 I want to now discuss an example of an integral domain that is not a unique factorization domain. This is a very important example to point out because to really appreciate what a unique factorization is, we have to figure out what a unique factorization isn't or I should say we have to figure out who's not a unique factorization domain. Now, to be a unique factorization domain, you have to be an integral domain, of course. So if we just come up with any ring that's not an integral domain, then it's not a UFD. So in particular, we should be interested in an integral domain that's not a UFD. So we have cancelation, we have zero product property, but we don't have unique factorization. And so I want us to consider the ring where we adjoin to the integers the square root of 3 times i, or you might think of it as we've adjoined the square root of negative 3 to the integers. Be aware that the Gaussian integers, we adjoined the square root of negative 1, but if we get a UFD in that situation, but if you adjoin the square root of negative 3, you don't get a unique factorization. Let me show you why that is. So of course, this ring, a typical element of that ring that's going to look like a plus b times the square root of negative 3, where a and b are arbitrary integers. So we can make a lot of arguments using the norm again, where if you have some element z inside of our ring right here, then same thing as before, the norm of z is going to equal the complex modulus of z squared. But in this situation, if my integer, I'm using the word integer loosely here. I don't mean a rational integer in this situation, but if my integer has the form a plus b times the square root of negative 3, then what we're going to see is the following. The complex modulus here means you're going to take the number times it by its complex conjugate a minus b times the square root of negative 3. When we foil this out, we're going to get an a squared. What else are we going to get? We're going to get a negative a b square root of negative 3. We're going to get a positive a b times the square root of negative 3. Clearly, those two numbers that I pause for a moment are going to cancel each other out. That's the whole point of using the complex conjugate. Finally, you're going to get positive 3 times b squared. Summarizing this, you have an a squared plus 3 b squared. When you take the norm of an integer in this domain, it always looks like a squared plus 3 b squared. That's different than with the Gaussian integers. They always get a squared plus b squared. Things are a little bit different now. Okay? A similar argument now applies that the units of this ring are going to have a complex modulus which is equal to, excuse me, they're going to have a norm which is related to the complex modulus. It's going to have a norm that's equal to 1. How do you make this thing equal to 1? Well, you can stick in plus or minus 1 in for a and then 0 in for b. That'll give you 1. But you can also try, if you tried to put in a 0 for a, you're going to have 3 times b squared where b squared is an integer. That doesn't ever equal 1. This ring has only 2 units, exactly plus or minus 1. Okay? Then consider the numbers 2, 1 minus root negative 3, and 1 plus root negative 3. I'm going to make the claim that these numbers are irreducibles, that they cannot be factored more than they already are. Let's prove that argument here. Let's first show that 2 is an irreducible over this domain. I'm going to be very careful here to use the word irreducible as opposed to a prime. We know that every prime number in a domain is an irreducible number, but the reverse direction is not actually true. This example will provide an example of such a thing. Let's first show that 2 is irreducible. Consider a factorization of 2. When we take the norm of 2, we end up with 4, 2 squared. If it has a factorization, that factorization will break up across the norm. We have v, excuse me, the norm of u, and the norm of v is equal to 4. Now, if this is a trivial factorization, one of the numbers u or v is a unit, and its norm will be 1, and then the other norm will be 4. It would be an associate. If we want this to be a proper factorization and a non-trivial factorization, then the only way you can factor 4 would be so that this is 2 and this is 2. We make that statement here. How can you solve the equation a squared plus 3 b squared equals 2? The answer is you can't do it because if you choose anything other than 0 for b, even if you take 1, you can end up with 3, which 3 is, you have at least 3, a number that's bigger than 3, but 3 is already too big for 2. So you have to have b equal to 0. But then that would imply that a itself is equal to the square of 2, which is not an integer. It's not even a rational number. So there is no solution. There's no integer solution to the equation a squared plus 3 b squared equals 2. We get a contradiction. So this shows us that 2 is an irreducible inside this ring. Now, just to talk a little bit more about number theory here, this is an example of a deophantine equation. We have an equation for which we only want to accept integer or maybe only rational solutions to it. That's a pretty big topic in number theory. So you can see why, as we're delving into these topics of commutative algebra and algebraic number theory, that those elementary number theory questions are very relevant here. The deophantine solutions to this equation, or I should say the lack of them, is evidence that 2 is an irrational integer inside a, excuse me, an irreducible integer inside of this domain. We could do the same game if you take the numbers 1 plus or minus the square root of negative 3, because it's very easy to show that their norm will also equal 4. And therefore, that gives us by the same argument that they have to be irreducible, because there's no way you could factor these numbers in such a way that one of the factors is in a unit. All right. What does this matter? Well, if you take the number 4, which 4 itself belongs to this domain, z adjoined the square root of negative 3, right? I can give you two different factorizations of 4. 1 is 2 times 2. And one of them is 1 plus the square root of negative 3 times 1 minus the square root of negative 3. This is a product of irreducibles, because 2 is irreducible. This is also a product of irreducibles. We showed earlier that these numbers are irreducible. So we have two different factorizations, which the two different prime factorizations here, right? Are they equivalent? Are these two different ones in fact, right? Well, if 2 is an associate of this, then it turns out this is not really a different factorization, at least not with regard to how we define unique factorization. My claim, of course, is that these are not associates of each other. Therefore, these are two different factorizations. So let's provide the details of this. Why do we have two different factorizations of 4? Well, we have, of course, that 2 divides 4, so you have this factorization. Does 2 divide this? Well, if these were associates, then this number would have to be 2 times 1, which is 2, which is not. Or it would have to be negative 1 times 2, which is negative 2, which it's not. Remember, there's only two units in this ring, plus or minus 1. So the only associates of 2 are 2 and negative 2. These are not associates. These are not associates. So this gives us two different factorizations. And so since we have two different factorizations of the number 4, and these are factorizations using irreducible numbers, then we have to conclude that we don't have unique factorization. So Z adjoin the square root of negative 3 is an example of an integral domain that doesn't have unique factorization.