 Welcome back. We are studying the algebraic integers in Q root D. Q root D is the set of all expressions of the form x plus y root D. So, these are all the numbers where we are looking at D to be a natural number which is square free. So, we have that Q root D is in fact contained in R. There is some theory which holds for D negative also. So, we will mention when we come to that, but mainly we will be considering the set Q root D where D is positive. We observed in our last lecture that the set Q root D is actually a field. You can add and subtract any two elements from each other and remain in the set Q root D. You can multiply any two elements and remain in Q root D. You can divide by a nonzero element and remain in Q root D. So, in some sense the elements of Q root D behave like the rational numbers. We have the addition, multiplication in rational numbers also and we can divide by any nonzero rational number to any other rational number and we are in the set Q. So, Q root D is a slightly bigger subset of complex numbers than Q and it has the same property that it remains a field. We then talked about what are called algebraic integers. So, we have the normal integers z sitting in Q and these algebraic integers they are the analog of z. And we saw the definition in our last lecture that these are the elements m plus n root D in Q root D that satisfy x square plus bx plus c equal to 0 with b and c integers. Now, we have observed that any element in Q root D is going to satisfy the equation of the form ax square plus bx plus c equal to 0 and here the a is clearly nonzero. Then we can divide by a and we get that x square plus b by ax plus c by a. This is also satisfied by the number m plus n root D. But here because a is nonzero and a can be any integer the elements b by a and c by a then need not be integers. Here we started with a, b, c being integers but if you divide by a then b by a and c by a need not always be integers. Therefore, any element in Q root D need not satisfy such an equation with b and c being integers. There are going to be some very specific such elements which will satisfy this particular equation. The set of elements which satisfy these type of equations are inside some very special set and if we denote Q root D by this letter k then the set algebraic integers k is denoted by this symbol ok. So, this is the set of all algebraic integers in Q root D where k denotes Q root D. It turns out that this is a ring you can add and multiply by any two such elements and you will be in the same set. And therefore, the main question is what are the elements by which you can divide. But even before we tackle that question the main question that we want to know is whether there is a nice description for such set of elements for these elements. Of course, the description is there given by the definition that any such element has to satisfy what is called a monic polynomial. Monic meaning the leading term the coefficient of the highest degree monomial is 1. So, the algebraic integers are the ones which satisfy monic polynomial over integers. This is one description but is there a better description. For instance for quadratic rationals we gave the definition saying that these are the ones which satisfy some a x square plus b x plus c equal to 0 where a b c are integers and something else b square minus 4 a c is a natural number and it is not a square and so on. So, these are the things that we had while defining a quadratic rational but then we also said that these are nothing but x plus y root m where x and y are rational numbers and m is a natural number which is non-square. So, this is some a simpler description for quadratic rationals. Is there a simpler description for algebraic integers in q root d? Yes, there is a nice very simple description. So, this description depends on whether d is congruent to 1 modulo 4 or not. When d is not congruent to 1 modulo 4 then the algebraic integers are of the form x plus y root d where x and y are integers. So, note first of all this is one very simple observation that we make that x being an integer is in ok and y root d if y is an integer is also in ok because any integer is going to satisfy a monic degree 1 polynomial. So, you can construct a monic degree 2 polynomial satisfied by the integer you may just take the square of that polynomial. Similarly, if I have y root d where y is an integer then we can again get a nice or you know even before we go to y root d you just consider that root d satisfies a degree 2 polynomial whose leading term is 1 namely x square minus d. So, root d is in ok x is in ok and this would imply that x plus y root d is an algebraic integer whenever x and y are integers. This is a very easy thing to see assuming that ok is closed under addition and products but this is a converse to that it says that everything in ok is of the form x plus y root d where x and y are integers this is when d is not congruent to 1 modulo 4. If you happen to have d to be congruent to 1 modulo 4 then this extra element 1 plus root d by 2 that is also an algebraic integer. Because 1 plus root d by 2 into 1 minus root d by 2 this happens to be 1 minus d by 4 which is an integer remember d is congruent to 1 modulo 4 and the sum of this element and its conjugate is also an integer. So, you can easily construct a degree 2 monic polynomial satisfied by the element 1 plus root d upon 2 when d is congruent to 1 modulo 4. So, just like the previous analysis 1 is an algebraic integer 1 plus root d by 2 is an algebraic integer so their integral combinations will give you algebraic integers. In particular you can write it as x plus y root d where x and y are of some certain type of half integers they are of the form m by 2 n by 2 where m and n should have same parity both are odd or both are even. So, this is the description for algebraic integers in both the cases. To remember this what you should note is that d can never be 0 modulo 4 because we are taking d to be square free d can never be 0 modulo 4 d should not be divisible by 4 any square free integer cannot be divisible by 4. So, the only possibility for d is 1 mod 4 2 mod 4 3 mod 4. If it is 1 mod 4 then 1 plus root d by 2 is an algebraic integer by the calculation that we have done. So, when it is 1 mod 4 it is an algebraic integer and in all other cases it is not an algebraic integer. You should only remember when 1 plus root d by 2 is an algebraic integer or not that will tell you whether you should look at d congruent to 1 mod 4 or 2 mod 4 or 3 mod 4. So, this is the description that we have for the set of algebraic integers the ring of algebraic integers. Now we want to know which of these are units. So, we want to know what are the elements x plus y root d whose inverse is also in the same set ok. We want to know what are the elements which are divisible in the ring of algebraic integers which are invertible in the ring of algebraic integers. That is the question that we want to study we call any such element to be a unit. So, these are m let me not use m n these are u plus v root d in ok such that 1 upon u plus v root d is also in ok these are the elements that we want to compute. So, this will then form a group of elements because whenever you have a unit then multiplied by any other unit you are going to get suppose alpha 1 is a unit and alpha 2 is a unit. Then alpha 1 inverse which is 1 upon alpha 1 is in ok 1 upon alpha 2 is also in ok. So, the product will give you that 1 upon alpha 1 into 1 upon alpha 2 is a product of 2 elements of ok which is also in ok it would mean that whenever alpha 1 is a unit alpha 2 is a unit the product is also a unit. And of course, you have that the element 1 is there in the unit 1 is always invertible and by the very construction a unit would mean that its inverse is there in ok. So, if you collect the set of all units in q root d these are called units but these are invertible only within ok. We are not looking at unit in the elements which are invertible in q root d those will be all nonzero elements of q root d. We are looking at numbers which are invertible in ok and as we have seen in the last slide these are of some very specific type these are of the type some x plus y root d where x and y are either integers or they are half integers. So, in some sense we are looking at a discrete set when you have x and y you know the analogy you should keep in mind is that if you plot the set of integers on the real line then you have 0, 1, 2, 3 or on the negative side minus 1, minus 2, minus 3 and so on. The distance between any two successive such elements is 1. What you have is that there is an epsilon such that the distance between any two is at least epsilon. For the set of integers you can take it to be 1. If you have half integers m by 2, n by 2 you allow m to be an integer then you have 0, 1 by 2, 1, 3 by 2, 2, 5 by 2, 3 and so on then your epsilon is 2 then the epsilon is 1 by 2. So, there is whenever you have a set of real numbers with the property that any two are apart from each other by some minimum distance then the set is called discrete. It is discretely divided, it is discretely situated, discretely located in the set of real numbers. On the other hand there is a set called compact set we will come to that later. So, what we have here is that our elements are discrete and these are also invertible. So, we are putting two very strong conditions on these sets and so as it turns out the units in Q root D can be explicitly described. There is a very important theorem by Dirichlet which will describe this. But before we go to that let me observe one thing for you that we have seen that the norm of u plus v root D is u square minus d v square which is always a rational number if alpha equal to u plus v root D is in okay. If alpha equal to u plus v root D and we take alpha prime which is alpha minus v root D which is also in okay for the description that we have given of algebraic integers. Then the norm of alpha which is u square minus v square D and alpha plus alpha prime which is to you these are in fact integers. There is a very nice and simple proof for this because given any such alpha u plus v root D we can construct a monic polynomial over integers satisfied by alpha. But if alpha is in okay then the monic polynomial should have integer coefficients the earlier was monic polynomial over rationals. So here the alpha should satisfy a monic polynomial over integers and then it turns out that u square minus v square D and to you must be integers. In fact this goes into the proof of the description that we have given in the last slide for writing down any algebraic integer in that particular form. So this is what we use. So we have that alpha alpha prime is an integer the norm is an integer and further we have that norm alpha into norm of its inverse is the norm of alpha alpha inverse which is 1. Both of these are integers so you have an integer dividing 1 in the set of integers and the only invertible integers are 1 and minus 1. If you have norm alpha to be anything other than 1 and minus 1 you cannot multiply to it by again an integer and get 1. Let me recall for you that these are all the elements u plus v root D in okay with the norm equal to u square minus v square D to be 1 or minus 1. These are precisely all the elements which are the units in the field q root D and here we have a theorem due to Dirichlet which describes the set of units completely. In fact the theorem of Dirichlet goes even higher. Here we are looking at elements which are of the form a plus b root D and so you have what is called quadratic. These are the elements which satisfy a quadratic polynomial. Dirichlet's theorem is a general theorem it is a very well celebrated result. It describes units in the set of in the field of algebraic numbers where you have a finite degree for the field. The field of algebraic numbers precisely mean that you are looking at the elements which are of the form a plus a 0 plus a 1 alpha plus a 2 alpha square plus dot dot dot plus a k alpha power k where that alpha satisfies a polynomial of degree n over rationals. You are looking at the ring of algebraic integers in that field of algebraic numbers and there we describe the units that is the theorem of Dirichlet but when applied to this particular specific case it gives a simpler description that whenever your D is negative then there are only finitely many units. So, this is an instance that we will talk about D being negative. Further if your D is positive then q root D has infinitely many units. So, this is a very nice dichotomy that we have whenever D is negative there are only finitely many units and whenever D is positive there are infinitely many units and we will see this in our now following part using only very basic theory. So, Dirichlet's theorem for quadratic fields can be proved by bare minimum. So, let us see what it involves note that the norm has to be 1 or minus 1. So, we are looking at something which is of the form u plus v root D where u and v are either integers or half the integers and then you are looking at u square minus v square D to be 1 or minus 1 this is the thing that we are looking at. In particular when you have D to be positive then the group of units in fact can be explained and it is isomorphic to this group called Z cross Z by 2 Z. So, this second statement is also something that we will see. Let us start with Dirichlet's theorem for D less than 0 we are going to describe all elements of the form u plus v root D in. So, there is the natural condition on u and v with the property that norm is equal to 1. So, we are looking at solutions to u square minus v square D equal to 1 and remember D is negative. So, we have u square plus e times v square equal to 1 we are looking at e equal to minus D. So, these are the solutions that we want to describe. Now notice that u and v are integers. So, if you have that your v is 0 then we get that u square is 1 which gives that u has to be plus or minus 1 these are called the trivial units. After all 1 is always a unit and minus 1 is also a unit. So, the element the units 1 and minus 1 will always come in the group of units whatever field you take and the then the ring of algebraic integers in that field you take 1 and minus 1 will always show up. So, they are called trivial units. We want to describe the non-trivial units. So, we will not consider this case. Therefore, we are going to look at v not equal to 0. If v is not 0 and suppose that your e is bigger than or equal to 5 remember we are taking e to be an integer our D and e these are integers D is negative. So, e is positive and moreover e is square free. Therefore, this is the same case as e bigger equal 4 but let us assume that e is now bigger equal 5 then e v square is bigger than or equal to 5 because v is an integer then e v square is strictly bigger than 1 because v is either an integer or it can be a half integer depending on what the discriminant does. So, v square will be simply possibly an integer or it can be a square of an integer divided by 4. So, you have if e is bigger equal 5 you are multiplying to a square of an integer by 5 by 4 and the v is non-zero. So, v square is a non-zero square therefore, the product is strictly bigger than 1. In that case we will never have a solution to u square plus e v square equal to 1. So, if you want to have a non-trivial solution so for non-trivial solutions for non-trivial units we must have D to be either minus 1, minus 2 or minus 3. These are the only possibilities for D to have a non-trivial unit in all other possibilities for D the only possible solutions are plus minus 1 which is a finite set. Remember what we want to prove is that whenever D is negative the group of units is finite that is what we want to prove. So, now let us take the case where D is minus 1. So, we are looking at solutions to u square plus v square equal to 1 u and v are now integers because minus 1 is not congruent to 1 mod 4. So, the algebraic integers for this D are of the form integer plus integer into root of minus 1. So, u v are integers and therefore, the only possible solutions are where u plus u is v plus minus 1 v is 0 or u is 0 and v is plus minus 1. So, these are the units which are plus minus 1 and plus minus i where i is a chosen square root of negative 1. So, we have an explicit description of units when D is minus 1 which is plus minus 1 plus minus i and now we go to D equal to minus 2 D equal to minus 2 or what is same as E equal to 2 then we are looking for solutions u square plus 2 v square equal to 1. Once again D is minus 2 it is not congruent to 1 mod 4. So, u and v are integers and the only solutions are where u is plus minus 1 and v is 0. If your v is non-zero then 2 times v square will take you beyond 1, u square is also positive. So, u square plus 2 v square equal to 1 has no solutions. Now, we come to an interesting situation where D is minus 3. Now, this is congruent to 1 mod 4. So, your u and v can be integers or they can be half of integers. So, when we are looking at solutions to u square plus 3 v square equal to 1 we write u as u 1 by 2 v as v 1 by 2 and take the 4 and put it on the other side. So, we are looking at solutions to u 1 square plus 3 v 1 square equal to 4 and u 1 v 1 are now integers and now you can easily see that this has solutions. If your v 1 is 0 u 1 has to be plus or minus 2 or you can have u 1 equal to plus minus 1 and v 1 is also plus minus 1. But of course, this does not give you the solutions when you have too many u 1's and too many v 1's. So, the set of solutions here is actually plus minus 1 plus minus 1 plus root 3 by 2 and plus minus 1 minus root 3 by 2. You get 4 solutions here and you get 2 solutions here. So, these are the 6 solutions and these are the only 6 solutions. In fact, we should note the following thing that the set of algebraic integers is a discrete set for d negative norm of u plus v root d which is u square minus d v square d is negative. So, this becomes u square plus e v square. This is actually the square of the modulus as a complex number. So, when you are looking at a discrete set of the set of algebraic integers and you are looking at solutions to the usual mod to be 1, then you are looking at the unit circle in the complex plane. So, you have on one hand the unit circle which is a compact set in mathematical language and then you have a discrete close subset of this compact set. And this is a slightly basic but on the advanced part of mathematics called topology that any discrete compact set has to be finite. So, this is the basic thinking behind this theorem that whenever you take such a field then the set of units in this has to be a finite set. So, we have described the set of units for d negative. We are going to describe the set of units for d positive in our next lecture and we will see how the continued fractions are used in giving the explicit description for these units. So, see you then. Thank you very much.