 So, in the previous module we have seen that these invariant A invariant subspaces have a critical rule to play in so far as our goal of getting this given matrix to a block diagonal representation goes. So, we have seen that in general if you have your underlying vector space v represented as a direct sum over i of subspaces such as w i such that each of these w i's is A invariant right. Then we are in for a special property which is that you try and find out a basis for each of these individual w i's stack them up together. They will form a basis for v and in terms of that basis now if you represent your operator A then it gives a structure such as a 11 a 22 until a k k right which is the best we can possibly do. So, this is our motivation behind searching for A invariant subspaces and we saw that one way of cooking up this A invariant subspaces through the kernels of certain special polynomials ok polynomials when their arguments become matrices the matrix A to be precise right. So, we have seen that you take any polynomial and you look at something like this this is not only a subspace but it is also an A invariant subspace right where of course f x is a polynomial right. So, this then motivates us to look deeper at these polynomials or dig deeper into the theory of polynomials to an extent and that is precisely what we are going to try and do now. But we will do well to remember that we are not studying polynomials in and of themselves without any motivation this is precisely going to serve as our motivation alright. So, with that in mind we will now launch into some properties of polynomials and see how these will help us in the long run. So, if I give you two polynomials of degree n and I ask you to check if the polynomials are equal what is going to be your strategy how you are going to verify whether the polynomials are indeed equal comparing the coefficients very good. But now suppose I have not told you what the coefficients are precisely alright, but I have told you some other property what other property would help you. So, for example, if there are two numbers nonzero numbers and you want to find out if they are equal what do you do subtract but that is one way divide yeah. So, you divide one number by the other and you see if it leads to unity. So, in case of polynomials it turns out that of course, comparing coefficients is one way, but if it so happens that you have been given say f 1 which is a polynomial and let us say f 2 which is also a polynomial and if I give you the condition in addition to this that f 1 divides f 2 what do I mean by that and f 2 divides f 1. So, when f 1 divides f 2 it means that this leaves this division leaves a remainder 0 what sort of a division is this I mean this is not the division that we do with numbers right this is the division of polynomial by another. So, just like we write for numbers we also have similar representation for polynomials. So, when we say f 1 divides f 2 you are trying to write f 2 x as some quotient polynomial yeah. So, only admissible variables here admissible objects here are polynomials otherwise you can do a whole lot of other things you can write rational functions that is not admissible right. So, some quotient times the divisor plus some remainder and our claim is that this remainder will be identically the 0 polynomial yeah. So, this is what we call when we were kids we learned this we call this the dividend this is the so called quotient just think about integers and it all goes through exactly the same manner and this is the divisor this is the remainder. So, when I say that f 1 divides f 2 it essentially means that the remainder must vanish. Now, in addition I impose the f 2 also divides f 1. So, what is it f 1 x can now also be written as some q prime x times f 2 x plus some r prime x where of course, because of the fact that f 2 also divides f 1 this must also go to 0 be the 0 polynomial identically. Now, both of these conditions are true can you not conclude that f 1 must be equal to f 2 if and only if yeah what else can be the possibility is it true. So, if I if I give you for example, x plus 1 and 2 x plus 2 do they satisfy this property. So, let us call this as f 1 let us call this as f 2 does f 1 divide f 2 f 2 divide f 1 they are not equal are they of course, not. So, now in addition if I impose and what else they are monic polynomials right that is very important what do I mean my monic monic means the coefficient of the highest degree of the polynomial is unity. Now, there is no way this this condition is ruled out because this one is not even monic. So, when I am talking about two monic polynomials such that one divides the other and the second also divides the first then they cannot help, but be equal is it not. So, just saying that these two conditions hold does not tell me that the polynomials must be the same, but only when I impose this restriction then they are indeed going to be the same right. So, that is another test for another equivalent way of saying I would not say it is a test because you do not really go about doing this while checking for equality of polynomials you actually compare coefficients, but this is an equivalent condition that will help us in many of our proofs. So, that is the reason why I wanted to introduce this notion of two polynomials be equal if they are both monic if they are not monic then you divide it out by the highest degree coefficient and if the highest degree coefficient is 0 then it is not even a polynomial of that given degree. You call something to be a polynomial of degree 5 only if the coefficient of x to the power 5 is non-zero if it is non-zero you can of course divide it out by the non-zero coefficient if it is not unity so that it becomes monic. So, you compare them after you turn them into monic polynomials yeah. So, then this comparison holds okay good. So, we will have to remember this as we go along because we will be using this. Now, the next thing I am going to claim is that f x is a commutative ring of course with identity. So, the identity being the constant polynomial 1 right. Can you check it is nothing to really prove here where does a commutative ring with identity vary from a field it is not a division ring which means it essentially does not guarantee that every element has a multiplicative inverse. So, take for instance x squared plus 2x plus 5. So, its multiplicative inverse would be 1 by x squared plus 2x plus 5 yeah, but this is not a member of this ring if at any non-constant polynomial will fail to have a multiplicative inverse within that ring. But all other properties you see which are there in the commutative ring the associativity the closure the commutativity yeah the existence of a multiplicative identity the additions and all those properties the additions and the multiplications and the distributivity everything else is imbibed in this the way we understand polynomials and their operations the additions and multiplications of polynomials right. So, there is absolutely no problem with any of those properties except for the multiplicative inverse. Now, of course, if it is a commutative ring it is also an integral domain. So, remember again the integral domain does not have necessarily the multiplicative inverse of every element, but what is true of an integral domain is that if you take two non-zeros then their product will also be non-zero. If the product of two elements is zero then at least one of them has to be zero that is exactly the property of polynomial that helps us to solve for roots right. When we take polynomials say x plus 1 into x plus 2 then we say that the roots are either minus 1 or minus 2 which means we are equating at least one of them to be zero. So, there is also an integral domain right. So, it is a commutative ring with identity it is also an integral domain okay. So, now for some perhaps some non-trivial ideas relating to commutative rings I will see how they reflect on polynomials. So, because it is commutative I mean you understand surely that commutativity has to do with the multiplication operation. It means whether you are multiplying from the left or whether you are multiplying from the right it makes no difference whatsoever right. So, now if it was not a commutative ring then we would have had to define in terms of the left operation or right operation, but because it is commutative we do not care about it instead we go ahead and define this object called an ideal okay. So, just pay very careful attention might seem like a very new notion and it is, but it is nothing fancy. So, suppose a which is a sorry as subset okay that is contain inside this alright that for f1 and f2 contain inside a f1 plus f2 of course this plus is as the addition is defined in the ring okay f1 and f2 also belongs to this. So, that is the first property and a second property for f contain inside this object which is a subset of the ring commutative ring with identity and g contain inside the ring itself f times g belongs to a then a is an ideal of so I am writing fx here you can just replace it with any general commutative ring okay it just so happens that fx is a commutative ring with identity as we have claimed and I ask you to verify that using all the properties that we have gone through many lectures back right at the beginning of this course. So, it will also serve as a good revision of the ideas. So, the point is this is exactly what an ideal is do you see a similarity with some other object that you have learned vis-a-vis let us say vector spaces or fields these are almost like similar in flavor to what we called a vector space when defined over a field. But when defined over a ring we do not have that luxury because of that one property that has gone on from that field right that field had a multiplicative inverse a ring does not. So, you cannot really construct a vector space out of it but at least we can create something like an ideal out of this ring okay what is it you take out take up any two objects from the ideal it is closed under addition yeah that is the first important property second is actually a little more interesting because this is quite straightforward to understand it is exactly like the vector space this one says you take any object from the ideal from the subset and the other object the second object you can pick out from the ring in general not necessarily part of the ideal. But once you multiply them they should form a part of the ideal right. So, to give you an example of what an ideal could be suppose a is defined as fx in a polynomial in the ring such that f of 5 is equal to 0 can you check that this is an ideal you take any two objects. So, you take f1 x and f2 x. So, let f1 and f2 belong to this subset first of all this is a subset of fx right from the definition okay take this. So, we know that f1 vanishes at 5 f2 vanishes at 5. So, what can we say f1 plus f2 at 5 is what f1 at 5 plus f2 at 5 is equal to 0 plus 0 is of course equal to 0. First property is verified the second property let gx belong to the ring any arbitrary object and fx belong to this ideal. So, this so called ideal we have not yet checked right. Then fx into gx evaluated at x is equal to 5 that is what we need to check for membership in the set A. So, this is what this is equal to f5 into g5. Now, I do not really care about what g5 is because f5 is already going to vanish anyway. So, this is just multiplication by 0 only leads to 0 right. So, this is an example of an ideal yeah let us give you two more examples one of course is very similar to this. So, this was example one just to get you comfortable with this idea. We will not be going into general ideals there are left ideals and right ideals when the ring is not commutative you have to define them as left ideals and right ideals we do not care about that because it is commutative ring. Let us take example two. So, this A is defined as fx such that fx can be written as qx times gx for a fixed gx and fx rather with fx comma qx sorry no we do not need that no fx is already there with qx is this an ideal yeah. In fact, it is a very special ideal this is said to be the ideal generated by g and represented as exactly like we did for the span yeah the same notation yeah. So, this is an entire set that can be characterized by this g so much so that I might even incorporate a subscript g here yeah let me do that because the g gets fixed once the g gets fixed this entire ideal gets completely defined by that g when you take any two polynomials f1 and f2 such that f1 is q1 g f2 is q2 g you add them it is q1 plus q2 g yeah you take any polynomial f from the ideal which is qx gx and you take any other polynomial say hx. So, the product is hx qx gx so h times q can be called as q tilde so that it is q tilde gx. So, again both those properties it is not very hard to check but I leave that to you to just write it up completely. The verification of the fact that this is an ideal and then once you know that this is an ideal I am just defining it to you that this kind of an ideal is a very special ideal it is called the ideal generated by this polynomial gx yeah all right let us take a third example such that fx is equal to summation psi i x g i x i going from 1 through k for instance with psi i x belonging to the ring the commutative ring and g i x being fixed polynomials what do you think this should be called by some analogy it is it is the ideal generated by the gis. So, this is also nothing but the ideal generated by g1x g2x until gkx all right that is what this is. So, you see that we can have an ideal generated by a single element we also can have an ideal generated by multiple of these right any questions up until this point about this is everything clear what have we done we have just defined polynomials yeah not define really you understand what polynomials are we have just told you certain properties that the ring of polynomials is a commutative ring with identity then we have just gone on to define this new object called an ideal and we will give you two three examples to show that we are not talking in out of thin air that there are indeed objects that are significant which come under this class of subsets of the commutative ring that are called ideals all right and then there is a special name for an ideal of this form which is an ideal generated by the singleton g and this is also an ideal generated by several of those fixed g's several fixed polynomials all right now we will next define something called so we understand ideals at this point all right. So, if I assume that we understand ideals we can take the next leap and say that there is something called a principal ideal okay so what is a principal ideal if ideal is generated by a single element it is a principal ideal okay so if you can basically fix up if you can fix up an ideal by just one element that means you are specifying an ideal by exactly one element and if I tell you that one element you know exactly what all fellows come inside that ideal right then such an ideal is a principal ideal right so clearly that example two we had was an example of a principal ideal just I am not going to answer it right away do you think this one is a principal ideal example three that we have on the board it is not okay okay we shall see we shall see okay it is a good point so some of you are saying it is not a principal ideal okay but at least you understand the definition of principal ideal the way we have described it right example two is a clear cut example of a principal ideal you in fact know that the g is indeed its generator next we have the principal ideal domain or a PID okay this is different from the PID you might have learned in a control theory course it has got nothing to do with it so what is the principal ideal domain so pay very careful attention if every ideal in an integral domain is a principal ideal then such integral domain is a PID yes I mean we take the domain from the integral domain but we call it a principal ideal domain so every ideal in that integral domain has exactly one generator or can be represented as being generated by a single element then such a integral such an integral domain is said to be a principal ideal domain you understand what this essentially implies if you are living inside such an integral domain where every time I give you an ideal you can tell me exactly oh this ideal you know is generated by this fellow and nothing more or less I mean it's uniquely characterized that entire subset inside that integral domain is entirely characterized by just that one element then it's quite straightforward right at least you think that it's a desirable quality to be had right it simplifies a lot of the successive steps that we want to do with it what steps we will come to that maybe much later perhaps not in this lecture but at least you do appreciate that living inside a principal integral domain has its advantages that's the main point see these are both nothing but definitions nothing to prove here that you agree this is a definition why did I define this because I don't want to have to say you know a horse is an animal with four limbs I would rather say a horse is a quadruped so I've defined a quadruped and then I'm defining a horse you see that's the reason we do we give these definitions not to make life messy but to make life simple because we don't want to write long things now imagine if I had not defined a principal ideal I would have had to write if every ideal in an integral domain can be generated by a unique element or one element only then such an integral domain is a PID there's a lot more words or phrases popped in there right on the other hand this I've defined principal ideal now based on my understanding of principal ideal I already know what this is so I'm now going to go ahead and define a principal ideal domain but that is not the most interesting thing these are just after all definitions the first non-trivial claim that we are going to make is this and we'll prove it in the next module but we will state this result so you agree that fx apart from being a commutative ring with identities also an integral domain so the claim is that this is a PID or a principal integral domain principal ideal domain I'm sorry principal ideal domain so fx the integral domain is a principal ideal domain okay that's the claim we shall be proving in the next module any questions on this so far apart from what the proof is like we'll have to hold on for a moment but any concept any concepts that you feel yes it's not necessarily unique but you can make it unique up to some we'll see that's up to some mechanism some to some property some operation we'll have to do and then we can make it unique at least in so far as polynomials are concerned we can do that but in and of itself it's not directly unique and you'll immediately understand why because you know multiplications of polynomials by integers or rational numbers or elements in a field do not leave the polynomial invariant it changes but we'll see shortly so that brings this module to a close