have imposed this condition that phi bar should satisfy this property that I compose it with phi with phi then I get phi but this will determine phi bar first so the uniqueness of the homomorphism will be obvious namely so if that phi equal to phi bar composed with phi in fact means that if I take phi bar of any element X so of any class 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 to use a written down formula for it and obviously we will use this 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 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 thus it follows so it means that this phi bar is well defined the image of the class depends does not depend on the representative and you know it's immediately to check that this is a ring homomorphism we have done similar things already a couple of times today we just check that if you take the sum if you take the sum if you take the product if you take the product this follows from the fact that phi is a ring homomorphism and then it is defined 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 you know of course we just every element in the image of phi is obtained this way we just take any we take the class for any class which is in the image of phi we just take the equivalence class of the thing that maps to it and this will map it under phi bar so really by definition and basically also by definition we have that the class of X is in the kernel phi bar if and only if phi bar is phi of X so that means if and only if phi of X is equal to 0 for a for 1 for every representative of X so this means if and only if so if and only if X is in the kernel which is if and only if class of X is in the kernel because we have just equivalence class I just take the class so maybe I can also finish this section by finishing with the homomorphism theorem which is a straightforward application it says like with the groups that if we have a subjective ring homomorphism with kernel I homomorphism of the quotient by I with what you are mapping so so let phi from A phi is a subjective 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 clear from the universal property so by universal property we have that so this phi bar is a so this map was subjective so phi bar is a subjective ring homomorphism what's the kernel so the kernel of phi modulo I so the 0 is I mod I so the kernel is just the 0 element of A mod I so that means it is also injective so so maybe so I have 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 we there was no argument which was not very easy so so it's 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 discrete and principle ideal domains so anyway we'll see each other on Friday