 So, now we talk about a few properties first I am going to make a claim every field is an integral domain how do we prove this what is the only property see both of them start with the structure of a commutative ring right. The thing is the field is a commutative division ring, but the integral domain is not a commutative division ring instead it has some nice property what is that that product of non zeros can never be 0. So, we have to see that if it is a division if it is a commutative division ring then that automatically implies that extra property which we embedded in the integral domain. So, suppose something is a field then obviously every element other than 0 has an inverse. So, here is a sketch of the proof I mean all these proofs mostly we will sketch. So, suppose a into b is equal to 0 if a is not equal to 0 then what can we say there exists a inverse right clear on this there exists a inverse such that a inverse a is equal to 1. So, let us hit this with a inverse from the left on both sides. So, we get a inverse a and we can put these brackets because of associativity times b is equal to a inverse times 0 ok. We have not still shown that you know there is no point in seeking and added a multiplicative inverse to the 0 element. Let us do a rough sketch of that proof as a sub part of this because we will have to now use the fact that this is going to be 0 seems funny right you multiply something by 0 you take it for granted it is 0 requires a proof as it turns out. So, how do we start that? So, let us take 0 times alpha say alpha belongs to some field ok. So, f stands for some field by the way you can also write f with the operations, but when the operation is clear from the context we just write the set itself. So, f is in this case our set along with some conventional operations that meet the criteria for a field. So, we will just write f every single time we want append it with those two binary operations and then put it in a bracket as a three tuple. So, this look I am just going to use only the properties that my algebra allows me to do. What I am going to write is this is 0 plus 0 and what property am I using here? The existence of the additive identity. So, 0 is equal to 0 plus 0 quite simply because 0 is it is everybody's additive identity. It is a unique additive identity for every element. So, it is also its own additive identity yeah. So, therefore, this is 0. Now, this implies that 0 into alpha is equal to. So, this I am going to now change what am I going to do? Distributivity right. So, this is 0 times alpha plus 0 times alpha. Now, because I know that there must exist an additive inverse to every element and because it is closed under multiplicative a multiplication. So, all of these elements also belong to the field. So, therefore, these are elements of the field and therefore, each of them must have an additive inverse. So, let me just put that additive inverse on both sides yeah minus 0 times alpha. So, this is basically the symbol for the additive inverse. Let us you know just write it in that manner plus 0 alpha is equal to plus and then I am again going to use associativity on this side. So, now an element added with its additive inverse gives me 0 back again. So, therefore, this is 0 this is also by the same token 0, 0 plus 0. But this is again because of additive identity please fill out the properties in between I have not written at which step which property I have used. I leave that to you to fill it out at every step what properties from that set of properties that I have written down a while back am I using at every step please note that right. So, this means that 0 acting on any alpha in the field will always result in 0. So, therefore, there is no point in seeking the additive the multiplicative inverse of 0 you see because no matter what 0 gets to act on it turns it to 0. So, that is why it makes sense to look for only multiplicative inverses of non-zero elements. So, therefore, harnessing that property here what can we say this is also 0 and this is 1. So, this is 1 times b is equal to 0 implying b must be equal to 0 proof is not complete it is a mere exercise now to assume as a second step that b is not equal to 0. And therefore, b inverse must exist and hit it with b inverse on the right on both sides and again proceed accordingly and conclude that a must be equal to 0. So, that is what is going to lead you to the completion of the proof. So, either a is equal to 0 or b is equal to 0 of course, it both as 0 then it is trivial to show that it is 0 yeah you could yeah. So, of course, that is that is the way I mean, but what I am meaning to say is that at this stage it is not yet complete you have to write a couple of lines more somehow justify that every time you start with something that is in a field yeah that field must also satisfy the properties of an integral domain. Unfortunately not every integral domain happens to be a field right that is definitely there. For example, you take the set of integers yeah there multiplicative inverses are not integers themselves except for 1 and minus 1 right. So, there still is some hope though it is not a complete lost cause because the second one says and this is where things start getting interesting every finite integral domain is a field that is interesting when you talk about the set of integers it is an infinite integral domain maybe that is the defect does that mean every for infinite integral domain is not a field of course, not there are infinite fields and by definition fields are integral domains. So, of course, infinite fields are also infinite integral domains. So, it is not necessarily true that infinite integral domains cannot be fields, but some infinite integral domains probably are not fields because of this assertion which says that if it is a finite integral domain then it must be a field. So, again what do we start with? We only start with what we know of integral domains to be true every other property is kind of overlapping except for the fact that in integral domain you have product of two non-zeros must be non-zero. So, in case of the finite what is so special that tells us that every element must have a multiplicative inverse unless it is 0 ok. So, let us try and see a again a quick sketch of a proof consider s is equal to say a 1, a 2 dot dot dot till a n because it is a finite field you see with some addition and multiplication to be an integral domain ok. So, it is a finite integral domain now the claim is that this must be a field which means that any object that you pick out from these n fellows here must necessarily have a multiplicative inverse that is essentially what this claim boils down to ok. So, further choose any alpha belonging to s and consider s till day to be the following set alpha a 1, alpha a 2, alpha a n. So, new set s till day alpha could be any of these a 1s through a n I am not saying which one it is, but it can only have finite number of possibilities because there are only n distinct elements in that set s. So, you take any one of those n distinct elements and call it let us say alpha and then you cook up a new set which is just the products according to the usual multiplication rules that are defined in this particular integral domain ok. Look at alpha a i minus alpha a j for i not equal to j ok. What I am going to claim is that this cannot be 0 ok. Let us see alpha. So, suppose alpha a i minus alpha a j is equal to 0 which implies that alpha a i minus a j is equal to 0 and I immediately I am saying have a contradiction. What is that contradiction? Let me just make one small little change here because I am not interested in the multiplicative inverse of 0. So, I am going to pick out this alpha to be a non-zero. So, if alpha is non-zero and this product is 0 and it is an integral domain what do I immediately have? That a i must be equal to a j. Why is that a contradiction? Because these are distinct elements, but if these are distinct elements then what have I just proved that the elements of s till day that I have enlisted here must also be distinct yeah. So, fill out that why is it a contradiction? I have said in words it is up to you to fill that out. So, this is a contradiction. So, therefore, this cannot be 0 is the assertion if this cannot be 0 then this means elements of s till day are distinct. How many elements are there in s till day and elements? How many elements are there in integral domain? So, there is the by the pigeonhole principle what can we say? One of the elements in s till day must correspond to the must correspond to which element am I looking for the multiplicative identity right. Therefore, one of the elements in s till day must be 1 and I am done because if it is so then I have actually explicitly found out which element which one multiplied with alpha gives me 1 and that is in itself the multiplicative inverse of alpha. So, no matter what alpha you pick from the set if you can carry out this multiplication operation you are sure to find out one element which takes you to 1 and therefore, alpha times a k is equal to 1 for sum k belonging to the set 1, 2, dot, dot, dot till n. Therefore, there exists alpha inverse is equal to a k in s which means that for nonzero alpha if you are assured that there is a multiplicative inverse this thing that we started with an integral domain a finite integral domain must necessarily also be a field. So, the only way an integral domain fails to be a field is if it is infinite, but that does not mean that if it is infinite it cannot be a field because infinite fields are classic examples of infinite integral domains right. But finite integral domains means it must be a field right ok yes because there are n distinct elements in s till day they are all distinct and these are all members of what because these are by these are by multiplication. So, these must also come from the set s and s has only n distinct elements. So, there are n distinct elements which must be equal to n distinct elements here and this set s must have the multiplicative identity yeah because it is an integral domain it must have a multiplicative identity. So, at least one of these fellows must map to that multiplicative identity and that is the fellow I mean that particular a k is going to be the multiplicative inverse of alpha which is all we needed in order to show that this is a field that is the that is just a symbol. So, one is depending on the operation that one can be. So, 0 can be equal to 1 if you define your operation in such a way if you call your addition as your multiplication of course, you cannot because there are difficulties there, but you know. So, that one is just a place holder do not look at one as I said square root of 2 and square root of 3 the ones examples that I gave you the other day right you should not read them as 1.414 or 1.732. Define your parameters or these objects according to the number system and the rules that we have given do not think of them in the conventional sense ok. So, hopefully this proof is clear right that is exactly what I have shown I have taken any 2 arbitrary elements in a still day call them alpha a i and alpha a j and I showed that if they have to be equal to 0 then a i must be equal to a j because alpha is not 0 it is an integral domain product of 2 numbers is 0 therefore, one of them must be 0 alpha cannot be 0. So, a i minus a j must be 0 if a i minus a j must be 0 then a i and a j are equal, but a i and a j by my claim which I have now erased in that set s are distinct elements of s. So, that is a contradiction right ok. Next we will also look at another interesting property now that we are dealing with these finite integral domains and we saw that their fields what kind of finite sets with these modulo operations do turn out to be integral domains and therefore, fields right. So, suppose z n is equal to 0 1 2 n minus 1 yeah and you define z n with addition modulo n and multiplication modulo n right. The claim is that this is an integral domain now once it is finite whether I say integral domain or field by this previous assertion is the same thing is an integral domain or field if and only if that is sufficient and necessary condition if and only if any guesses and if you have encountered this n is a prime number n is prime. We will devote some time to this ok. We will try to see this at least one part of it in two different ways ok and that is going to be important because much later in the course we shall be using one of those tools that we develop in this proof yeah. So, one part will be quite straightforward if you take it to be composite what needs to be shown is that it cannot be an integral domain or a field and that should be very apparent. If n is a composite number then n can be factorized in more ways than one, one way is obviously n times 1 yeah, but if n is composite there must be some other factors also. So, if you take these two other factors these are nonzero factors right and these are nonzero factors and yet when you multiply them they lead to n. So, n modulo n is 0. So, you already have nonzeros multiplying to give you 0. So, if n is composite then it is not an integral domain and one part is done. So, n being prime is a necessary condition. So, that only if is done right. So, let us just look at the only if part I just write it formally suppose n is composite there exists this implies that there exists q and s such that n is equal to q s ok let us say q and s belonging to let us call this set as s oh z we have already given it a name sorry my bad is already z n yeah. So, q and s if it is composite you must have this property yeah. Therefore, when you look at the multiplication operation q s modulo n is equal to 0, but q and s are not equal to 0. Hence not an integral domain this is clear right it cannot be an integral domain if it is a composite number. So, if you are looking for a finite field subject to these operations there is no way that if you choose n to be a composite number it is going to be a field or an integral domain, but the other part still remains to be shown. So, here it goes where should I erase now maybe you know these assertions well. So, suppose n is prime consider any alpha belonging to z n and look at the following set alpha, alpha squared alpha cubed till some alpha m alpha n m plus 1 likewise not saying this is necessarily finite what will happen if you keep looking at these this is cooked up by successive multiplications which are well defined yeah. So, we are at least assured that this multiplication is well defined there is no ambiguity what are we guaranteed because it is a finite set what can we say. But if we are going on taking higher and higher powers. So, for some l comma k we must have a sorry alpha to the l is equal to alpha to the k with k minus l greater than 1 think about it can it be equal to 1 can you have something alpha to the power something yeah and then you multiply it with alpha and then it is equal in the modulo sense it is a prime number remember alpha is a number that smaller than the prime number you just multiplying this alpha multiple times and you are dividing it by P to take the modulo the P is irreducible because it is a prime number. So, factors of alpha even if the even if alpha is not a prime number which it did not be but the factors of alpha cannot combine to give you P. So, that is also indivisible right it is irreducible yeah. So, therefore, if you just enhance k by 1 from l right if you just take k is equal to l plus 1 can it just give you say a number and that number multiplied by another number smaller than the prime number modulo P the prime number can they be the same they cannot right. So, it must be more than 1, but at the same time you will run out of possibilities because there is only a finite number of fellows here yeah you are you are looking at a potentially infinite set. So, I mean at most beyond n entries you are likely to repeat you are you are hard to repeat right. So, this must be true which means that we have taken that k is greater l is smaller. So, alpha to the l 1 minus alpha to the k minus l is equal to 0. Again I should probably just clarify a bit here because obviously we are looking for non-zero elements yeah. So, then what happens from this we use the same tool as before reminiscent of what we have done just a while back what do you know about this can this be 0 modulo P modulo any prime number this alpha is smaller than the prime number and you just going on multiplying it multiple times multiple times and then taking its division I mean it is remainder with respect to division by the prime number. Just if you if you have difficulty in figuring this out take prime number say 5 or 7 whatever comes to mind. Take any number smaller than that prime number say 4. So, you are taking 7. So, you have numbers from 0 to 6 and you are taking 4 you keep multiplying 4 multiple times can you divide it by can you divide it by 7 and get 0 because the multiples of 4 I mean take powers of 4 they will never contain a factor 7 it cannot be 0 right. So, this cannot be 0 right. So, let us continue here, but alpha to the l is not equal to 0. So, I have I hope I have justified that why therefore, what do we say 1 must be equal to alpha to the k minus l minus 1 times alpha and we are done because this remember k minus l is greater than 1. So, k minus l minus 1 is greater than 0. So, this is not a unit yeah this means that alpha inverse is equal to alpha raised to the k minus l minus 1. So, anytime you pluck out a non-zero element yeah from this when n is a prime number you are guaranteed to have its inverse within that set right. Remember this is not just raised to this power, but you have to also look at it modulo p that is the important thing because all these operations are ok I have called it n. So, modulo n where n is a prime number right. So, if it is not prime it cannot be a field if it is prime every element must have a multiplicative inverse. So, it must be a field right there is another way in which we will prove this latter part, but that is again part of the next module.