 So we were, we wanted to finish group theory today. We were talking about the CELO theorems, and the only thing that was missing was the proof of the third CELO theorem. Ah, so I'm supposed to, okay. So we had, so the third CELO theorem said the following. So if, again, we have that, we have a group such that the number of elements is n which is equal to p to the m times r. And we assume, as usual, that p does not divide r. So then, so we look at, we want to know how many p-CELO subgroups are there. And so the statement is if let S be the number of p-CELO subgroups. Then, we have first that S divides r. I will now kind of always use this notation for divides, no? And S is congruent 1 modulo p. So if you divide S by p, the rest is 1. So we don't precisely know what the number of p-CELO subgroups is, but we get some good estimate we have seen in some of the applications. That is sometimes when numbers are not too big, for instance, allows you to conclude that the number of p-CELO subgroups is indeed 1. We had some cases where this was the case. Okay, so now we want to come to the proof. So we had them a corollary to, so proof. So a corollary to the second CELO theorem told us that all p-CELO subgroups are conjugated. All CELO subgroups are conjugated. Now, you remember this, the statement was much more precise. Contrugated anyway, so I may be right in such a way that I can actually read it, although. So they are conjugated to each other. So I can say they are conjugated to one given one to one p-CELO subgroup, which I call H, so I've kind of picked out one. So we have that G acts on these p-CELO subgroups by conjugation. So by the stabilizer under this conjugation action of H is the normalizer of H. So that means that the number of CELO subgroups, which is just the number of elements in the orbit of H. So it means that S is equal to the number of elements in the orbit of H is therefore equal to the index by the orbit stabilizer theorem to the index of NH in G. Now NH is a subgroup of G and H is a subgroup of NH. So H is a subgroup of NH and NH is a subgroup of G. So therefore we find that the index of NH in G, which is our number S, divides the index of G in H. Namely, by an exercise we know that we just have that this number, the index of G in H, is just a product of this and the index of NH of H in NH. So this was an old exercise in the first session that G H is equal to G NH times NH H. Well, and so that's it because the index of the p-CELO subgroup is precisely R, because the group H has p to the m elements. So this proves the first statement. So this first statement is basically obvious. Now we have to show that S is congruent to 1 modulo p. It's also not very difficult. So we look at this set, so first we look at the set H1 to HS is a set of p-CELO subgroups. Now, we know that G acts on it by conjugation and the subgroup, if I take a subgroup of G, it also acts on it by conjugation. So I can take H to act on, you know, also the subgroup H of G acts on this set of by conjugation. So G acts on the set by conjugation and therefore also H. So we can decompose this set into orbits for the action of H. We know that if you take the action of G, there's only one orbit because they are all conjugate. But H is a subgroup, so there might be more. And I will, so these are, we know that, so one of these CELO subgroups is after all, the group we chose, so the first one we call H. If H was one of our p-CELO subgroups. So we have this, so we want to know how many, you know, we want to see that S is congruent to one more p. So we want to see that, basically we want to see that there's one orbit which has one element and all the other orbits have a number of elements divisible by p. So we can look at the orbit of H. So obviously, so the orbit, I'll first say, the orbit of, do I really, you know, of H i, consists only of the element H i itself, so only of one element, if and only if H is contained in the normalizer of H i. The normalizer of H i are all the elements such that if I conjugate with it, I get back H i. So and you know, we want to, we look at the conjugation with elements in H, so they won't do anything to H i if and only if H is contained in the normalizer of H i. So now we can, so in this case, I say that both H and H i are pseudosubgroups of the normalizer of H i, because the normalizer of H i is a, you know, is a subgroup of G. So the maximal power of p that divides the number of elements can be at most this m, but H is contained at it, so it is m. So the pseudosubgroup of it will have p to the m will be this h. There cannot be a bigger p group inside n of H i, because, you know, it must be contained in G, the number of elements. So maybe the number of elements in n of H i divides the number of elements in G, because it's a subgroup of G, so thus which is equal to p to the m times r. So it follows that the largest power of p which can be contained in this is p to the m. On the other hand, we do have a subgroup of order p to the m contained in it, because h is, in under our assumptions, h is contained in it, so therefore this is the p0 subgroup. But if this is the case, so H i, by definition of the normalizer, H i is a normal subgroup of n of H i, you know, because the, you know, n of H i is just all the, so just if you look at the definition, it's a normal subgroup. So all the p0 subgroups of n of H i are conjugate to each other in this n of H i, so and, but this one is a normal subgroup, so it means all p0 subgroups are equal to this one. So it means H i is equal to H, a conjugate, now in n of H i, it follows that, you know, as it's a normal subgroup, it cannot be conjugated to anything else than itself, it follows that H is equal to H i. So under the assumption that the orbit of H i consists only of H i, we find that H i must be equal to H, okay. On the other hand, if we look at H, obviously the orbit under conjugation by H of H is also H itself. So we find there's precisely one orbit which has only one element, so thus the orbit of H is the only one with only one element. And now we will want to, you know, so by the orbit stabilizer theorem, so if we have any out, for any other, for any other H i, and by the orbit stabilizer theorem, we have that the number of elements in the orbit is a divisor of the number of elements in the group that acts, which is H. So the number of elements in the orbit, so under, so H acts on this, the under of elements which lie in the orbit under conjugation of H on this, so number of elements in the orbit. So the number of elements in the orbit is a divisor of the number of elements in H, which is P to the m. So and it's a divisor which is not one, because we know that the only case when the number of elements in the orbit is one is when this pseudosubgroup is H itself. So it follows thus the number of elements in the orbit is divisible by P, no, and this is not one. It's different from one, okay. So thus this number is divisible by P, so we find we can decompose the set H1 to Hs into disjoint subsets, namely the orbits, so thus H, so this thing H1 to Hs is a disjoint union of orbits. So such that one of them has one element and the order of all others is divisible by P, have a number of elements divisible by P. So it follows that the number S, the number of elements here, number S, leaves rest one if one divides by P, thus S is congruent to one. So it's not, okay, so that's, that is it. So it's not, this one is in some sense is not so difficult but maybe slightly tricky to take this action by H, but in some sense we just use the standard results of a stabilizer theorem a few times and there was nothing really fancy going on. And maybe kind of using that, if you have a normal subgroup then it's fixed by the conjugation and whatever, okay. So maybe that's now it for the, for the CELO theorems, are there questions about this? I should say that you are looking, anyway, so I mean I can also, if it's not clear I can also repeat some part but in some sense it's really an elementary argument, it's only a bit more complicated than what we had in the beginning. So the CELO theorems were certainly the most complicated part of the group theory. I mean mostly because we somehow put everything together that we learned until now and kind of expected to be able to use it freely without, kind of, okay. So maybe that's now enough if there are no further questions, I will start with rings. So basically in this group theory there was a certain acceleration, it got slightly more difficult towards the end and with the rings we start again by zero and it again starts quite easily. We just give the basic definitions of rings and the first properties. So basically, from school you kind of know the integers and you kind of know that you can add them and you can multiply them. And you can also take, you have a neutral element for the addition and the additive inverse in the integers. But you don't always, you don't necessarily have a multiplicative inverse of every element. So a ring is something which kind of has this property that you have some set where you have an addition and multiplication and the addition forms a group and the multiplication at least is a little bit compatible with the addition. And so that some of the properties that you're used to from the integers still hold most of them. So let's just give the definition. So a ring is a non-empty set R together with two binary operations. Operations, which just means maps from R times R to R. So plus from R times R to R AB is sent to A plus B. And times from R times R to R AB is sent to A times B. Such that with the addition R is a commutative group. So also we have a distinguished element and an element C0 in R. Such that if I take R together with the addition, this is a commutative group like the integers with addition with neutral element 0. And we will always know for this additive group, we use the additive notation. So the inverse of an element is minus A and so on. This is the first thing. And then we have this operation and it should be somewhat compatible. And this is reflected first, the multiplication should be associative. And second, it should be compatible with the addition, which means there's a distributive law. So such that maybe, so this is the first statement. The second is we have the multiplication associative. So that means for all A, B, C in R, we have A times B times C. This is the same as if I do it in the other order. And we have this compatibility between additional multiplication, which is the distributive laws. Namely, we have A times B plus C is equal to A times B plus A times C. And say B plus C times A is equal to B times A plus C times A. So I don't here, for the moment, I do not assume that the multiplication is commutative. So this is it. And so this is the definition of a ring. I do not require that I get a group with a multiplication. And I do not require that multiplication is commutative, but we have these laws. And I call R with just operation plus. I call is called the additive group of R. And I call 0, one can also call the 0 element of R. And I will, as usual, from now on, I can also just not write the dot for multiplication, as I did also for the multiplication in groups. And obviously, the typical example is that if I take Z, together with the usual addition and the usual multiplication, this is the ring. And the 0 is the 0 of R. Maybe I don't need that now. So now that we have rings, we can define subrings. So like with groups, if you have a ring and you have a subset of a ring, it's called a subring. If by restricting the operations to this subset, you again get a ring. So practically, that means, so to check it, this means the definition that R be your ring and A subset R subset. It is a subring of R. By definition, if the following holds first, we have that if I take A plus, so A is a subgroup of the group R with the addition. Now we know how one checks for a subset to be a subgroup. For instance, for every element, A and A minus the element is contained for any two elements, A B and A A plus B is contained. And the second is that it is closed under multiplication. So for all A, B, and A, we have that A times B is an element in A. And then obviously, by definition, if I take A and give it the restriction of the addition on R and the restriction of the multiplication to A, this is again a ring. So a typical example, a simple example, is for instance 2z, which we had before. Set of all 2n such that n is in z is a subring of C. So we'll later restrict attention to commutative rings. So what does it mean for a ring to be commutative? We have assumed in the definition of a ring that the addition is always commutative. So the ring is commutative if the multiplication is also commutative. It's called commutative if the multiplication is commutative. So if A times B is equal to B times A for all A, B in R. As I said, the addition is anyway commutative. So for instance, the integers are commutative ring. Of course, multiplication is commutative. In the integers, we also have a neutral element for the multiplication. If you multiply anything by 1, we get back the element. So this would be such an element is called a unit element. So definition, an element 1, I noted by 1. But in some cases, it would be not the number 1, but just something called 1, which is an element R, which is different from 0. So we require this element to be not the 0 element. It's called a unit element. So in a moment, we will also talk about something which is called a unit. That's something different. So a unit element is the element 1. It's called a unit element if it's a neutral element of multiplication. So if for all elements A and R, we have 1 times A is equal to A times 1 is equal to A. And such an element does not always. And so it's easy to check. I mean, you can do it. It's like for an easy exercise to see that in this case, if such an element exists, it is unique. And check that. And if R does contain such a unit element, one calls it a unitary ring or a ring with 1. So I'm only telling you definitions. So nothing at all is happening. There are all these words which eventually you will have to master. But obviously, I have not done anything. So if R contains a unit element, then it's called a unitary ring or a ring with 1. I usually just use that or a ring. So we can see that, for instance, the integers Z contain a unit element, namely the 1 in Z. And 2Z does not contain a unit element, because whatever you multiply any element here with, you always get something different. So later, we will mostly be interested in commutative rings with 1. Can you just kind of check it? So let's look at at least one case of a non-commutative ring example. So we can look at, say, so that R is the real numbers. We look at, say, M2 times 2 of R, which is the set of all 2 times 2 matrices with coefficients with entries in R in the real numbers. So the elements look like A equal to A, B, C, D, where A, B, C, D are real numbers. And I claim this is a ring. So the addition in the ring is just the component-wise addition, as usual. So A, B, C, D plus A prime, B prime, C prime, D prime is, as usual, A plus A prime, B plus B prime, C plus C prime, D plus D prime. So you just add them as vectors. And the neutral element for the addition, obviously, is 0. So 0 element is the matrix where all entries are 0. And the multiplication in the ring, you take the matrix multiplication. Maybe I don't know multiplication equal to the matrix multiplication, like you learn in high school or in the first semester. So if I would do it for the same thing, you would have for multiplying the same things with each other, you get A times A prime plus B times C prime, A times B prime plus B times D prime. And then you have here C times A prime plus D times C prime and C times B prime plus D times D prime. So the usual multiplication. And as usual, this is actually a ring with 1. So M2 times 2R is a unit ring. And the unit element is the identity matrix with unit element, the standard identity matrix. OK. And maybe I can also come back to these integers modulo k. I want to use them. Also another example. So if k is a positive integer, I had introduced the set. So zk was supposed to be the integers modulo k. So this is just a set. It's just a set 0, 1, 2k minus 1, as we had before. We had defined the addition by saying that in this group, we have, for the moment I write the addition differently, n plus m. It's just we take the sum n plus m. And then we take the rest after division by k. And we can, in the same way, define also the multiplication, namely the product of n times m. It's just we take the product of these two numbers in the integers and we divide by k and take the rest. And you can check that this is, again, a ring, a commutative ring because the multiplication in the integers is commutative, obviously, ring with 1. So the 0 element is element 0. And the unit element is element 1. Anyway, this is quite standard. And I think this you might also have learned in high school. OK, or at least we had it for the addition. It works the same way for the multiplication. So we make some trivial remarks that some of the obvious rules that you have when you work with integers also hold in any ring. So just, for instance, in the integers you know if you multiply anything by 0, then you will get 0. And this also holds in any ring. Then if you multiply anything by minus something else, get minus something else. So minus a times b is equal to a times minus b is equal to minus a times b. And also say if you take this twice, you get this. And if you multiply, so if r is a ring with 1, so r is a ring, so let r ring. And now I'm above the line, but it's not. And a b in r. So if I assume that r is unitary, so if r is a ring with 1, then we have if you take minus 1 times a is the same as a times minus 1 is equal to minus a. So these are all things that you certainly have seen maybe even before you came to high school, but now we are talking about an arbitrary ring, not about the integers. So I just want to prove them. It's all kind of obviously trivial, but it's some exercise in playing with these axioms like we did for groups. So if I take a times 0, now 0 is a neutral element for addition. So that's the same as if I do it twice. I added 0 plus 0 is 0. By the distributive law, it's a times 0 plus a times 0. Now the addition is a group. So we can subtract a times 0 from it on both sides. So it follows that 0 is equal to a times 0. And obviously with the other one, you can imagine how to modify the proof. And for second statement, it's quite similar. So we want to show, for instance, that if we take minus a times b, that this is the additive inverse of a times b. So we have to see that if we add this to a plus b, we have to see we get 0. Now again, we can use the distributive law. So this is minus a plus a times b. And minus a plus a is 0. And so we find that this is the inverse of that. And the other one is similar. Now if you, the third one is just applying this twice. You do first do it here. This is minus a times minus b. And this is minus minus a times b. And minus minus is just a. And while the fourth one is actually essentially just a special case. So if I take minus 1 times a, this is according to what we said here, this is minus 1 times a. This is minus a. So this is not very difficult. So now we want to introduce a couple of further properties that rings sometimes have. So one nice property that the integers have is that it does not happen that you have two elements which are non-zero and you multiply them and get 0. And so a ring which has this property, say if it's a commutative ring, which has property is called an integral domain. So we can introduce this property. And we will later mostly be interested in integral domains. So definition, so as we assume now, we maybe just look at commutative rings. So let R be a commutative ring. So an element, say b, say a and R, and the element is not supposed to be the zero element, is called a zero divisor. Well, in some sense, if it is divisor of 0, so that if you can multiply it with another element to get 0. So if there exists an element b in R and this element should also be different from 0, such that a b is equal to 0. We know that this is something which cannot happen in the integers, for instance. And a ring which does not have zero divisors is called, so commutative ring with 1 without zero divisors is called an integral domain. Zero divisors is called an integral domain. So somehow it says it's very much like the integers. And as we know, it's obviously clear that z is an integral domain. So if we are in an integral domain, we can cancel factors from products, like we can in groups, I mean, trivially. So that means we have the cancellation property. So if we just have need that we have no zero divisor, but I say integral domain because I don't worry about other cases. So let R be an integral domain. So if we have elements, if we have a, b, c in R such that a, b is equal to ac, then such that then b is equal to c. So we can cancel common factors. Well, and this is quite obvious. So we assume that a, b is equal to ac. So we can bring this to the other side. So we have a, b minus ac is equal to zero. It's the same as a times b minus c. So now a is non-zero, and we have a product of two elements so that we get zero. As we are in an integral domain, one of the two elements must be zero, and a is non-zero, so b minus c must be zero. Yeah, and so I failed to make the assumption that a is different from zero. Such that a is different from zero, and then b is equal to c. B minus c is equal to zero, and thus b is equal to c. So let's see what the few rings that we have seen so far, which of them are integral domains in which are not. So we have seen that z is an integral domain. Then if we look at, say, z6, so integers, modulo 6, this is not an integral domain because if we take the product of two, now I write the usual multiplication. Two times three in this ring is equal to what you do if you take six and take the rest dividing by six of that, which is zero. So we find actually two non-zero elements whose product is zero. And on the other hand, so one could think that all these z n's are not integral domains, but so if p is a prime number, then we can take, so if I take any element, so if n is different from zero, so if p does not divide n and p does not divide m, then sp is a prime number, but also does not divide the product. So that means that if I have elements n and m in z mod p, so thus if n is different from zero, also m in z mod p in zp, then n times m is also different from zero in zp. So we find that zp is an integral domain. OK, so I had introduced these, so in a group, we have that every element has an inverse. In the additive group, obviously you have always a negative of any element because that's a group. But in the multiplication does not form a group, so we do not require that every element has an inverse. But those elements which do have an inverse, we call the units. So the units in R, so which is a ring with one, are elements with a multiplicative inverse. So that means, in other words, so an element A in R is called a unit if there exists an element B in R, such that A times B is equal to B times A is equal to 1. And so the set of units in R is denoted R star, and now the tiny. So it's easy to see in group theory that if an inverse exists, it is unique, and we denote the inverse then by B to the minus 1. So B will be unique, denoted A to the minus 1. You can check that it's unique. So for instance, as an example, what are the units in Z? So these are the elements which have an inverse for the multiplication. So this is only 1 and minus 1. And whatever if we take the rational numbers, then the inverse, the invertible rational numbers are precisely the rational numbers minus the element 0. Every rational number which is not 0 is invertible. So in these cases, it's easy to see that this set together with the multiplication forms a group. 1 times 1 is equal to 1. 1 times minus 1 is equal to minus 1. Minus 1 times minus 1 is 1. And this indeed is always the case, the units form a group under multiplication. Well, it's almost trivial, but still for some reason I call it a proposition. So let R be ringed with 1. Then R star with a multiplication is a group. And this group will be called the multiplicative group of R. This is essentially, again, trivial maybe as another stupid exercise in the definition. We have to see what we have to check, proof. So we know that the multiplication is associative. So it's also associative if we restrict it to R star. By definition, every element in R star has an inverse. So first, we have the element 1 is in R star. So we have a neutral element. And for A in R star, we also have A to the minus 1 is in R star. So all the actions of a group are fulfilled, except possibly that the product of two elements lies in the group. So let A and B be in R star to see AB is in R star. Well, and that's also not so very difficult, because if we take AB times B to the minus 1, A to the minus 1, well, then this will be A times B, B to the minus 1 to A, A to the minus 1 is equal to 1. And the same way, B to the minus 1, A to the minus 1, AB is equal to 1. So we see that AB is a unit if A and B are units. So AB is a unit. And thus, we have seen that the units form a group. And as I said, this group is called the multiplicative group of R. So I just remind you of the notation that I introduced in the very beginning. So if A is an element in R and N is in Z, we can define N times A, which if N is bigger than 0, is A plus A plus A N times. And we can also define A to the N, which would be the multiplication. No, that's not. So this is if N is positive. So anyway, we can define this. So if N is bigger equal to 0, say bigger than 0, or A is a unit, then we can also define A to the N, which if N is bigger than 0, it's just A multiplied with itself N times. And otherwise, we have to take the inverse and then take it to the minus that power. So this we had introduced before. And we have the usual rules that I also mentioned the other time. So we have, for instance, that N plus M times A is equal to NA plus MA. And N times M times A is equal to N times M times A. And also the same for the powers A to the N times A to the M is equal to A to the N plus M. And A to the N to the power M is equal to A to the N times M. So all these statements are very easy to prove by induction. And I already mentioned them for groups before. OK, so now how much time? So now we want to come to somehow, maybe for us, the most important example offering, or in some sense of a construction of one ring from another, which is the polynomial ring. So if we are given a ring, we want to consider the ring of polynomials with coefficients in the first ring. So I mean, from high school or university, you know about polynomials. So you would say, for instance, in school, you have a polynomial is, say, f equal to, so f of x, you would say, is equal to sum i equals 0 to N, ai x to the i, where, say, ai is an integer, is a real number. And you would view this f as a function from r to r. And you view it as function f from r to r. So to sum element b, you send b to f of b, which is sum i equals 0 to N, ai b to the i. So this is certainly a quite useful viewpoint. But we actually want to see the polynomials more formally. For us, a polynomial would just be, in some sense, a formula. So just these symbols, this thing, is the polynomial. I don't care about the function. For us, a polynomial will be a formal expression. So the polynomial is fixed by saying what these numbers ai are. So let me give the definition, which is strictly speaking only 99% precise, but it's much simpler than that. So definition, let r be a ring. Don't need the one I expect. So a polynomial f, f with coefficients in r is the following expression. So f equal to, say, sum i equal 0 to sum number N, a0, x to the ai, x to the i, for sum n bigger equal to 0, and ai are elements in r. But it's not quite the truth, because if two polynomials only, if these numbers N happen to be different, but at the end, we just have some 0s, they still are the same polynomials. So let me write this down. So the two polynomials f and g, bj, x to the j are equal. And we will just write f is equal to g if, so we have these two numbers N and M. We can assume that one of them is bigger than the other. So assume maybe that M is bigger equal to N. So if, so assuming M is bigger equal to N, we first have that bj is equal to 0 for all j bigger than N. So the extra terms that we have here all have coefficient 0. And ai is equal to bj for 0 smaller equal to i, smaller equal to N. So two coefficients are equal. If two polynomials are equal, if these numbers here are equal, and some extra 0s don't count. So if this, in some sense, so we denote rx. So actually, polynomial f in the variable x with these coefficients. Denote rx, the set of polynomials in x with coefficients in r. So we can see, so the polynomial say a0 equal to, so a0 times x to the 0 is identified with the element a0 in r. So you could say we have a map from r to x, which sends a0 to a0 times x to 0. But anyway, with this, we can view r as a subset of our x. So with this, we have that r is contained in rx. The ring r consists just of those polynomials, which have only, where the only non-zero coefficient is a0. So these polynomials are called constant polynomials. So r in rx is the subset of constant polynomials. So I wanted to, in the beginning, I stressed that for us polynomials are not functions, but they are these formal expressions. They are just such expressions with this identification. And they are not just viewed as functions from r to r. So the polynomial equals sum i equals 0 to n ai x to the i in rx is not the same as the function, which sends an element b and r. So from r to r, which sends b to what one would call f of b, which is sum i equals 0 to n ai b to the i. For instance, one could look an extreme case. So for instance, if r is equal to z2, so r has only two elements. So the total number of functions from r to r is four, four functions from r to r. That's all. But there are infinitely many polynomials with coefficients in r. You can just put any coefficient, any set of finite many numbers gives you a polynomial, and they are all different as polynomials. But infinitely many, but there are infinitely many polynomials in rx. OK, still 10 minutes. So now, in a moment, I've just introduced them as a set. These polynomials now I have to define the ring structure, obviously. So we want the polynomials to form a ring. And so we have to define what the addition and the multiplication of polynomials is, finished. So we define a ring structure on rx. So we have to say how we want to add and to multiply polynomials. So if we have f sum i equals 1, 0 to n an, x to the n, and g i equal. So we can assume that this number is the same, because we can always add 0 times x to the i for the larger ones, because it's the same element. So for this, we can define the sum. So we define f plus g to be how you learn to add polynomials in school. Namely, this is sum i equals 0 to n a i plus b i x to the i. So you just add these coefficients. I should maybe say that a i will be called the coefficient of x to the i in f. And for the multiplication, it's a bit more complicated. You just write down the formula that you get if you do what you learn in school, how you multiply it out by using the distributive law. So this would be the sum over all k, say in this case from 0 to 2n. Take the sum over all numbers i and j such that i plus j is equal to k a i e j x to the k. So this looks a bit more complicated, but it's just what you're used to. Namely, here this means, in other words, we have that if I take a times x to the i times b times x to the j, this is defined to be a times b times x to the i plus j. And then you formally apply the distributive law to this sum. And then this will give you this formula. And with these definitions, one can check that we get a ring. So clearly, we have that the element 0 in R, which is the constant polynomial 0. So all the coefficients are 0. It's the neutral element for the addition. And we get, so you can check, is a ring. So if R is commutative, then also x is commutative. That's obvious if you look at the definition here. No, it's just switch them and get the other definition. And if R is a ring with 1, then Rx is also a ring with 1. And the unit element will be just the same. It is just the constant polynomial 1. So if you look at the definition here, if just one of them is 1, you just get back the original polynomial. If you say f is equal to 1, in this multiplication, you just will get back g. And we also see R is a subring of Rx. So if we look at the multiplication in Rx and restrict it to constant polynomials, it's just the multiplication in R. If you look at this definition, then all this Hocus Pocus Blue vanishes just the sum from k from 0 to 0. And if this is f is equal to a, g is equal to b, this is just a times b. So there's nothing going on. And so we also want to maybe use some kind of simplified notation. So if say we have 0 times x to the 0 plus 2 times x to the 1 plus 3 times x to the 2 plus 0 times x to the 3 plus x to the 4, 1 times this. So we would just write this, write only the terms which matter. So we write this as you would do in school as 2x plus 3x. x squared plus x to the 4. Just we don't need to keep the terms for 0. And we don't need to keep the 1. OK, now we're still two minutes. So maybe so the final, let me just state it. So finally, if the ring r is an integral domain, then also x will be an integral domain. And the units of rx will just be the units of r viewed as constant polynomials. So first I introduce the degree of a polynomial definition. Maybe I stopped there. Let f equal to sum i equals 0 to n ai x to the i view polynomial. And we can always do it in such a way that the highest for the highest for that the an is non-zero. Because if we have any further zeros, we can forget them. So with an different from 0, then the degree of f is, so it's denoted, the degree of f is equal to this number n. So if it's a coefficient, it's the highest power for which the coefficient is non-zero. And we call an the leading coefficient of f. So the coefficient of the highest power which is non-zero. And so I wanted to say, so remark, so assume r is an integral domain. And we take two non-zero elements in r. So two non-zero polynomials. So they are not equal for the constant polynomial 0. Then first, f times g will be non-zero. And in fact, the degree of f times g is equal to the sum of the degrees of f and g. And the second statement is that rx is an integral domain. OK, this is kind of trivial because we have just said if f and g are two non-zero elements, then the product is non-zero. So that means it's an integral domain. So this is just a reformulation of part of one. And the third statement is that if I take the units of rx, this is the same as the units of r. So that means a polynomial can be a unit for the multiplication only if it's a constant polynomial. So it has only terms of degree 0. And these are a unit. OK, so this is all very simple, and then I stop. So we can write f, so we are doing 1, can write f as sum i equals 0 to n ai x to the i with an different from 0. And g sum j equals 0 to m dj x to the j is bm different from 0. So that means the degree of f is equal to n, and the degree of g is equal to m. And so we had this formula that f times g was sum i equals 0 to n, say ci x to the i, so k, ck x to the k, where ck was equal to the sum i plus j is equal to k ai bj. Yeah, OK. Now, so the highest possible thing that occurs is n plus m. But if you look at the case n plus m, the only possibility that i plus j is equal to n plus m is that i is equal to n, and j is equal to m. And then we get therefore that c, n plus m, is equal by this formula to an times bm. So it follows, and this is non-zero because these were both non-zero, and this was an integral domain. So therefore, we find thus f times g is indeed different from 0, and the degree of f plus g is n plus m, which is equal to the degree of f plus the degree of g. And then finally, we can do this, how much more, OK, you can do this thing with the units. So if, say, a in r is a unit, then it will be unit in rx, because you can always just multiply with the inverse in r, and this will, if you multiply a polynomial, and you will get back the element 1. So there's no problem with that. We have to see that these are all the units. So let f be unit in rx. So it's actually like this. So then by definition, this means there exists an element g, the polynomial g, such that f times g is equal to the constant polynomial 1. So constant polynomial 1 has degree 0, and we have seen that the degrees add up. So the degree of f plus g, of f times g, which is the degree of the polynomial 1, and this is 0, is equal to the degree of f plus the degree of g. So it follows that both the degree of f and the degree of g is equal to 0. But a polynomial of degree 0 is just an element of r, and they are still inverse to each other, so f is a unit. So maybe that's enough for now. I'm sorry I went slightly over time, but sometimes also finished earlier. OK, we see each other next week after this holiday. Any question?