 So far, we know that all fields with p elements are isomorphic to zp, and there is only one finite field with four elements. So we might ask, can we find a finite field with six elements? Well, we can certainly try. So let's assume our elements are 0, 1, a, b, c, and d, and we'll start by trying to find our multiplication table. So again, we know the product with 0 and the product with 1, and by the uniqueness of products, we know that a, a must either be 1, b, c, or d. So if a, a equals 1, then a, b is either b, c, or d, but a, b can't be b because if it did, a would be the multiplicative identity. So a, b is c, or a, b equals d. Suppose a, b equals c, then a, c would have to be either b or d, but if a, c equals b, then a, d would have to be d, which is impossible. So a, c must be d, and a, d must be b. Well, that worked out pretty well. Let's go on. If a, b equals c, then b, a has to be c, and again, b, b can't be 0, b, or c, so it could be 1, a, or d. Suppose b, b is equal to 1, then b, c can't be 0, b, c, 1, or d, so b, c must be a, and we have a problem because we have nothing left for what b, d could be other than d itself, but it can't be. So b, b can't be equal to 1. Well, what if b, b is equal to a? Then again, b, c can't be 0, b, c, a, or d, so b, c must be equal to 1, but again, that means that b, d would have to be d, and we can't have that. And in fact, no matter what choices we make on the values of the different products, we can't form a consistent multiplication table. And so it's not possible to form a finite field with six elements. Now, you might notice that it's a lot of work constructing the multiplication table, and it's not clear we'll be able to do it. So let's consider a general approach. Consider one, the multiplicative identity in a field f. We generate additional elements by repeatedly adding 1 to itself. And for convenience, we'll designate 1, well that's 1, 2, well that's 1 plus 1, 3 is 1 plus 1 plus 1, and so on. And remember, there are only so many symbols. These 1, 2, 3 have nothing to do with our familiar 1, 2, and 3 from arithmetic. Are the result of repeatedly adding the multiplicative identity of the field to itself? This allows us to introduce the following definition. Let 1 in R be the multiplicative identity in a ring R. The characteristic of R is the least positive value for which the sum of N1s is equal to 0, the additive identity. Now, it may be that there is no least N, and in that case we say that R has characteristic 0. Since every field is a ring, then we can also talk about the characteristic of a field. So this leads to an important result. The characteristic of a finite field is a prime number. Well, let's actually prove this. Since the field is finite, there is necessarily a least N for which 1 added to itself N times gives us 0. I'm not sure what they have to do with that statement. Suppose N is not prime. It's a product of P times Q. So N is the sum of PQ1s, and we can get a sum of PQ1s from the product, the sum of P1s times the sum of Q1s. But under our assumption, the sum of N1s is 0, and with our notation, the sum of P1s is P, and the sum of Q1s is Q. Which means that 0 is the product PQ, but a field can't have 0 divisors. And so the characteristic N can't be a product of numbers. It must be a prime number. And what this means is that every finite field has a subfield isomorphic to ZP. Specifically, that's the subfield whose elements are 1, 2, 3, and so on, all the way up to N minus 1, plus the 0. Since this is a subset of a field, F, we only need to verify closure and the presence of the inverses. So here's a quick sketch of the proof. Suppose P and Q are in the set. Remember that our definition P is the sum of P1s, Q is the sum of Q1s. So P plus Q and PQ are the sum of a whole bunch of ones. As for showing the inverses present, what's up with these pigeons? Now remember, the only field with P elements is ZP. So our subfield is isomorphic to ZP, and we'll abuse notation and call this subfield ZP. So let's think about extension field. Suppose X in our field is another element of the finite field with characteristic P, but X is not in ZP. So before I get my mathematician card torn up, a reminder that when we say X is in ZP, we really mean that X is in F prime, where F prime is a subfield of F and F prime is isomorphic to ZP. So if A0 and A1 are in ZP, then A0 plus A1x is in F. And if the As are not equal to the Bs, then A0 plus A1x is not equal to B0 plus B1x. In other words, these elements are distinct. And so we know that F has at least P squared elements. But wait, there's more. Suppose X squared is not this linear combination for any A0, A1 in ZP. Remember there are only so many symbols. X is a specific element of our field and not an unknown value or a placeholder. So I take three elements of ZP, then my linear combination A0 plus A1x plus A2x squared form distinct elements of our field. And ZOR field has at least P cubed elements. And Lather rins repeat. If F is a finite field, it has P to the n elements for some prime P and some whole number n. Consequently, it's impossible to find a finite field with six elements. Our attempt to do so was doomed from the beginning. But we could solve the problem of, say, finding a finite field with two to the third eight elements. Now we could try to produce this field by creating the multiplication and addition tables, but as you've seen, that's hard work. And so the natural question to ask is, is there an easier way? We'll take a look at that next.