 So remember a field with a finite number of elements is called a Galois field. And so the question you've got to ask yourself is, self, what does a Galois field look like? So remember the key properties for a field, plus and times are commutative. There's an additive identity, which we call zero. There's a multiplicative identity, which we call one. Every element has an additive inverse. Every non-zero element has a multiplicative inverse. For a, b, and c in our field, a times b plus c is a, b plus ac. And there's a few useful propositions. One, the identity uniqueness, the additive and multiplicative identities in a field are unique. And there's product and sum uniqueness. If b is not equal to c, then a plus b is not equal to a plus c. And as long as a is not our additive identity, a times b is not equal to a times c. So let's find the addition and multiplication tables for a field of two elements. Since a field needs to contain an additive identity zero and a multiplicative identity one, then our field needs to contain the elements zero and one. And remember we've shown that if a field contains more than one element, then the multiplicative and additive identities must be distinct. And since they have to be distinct, our field contains just the two elements zero and one. So let's produce that multiplication and addition table. So for addition, because zero is the additive identity, we know that zero plus zero is zero, and one plus zero is one, and zero plus one is 42, um, one. Now, because this is a field, one must have an additive inverse. We have to add something to one to get zero. Well, we've already determined that if we add zero, we don't get zero, we actually get one. So the only other choice for the additive inverse of one is one itself, and so we must have one plus one equals zero. We can form our multiplication table in the same way, because zero is the additive identity, we know that zero times zero, one times zero, and zero times one are all equal to zero. And because this is a field and one is not the additive identity, then we know that one must have a multiplicative inverse. So one times something has to give you one. Again, the only choice we have for our something is one itself, and so we know that one times one must be one. And that gives us the addition and multiplication table for a field with two elements. Now, we know that the integers under addition and multiplication mod n form a field with n elements if and only if n is prime. So the real question is, are there other fields with a prime number of elements? And in fact, no. And as it turns out, suppose a field has p elements where p is prime, then the field is isomorphic to zp. So the fields with a prime number of elements aren't very interesting, we know all of them. So the important question is, what if you don't have a prime number of elements? Or let's find the field with four elements. So our elements have to be the additive identity zero, the multiplicative identity one, and two other elements, which in a fit of creativity, we'll call a and b. And we'll set up our addition and multiplication tables. Now, since zero is our additive identity, we know that zero plus anything or anything plus zero will be unchanged. So that gives us the first row and first column of our addition table. We also know that for all a, a times the additive identity is going to give us the additive identity. So that gives us the first row and first column of our multiplication table. But wait, there's more. Remember that one is our multiplicative identity, and multiplying by one doesn't change anything. So that actually gives us the second row and second column of our multiplication table. And since we know more products than sums, we'll complete our multiplication table first. So looking at our multiplication table, we need to find a, a, a, b and b, b. So remember, products have to be distinct. And since we know a times zero is zero and a times one is a, then a, a can't be zero or a. It must be b or one. But if a times a is one, then a times b has to be b, and that's going to make a the multiplicative identity. But the multiplicative identity is unique, and so a can't be the multiplicative identity. So a times a can't be one. It's got to be equal to b, which means that a times b has to be the identity. And since multiplication in a field is commutative, b times a is also equal to the identity. And since the products must be unique, b times b must be equal to a. Now addition is harder. We need to find one plus one, a plus one, b plus one, a plus a, a plus b, and b plus b. So the uniqueness of sums says that a plus one could be zero, a, or b. So let's think about this. If a plus one is equal to zero, let's consider the product b times a plus one. Well, if a plus one is zero, that's really just b times zero. And remember, in a field the distributive property has to hold, so b times a plus one must be b times one plus b times a. Now b times zero is zero. b times one, well, that's b. And b times a, well, that's going to be one. And so we know that b plus one is zero. But then a plus one is equal to b plus one, and so a is equal to b. And since a and b are supposed to be elements, this can't be true. And so whatever a plus one is, it can't be zero. What if a plus one is equal to a? Well, if a plus one is equal to a, then one is the additive identity, which is impossible. So a plus one can't be a, and this leaves only a plus one equal to b. What about b plus one? Well, b plus one is either zero or a. What if b plus one, we have again, a times b plus one, and by commutativity, we have b plus one equal to a plus one, and so a equals b, which is impossible. So b plus one has to be a, which means that one plus one has to be zero. So what about a plus a? Either a plus a is zero or a plus a equals one. If a plus a is equal to one, then again, the product b times a plus a, that'll be, but since our additive identity has to be unique and it's not b, this is false. So a plus a must be zero. And again, the uniqueness of the sum means that a plus b is one, and b plus b has to be zero. Now, one important thing to notice here is that once we decided what our additive and multiplicative identities were, then we had no choice in how we assigned these sums and products, which means that this is the only field with four elements. Any other field is isomorphic to this one.