 So on our lecture series math 42 30 we've been focusing recently on the topic of fields particularly field extensions We want to utilize the tools we've developed for field extensions to focus on the idea of finite fields So let's say a few things about that So first of all recall that if a ring has a characteristic in remember in the characteristic of ring is how many times does it take for the 1 element to be added together to give you back zero right so if this happens in times That is one plus one plus one plus one plus one for which we can define this to be in times one in that situation So you add one together in times if you can add one together in times and get zero We call that to characteristic of the ring because then it has the property that for all elements of the ring are We have that n times r will likewise equal zero that that that so the characteristics pretty important In particular though if a ring has characteristic in that means that the ring Z mod out NZ can be embedded inside of r and the basic idea is you take you take the one element of Z mod in and identify it with the unity of the ring are and therefore you've embedded Z mod in Into r and r doesn't even have to be a commutative ring for that to make sense so every every ring a characteristic in has a Ring that looks like the integers now I mean it could be the integers mod in right there could be some reduction module in going on there But every ring has a every ring with unity has a subring that looks like the integers are the integers mod in now in particular When your characteristic is zero That means there is no positive number such that one plus one plus one plus one plus one plus one and times We'll ever add up to be zero and we call that characteristic zero now Why do we do that? Well, that's because when you look at the ring Z mod zero Z You you mod out zero. It doesn't do anything as a ring. That's isomorphic to the integers themselves And so we call that characteristic zero because a ring of characteristic zero will contain the ring Z mod zero, which is just Z itself So rings of characteristic zero will contain an isomorphic copy of the ring of integers Z This is even true of r is a non commutative ring. It just has to have unity And so in particular if you're a ring of characteristic zero, you must be infinite because you have an infinite subset So as we try to focus on Finite fields that means we have to look at the situation where we have a positive characteristic So when we consider finite fields, we're gonna consider fields of characteristic P now We which of course is a positive number It's all we call it P because for a for a characteristic of a field Actually the characteristic of a domain if it's not zero then it has to be a prime number because of your characteristic was Composite like two times three. You could actually produce proper divisors of zero. That's not allowed in a domain That's not allowed in the field. So finite fields always have prime characteristic now. There is one finite field I should say there's a family of finite fields. We already know So if you take for example Z mod P, which we often denote as Z sub P, right? That's the same notation we used before some people like to write it as Z mod PZ some people just write Z sub P While some people might mean different things in this lecture series. That means the exact same thing Zp is just the ring of integers mod P. This does give us a finite field it's gonna have order P of course and We claim up to isomorphism This is the only field of order P and thus we call it the field of order P We call it Fp and by the end of this video will actually prove why there's only one field Up to any finite order. Okay, so if you pick different primes You get different fields of order P Can we get other fields than that? Well, the good news is yes by Croninger's theorem. We can build new Finite fields by using irreducible polynomials over Zp adjoin X So if I can find a polynomial say P of X Let's not use P since P already has a different meaning here. So let's take the polynomial Q of X which belongs to Fp adjoin X right here, and if you're not used to the notation yet We'll still call it Zp right Zp adjoin X like so and let's suppose This is an irreducible polynomial and suppose that the degree of Q is equal to n something like that then if you take Since this polynomial is irreducible if we take by Croninger's theorem the field Zp adjoin X and we mon out by the principle ideal generated by Q. This is going to give us a field Let's call this field F sub P to the n like so. All right This is going to give us a polynomial whose degree Fp to the n right here over F is going to be in all right If you take the root of an irreducible polynomial and adjoin it to a field That will then extend the field by the degree of the polynomial So we call it the degree of the extension there. So this will be degree n For which then I want you to unravel that for a second and think about how big is that set So you can get it you can get a field extension of Zp because that's that the base field the base field Here is Zp should have written that down there We're extending the field Zp by Indimension and I claim that the only possibility you can get there is P to the end And so this leads us directly into our proposition right here Suppose that F is a finite field of characteristic P Then the the card melody of this field which we often refer to as the order of the field is going to be P To the end where P is of course that same prime number and in is any natural number Where you could include zero because the zero the the That as you can just take a field with one element Which would be kind of weird it would just be zero. We typically don't want to think of that as a field Because typically you want to separate it with zero and one but with the exception that zero and one are Distinct it does satisfy all the field axioms So typically we are thinking these as positive integers in that situation P to the first P squared P cubed Etc. Etc. Well, so we get a field of order P to the end for every possible order you want and so like mentioned above If you have a field of characteristic P, then it'll contain an isomorphic copy of Z mod P Okay, this is true. Whether you're an infinite field of characteristic P or a finite field of characteristic P Therefore we can view our field F as a VP vector space That's that's something we can do if a field contains a subfield it will be a vector space over the base field So F is a VP FP vector space All right, remember FP and ZP are the same thing FP is generalizing the notion of ZP we're gonna see that and we do we don't want to call it in general ZN because that does mean the integers mod and Which is not what we're doing. We're talking about the field of order such and such So if we consider the degree of this thing so we have a field That extends the base field the prime field FP there and so the dimension of That the degree is mentioning the dimension of F as a VP vector space Let's call that dimension N. We mentioned earlier by Chronicles theorem if we can find irreducible polynomials We then can construct field extensions of degree N if there's an irreducible degree in polynomial and by accounting argument, there's always a irreducible polynomial over ZP of degree N But that's the topic we can handle some other time here. So in particular F as a vector space is isomorphic to the set FP to the end where FP to the end here This is the set of column vectors Within coordinates, but each coordinate has P possibilities You have 0 1 2 3 up to P minus 1 so because these sets are isomorphic as vector spaces I'm not claiming that they're isomorphic as rings because does this thing even have a ring structure? The typical ring structure you put on to that set is not a field, but they are isomorphic as vector spaces So there does exist a linear bijection between the two that bijection says the sets have the same card Nality which gives us this right here But as FPN is just the Cartesian product of FP n times This gives us that the card Nality that set is the card Nality of FP to the nth power But FP it's a set of P elements that gives you P to the end. So every finite field Will have order P to the end and by Kroniker's theorem assuming we have enough irreducible polynomials We then can produce a finite field of every possible order so we get a we get a finite field for every Possibility that's allowed to us. There is a finite field for every power of a prime But why do I keep on saying the the the why is there only one? We need to make an argument that they in fact are one and the same thing Now in order to do that I need a little helper proposition first And this is what's commonly referred to this proposition is commonly referred to as freshman Exponentiation and this is actually a property that's true for commutative rings with unity who have characteristic P It doesn't necessarily have to be a finite field, but a finite fields included in that So if R is a commutative ring with unity and the characteristic is P Then it turns out that if you take a plus B to the P in power This is the same thing as a to the P end plus B to the P end And this is why it's called freshman Exponentiation because commonly if you have a college freshman when they join a class like college algebra math 1050 at Southern Utah University They typically want to distribute exponents across addition and it makes sense multiplication distributes over addition and Exponents is just iterative multiplication. So doesn't it make sense to distribute? Repeated multiplication over addition of the conflation makes sense, but it's not a valid Exponential rule at least not over the real numbers for or the complex numbers Which is pretty much always where a college algebras is doing their algebra But over a commutative ring with unity of characteristic P It turns out you can distribute prime powers when that prime is the characteristic And that's how we're gonna do this We're gonna do this by induction Because if you can distribute a if you can distribute the p-th power Then you can distribute the P to the n-th power because you just distribute P then the second P Then the third P then the fourth P up until the n-th P So by induction it suffices to prove that a plus B to the P is equal to a to the P plus B to the P That is that the prime the characteristic can distribute So you can't distribute every exponent But if the exponent is the characteristic of your ring you can distribute it and this is a consequence of the binomial theorem For any commutative ring the binomial theorem is valid and you can use the exact same proof that you use to prove The binomial theorem it's a combinatorial argument, but it requires commutivity of the ring Distribution and the ring axioms that's all it is so the binomial theorem is a valid for any commutative ring And so therefore the shell this shows us that a plus B to the P is equal to the sum or k ranges from 0 to P And which case we then get the binomial coefficient P choose k Which is a reminder This is by could by definition equal to P factorial over k factorial times P minus k factorial One interesting thing about these binomial coefficients that this is always an integer This belongs to the set of integers in particular. It belongs to the natural numbers right In which case the natural numbers can be sent into any ring Because we can send the integers into any ring right now there might be reduction mod P going on here Of course, but if you have since the binomials the binomial coefficients are integers We can identify that integer with the integer mod P And so it makes sense to talk about a binomial coefficient in this ring R So you have the binomial coefficients times a to the k times B to the P minus k like so And so this this this would hold because of the binomial theorem which is valid for R Now look at this number right here P to the K as an integer so not reduced by P yet This number P choose K will be divisible by P For all of the binomial coefficients P choose K as K ranges from zero to P with two notable exceptions It won't be divisible by P when K is equal to zero or when K equals P Because in both of those cases the binomial coefficient is equal to one and one will not reduce One's not divisible by P and thus it won't reduce to zero when you mod out by P But all the other binomial coefficients will so when you look at this expansion This expansion all of the binomial coefficients are divisible by P and therefore are congruent to zero mod P With the only exception being the first one P choose zero for which you're gonna get a to the P power times B to the Excuse me. This is going to be a to the zero power B to the P minus zero power And then you'll get the last term you're gonna get P choose P Which is going to be a to the P times B to the P minus P And that of course reduces since the binomial coefficients are one and anything raised to the zero power is likewise one This will simplify of course to be B to the P plus a to the P Which is exactly what we claimed it to be thus proving this very important proposition freshman exponentiation So we're gonna use this all the time when we work with fields of characteristic P in particular for finite fields This freshman exponentiation applies So now let's get to the heart of what we want to prove about finite fields that they're unique up to their order Suppose that f is a finite field and suppose that its order is P to the n now This sometimes leads to annoying subscripts and superscripts nested inside of each other Which really doesn't really to great notation Oftentimes when people talk about finite field since the finite field has always an order P to the n Oftentimes that's abbreviated as just Q. So Q is a power of P where the exact exponent n is somewhat irrelevant So it'll often be suppressed inside this Q right here. So f is a field a finite field of order Q Which is a power of P then we claim that f is a splitting field of the polynomial x to the Q minus x So this is a polynomial viewed over the field f P So if f is in fact a splitting field We showed previously when we worked with splitting fields that all splitting fields are unique inside of an algebraic closure When we apply this to finite fields in particular this says that all fields of order Q are unique Inside of a fixed algebraic closure of FP here. All right, so since finite fields are always splitting fields They're gonna be isomorphic to each other Which as that their order is a is an invariant of isomorphism This is gonna give us that the field of order Q is isomorphic to every field of order Q because every field of order Q is The splitting field to this polynomial and the proof basically is a generalization of Fairmask little theorem Remember what that says Fairmask little theorem tells us that if we have a number a and you raise it to the P minus 1 power This is congruent to 1 mod P where P of course is at prime Now if you times both sides of this equation by a to the P This tell us excuse me times both sides of the equation by a you're gonna get a to the P equals a Mod P when you look at the first equation you do have to have the exception that a doesn't equal zero But in the second case Because Fairmask theorem applies this for every non-zero value. It's also true for zero two zero to any power is equal to zero So Fairmask little theorem basically says that if you take an integer and raise it to the P power That's congruent to itself mod P That's basically what we're gonna do right here. So we want to argue that in your finite field F If you take any element of F and raise it to the Q power you get back that same element So let's first start off a zero because it's the only element of the ring That's not a unit and so it does have to be treated separately for that reason So if we take zero raise to the Q power that gives you zero no big deal All right, so let's choose any other element of the field Because this element is non-zero it necessarily is a unit. So you belongs to F star right here This is the set of units of the ring Which because F is a field F star necessarily is a finite group It's finite because F is finite. Of course if you take away an element. It's still finite Why is it a group? Well, clearly it's associative because multiplication is associative It has an identity because the ring has unity it's a field But every element has a multiplicative inverse because it's a field, you know except for zero But if you take away zero that means we have a finite group F star Now by Lagrange's theorem For finite groups here if you take an element of a group and raise it to the order of the group That's always going to give you one right here. Now. What's the order of F star? Every element of a field has a is a unit has an inverse except for zero So the order of F star is just the order of F take away one element Because that's what F star is it's just F take away zero. So the order of F star is q minus one So we get that you to the q minus one is equal to one And if you if you multiply both sides by you you end up with u q equals you just like we did before In fact in when we talked about fair muscle theory in group theory We actually proved it as a corollary of Lagrange's theorem We're generalizing that argument to show this property is true for every field So we've now argued here is for a finite field if you take any element of the field and raise it to the q power You get back that element. That's true whether you have a unit or whether you have zero Although we need different arguments for both of them All right So take an arbitrary element alpha that belongs to the field and consider our polynomial x to the q minus x We then call that polynomial f and that's the polynomial. We care about this sits over the field zp Like so what happens if we evaluate f at alpha? Well, you're gonna get alpha to the q minus alpha But by the previous argument fair muscle theorem you're gonna get that alpha to the q is just alpha alpha minus alpha is equal to zero So alpha is a root of the polynomial f and as f was an arbor as alpha was an arbitrary element of f This shows us that literally every element of f is a root of this polynomial f right here Okay Now as f of x has at most q roots because it's degree q We see that these are all of the roots of f of x there can't be any other ones because we already accounted for Q distinct roots of alpha and there's that most q So we so we see that f contains every root of f Of little f and therefore f of x splits over our field f That doesn't necessarily make it a splitting field right because to be a splitting field your polynomial is to split Yes, but it has to be the smallest field that that happens, but of course Since the roots of f of x are exactly the elements of the field f I also should point out that if you take the base field f p and a join a larger field to it That just gives you back the field But as we discussed with splitting fields If you take the base field and a join all of the roots of a polynomial that Constructs the splitting field and therefore f is the splitting field for f of x right here and so as this holds for any field of order q The previous theorem that we have proven in this lecture series, which we numbered it as 21.2 0.3 which told us that Splitting fields exist and are unique in a fixed algebraic closure This then finishes the proof and shows us that all finite fields are unique Up to isomorphism not even isomorphism. They're unique up to equality inside of an algebraic closure Now if you have different algebraic closures those different algebraic closures will be isomorphic to each other and the resultants Splitting fields in them will be isomorphic to each other. So this then gives us the result We're looking for that all finite fields of the same order are in fact isomorphic to each other Now we'll talk some more about finite fields in the next lecture of this lecture series lecture 30 But we're gonna call it close for this lecture 29. Thanks for watching If you learned anything from these videos, please like the video subscribe to the channel to see more videos like this in the future Of course, and if you have any questions, please post them in the comments below and I will answer them as soon as I can Thank you. Bye