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Integral Domains Part 2

welcome back to this next video in which we are discussing integral domains okay so in this next video what I want to do is tell you about two more basic but important theorems about integral domains okay so the first one is that if you take an integral domain and you produce the polynomial ring over that integral domain ie with coefficients in that integral domain then that new polynomial ring over the integral domain that you've produced will also be an integral domain okay so let me write this out here then so let's say D here is our integral domain so I'll put D is an integral domain and I'll abbreviate integral domain down to I D here okay then my claim is that if we generate the polynomial ring with coefficients in that integral domains if we generate D adjoin X here or d square brackets X here this is an integral domain so DX is an integral domain as well okay so this being true implies that this is also true okay so why is that going to be the case then okay so if D here is an integral domain then it's the case that if you multiply two nonzero elements you will not get zero back again so what I want to prove to prove that this is an integral domain is this is that if I take two nonzero polynomials so that say I take ax and BX okay which are both nonzero polynomials that say ax and BX are both nonzero polynomials from my polynomial ring over the integral domain D okay then I want to prove that if I'm multiplying these two together it's also going to be a nonzero polynomial okay and I'll just put actually here that it's not the zero polynomial so it's an element of the ring of polynomials subtract the zero polynomial so both of these are not the zero polynomial and I want to now prove that if I multiply these two two in the ring of polynomials if I take X and I multiply it by B of X okay I want to prove that this is not equal to the 0 polynomial okay so how can I do this well the important point here is that if these two polynomials are not the 0 polynomial then they have some leading coefficient that is not equal to 0 okay so for ax here then let's say that the leading coefficient is a m and it's in front of X to the power of n ok so there has to be some leading coefficient which is not 0 okay so a n is not equal to 0 in the integral domain D ok so here is our leading coefficient which has well which is not the 0 okay which is not equal to 0 and of course it could be the case that it's a constant polynomial in which case the leading coefficient will just be a constant term but at the very least that constant term cannot be equal to 0 otherwise it will be equal to the 0 polynomial and the same is true of BX here B of X will have some leading coefficient which will have as BM here and that'll be in front of X to the power of M here again BM will not be equal to 0 in our initial integral domain here now when I multiply these two polynomials together the leading coefficient now of the answer here a of X multiplied by B of X will be am x BM where they're multiplied together in the integral domain and then that will be in front of X to the n plus M here so this will be our new leading term basically and this will be our new leading coefficient here a n times B M now if both of these are not equal to 0 in the initial integral domain which was my initial assumption because they were the leading coefficients which were not equal to 0 then we know that when we multiply them together because we're working with an integral domain we'll get some nonzero elements okay so a M times B M will no longer or well will not equal 0 because we were working with an initial integral domain and that means that this I'm not going to be the zero polynomial because it's leading coefficient here is not equal to zero okay and therefore it cannot have all coefficients equal to zero okay so when you multiply together two nonzero polynomials in this ring of polynomials over an integral domain you will not get the zero polynomial back again you'll get a nonzero polynomial back again so indeed this ring of polynomials over an integral domain is also an integral domain and of course I should stress that when you take a commutative ring this integral domain will be a commutative ring when you take the ring of polynomials over a commutative ring you get a commutative ring back again and also if this was the non zero ring this will not be the zero ring that I ever okay so the other two conditions you know translate onto this ring of polynomials as well so that it does indeed satisfy all three conditions of an integral domain okay so that's very limber one that I wanted to tell you about the next thing that I want to tell you about is a little bit of a further investigation to something that I pointed out in the previous video now in the previous video I showed you that the multiplication table of an integral domain is special because all of the rows and all of the columns that correspond to non zero elements in the integral domain have to contain no repeats okay they cannot contain an element of the integral to claim more than once okay now in the case that the integral domain was finite had pointed out that that would mean that every single row and every single column corresponding to a nonzero element in this multiplication table would have to contain every single element of the domain once and only once okay that didn't have to hold true if it was an infinite integral domain now I'm going to investigate that further because what that's actually going to mean is that if we have a finite integral domain so if we have a finite integral domain D then I claim that a finite integral domain is actually equal to a field so the finite integral domains are all fields there is no such thing as a finite integral domain that is not a field okay if you're an integral domain in your finite then you are also feel this does not hold true that infinite integral domains if you've got an infinite integral domain then it is not necessarily a field example being the integers the integers is an infinite integral domain but it is not a field it does not have a multiplicative inverse for two for instance okay so why is this the case well this is really really simple to understand I have now said that if we for instance well actually let's just say what we need to prove here we need to show that if I take a which is an element of my integral domain and I'll just move this up a bit so if I take a which is an element my integral domain and a is not equal to the zero elements and also equal to the additive identity so I take some arbitrary element in my integral domain that's not the additive identity I need to show that there's also a multiplicative inverse so there is one over a which is in the integral domain such that a times 1 over a is going to equal the multiplicative identity 1 so I have to show that there exists another element in the integral domain which multiplies by a to give 1 now why is that going to have to be the case well I said it basically if we look at the row corresponding to a here each will have to contain every single element of the finite integral domain once and only once there's no other option it can't contain repeats that's what we proved here ok and since you've only got a finite number of things you've got the exact number of slots in this row that you've got things in the integral domain everything must appear once so in particular one must have to appear at some point and then you just ask well what is it that I multiplied by a well with a rather to get 1 and that of course will be our multiplicative inverse for a so that will have to exist some element which I can multiply with a to get 1 basically if we're working in a finite integral domain and that element will now be the multiplicative inverse of the element a and that applies that absolutely all nonzero A's that you can come up with from the integral domain it only doesn't work for 0 because of course we know that this fact about the rows and that doesn't hold true when we're talking about zero okay so that's the simple explanation is to one I if you have a finite integral domain that has to be a field okay it's because all of the rows and columns corresponding to elements that are not equal to zero in the integral domain have to contain all elements once and only once if we're working in a finite interval domain okay if we're working in an infinite integral domain I will stress again this does not have to apply and the reason is that saying that you can't have repeats in one of these rows or one of these columns it doesn't mean that every element has to appear in the row or column okay because if you've got the infinity and infinite number of elements then of course infinite set Theory comes into the play here and infinite set theory is very different from finite set theory okay so it's only when we're working with a finite integral domain that we can then show that it has to be equal to a field okay so finite integral domains are all fields okay that's theorem number two and with that I will end this basic introductory video on integral domains

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