 In our discussion of polynomial rings, it's very appropriate to mention that you can have a polynomial ring with multiple indeterminates. You can have more than one. As polynomial rings can be formed over any ring, it's possible to construct a polynomial ring when the coefficient ring itself is a polynomial ring. So, if R is a ring, then R joined X is a polynomial ring, but that makes it a ring and therefore we could join a new indeterminate because all that we require is that Y be some symbol, some number that is not inside of the set, R X whatsoever. So we could join a second variable Y to the ring R X. And so it does take a little bit of an argument, but it is straightforward to prove that the ring, the polynomial ring with the variable Y whose coefficient ring is the polynomial ring R joined X, this is isomorphic as a ring to the polynomial ring where your indeterminate is X and your coefficient ring is R joined Y. And that isomorphism is very, very straightforward. I'm gonna leave it as an exercise to the viewer to prove that in fact these two rings are isomorphic to each other. And since they're isomorphic to each other, it doesn't really matter which representation you use. And as such we define the polynomial ring with two variables R joined X and Y to be then the ring that represents these two isomorphic rings. Pick whichever one you want, it doesn't matter. And so this is the ring of polynomials over R with two indeterminates. And by induction we can repeat this process because this is a ring, so we could add on a new variable and that's a ring, so we could add on a new variable and we could keep on going, going, going, going. And since we have all these variables it's probably just easier to call them X one, X two, all the way up to X in. And so this is then the ring of polynomials with coefficients coming from R with indistinct indeterminates which we do require that these are different elements they're not the same element and we do allow them to commute with each other. So when we talk about these polynomial rings X Y equals Y X, X I times X J is equal to X J times X I. With these polynomial rings, these multivariance polynomial rings the variables do commute with each other. Otherwise we might have a frustration when it comes to this isomorphism that should be stated. And that's important to state because I said earlier that with these indeterminate elements we're not assuming any algebraic relations there is one exception. I mean, other than fact that anything that's a consequence of the ring axioms we have to accept but the one exception is we do consider these indeterminate elements to be central. That is they commute with everything in the ring they can they commute with coefficients and they commute with other indeterminate variables as well in this situation. And in particular, if R is a commutative ring then this polynomial ring will be commutative if and only if if R has unity then this polynomial ring will have unity and that's an if and only if statement. If this ring R is a domain then the polynomial ring will be a domain if and only if again, this is just a consequence of induction. Because as you add on each new variable that's by induction a commutative or with unity or domain and therefore adding the new one keeps on going there. Now, I wanted to end this in this short video about multivariant polynomial rings by mentioning the very famous Hilbert Basis Theorem. We're not gonna provide the proof of that that goes beyond the scope of this lecture series but this is a very important result to state here that how does this multivariant polynomial ring how is it affected by the no-theorem condition so the ascending chain condition? If R is a no-theorem ring then it's true that R a joint X is a no-theorem ring and therefore by induction so would be R join X one X two a join X three a join X four and so if your coefficient ring is no-theorem then your polynomial ring is no-theorem as well as you put on any number of variables and all of these statements I'm referring to a finite number of variables if you were to have an infinite number of variables induction exact doesn't exactly work anymore you have to use some type of trans-finite induction and therefore some of these results might not be true so we will only consider the situation when we have a finite number of variables we're also assuming that the variables commute with each other one can actually construct a notion of a non-commutative polynomial ring where the variables don't commute with each other but again those are topics, very important topics in ring theory non-commutative ring theory to be specific that we are not gonna delve into in this lecture series