 Okay, I will start. I hope that everybody will be there soon. So we had started to talk about fields and field extensions. And so we finished by talking about, so we had done the degree theorem and introduced a degree of field extension and proved the degree theorem. And then we finished by studying simple algebraic extensions. Maybe I state the last theorem that we stated and started to prove. So theorem, so we are in the situation that k, large k is a field, let's say it is a field and small k in k is a subfield. So large k is a field extension of small k. And so we take an element a in k, which is algebraic over the smaller field. And we want to look at the simple algebraic extension generated by it. So we assume the minimal polynomial of a polynomial f a. And then we have the following statements. First, the simple extension generated by a can be written is isomorphic to the polynomial ring in x divided by the ideal generated by the minimal polynomial. And second, the degree of k a over k is equal to the degree of the minimal polynomial. And in fact, if the m is equal to this degree, then the elements one a, a squared and so on until a to the m minus one are basis, k a. So basis of k as a k vector space. Of k a as a k vector space. Okay, so we had basically done the proof of one. Maybe I can go through it very quickly again. So we have the evolution homomorphism from the polynomial ring in x, which maps to k a, which sends any polynomial g to g of a. Obviously this goes to the field generated by a because we just have polynomials in a in the image. So this is a ring homomorphism. And our definition of the minimal polynomial was that the kernel is the ideal generated by f a. So if we take l to be the image, then we know for one thing that, so this kernel is an ideal in the principle ideal ring k x and principle ideal domain to x. So the kernel is a maximal ideal. So f a is a maximal ideal in the principle ideal domain k x. And thus we have that the quotient k x model of f a is a field. And by the homomorphism theorem, the quotient is isomorphic to l by the homomorphism theorem. No, if you have a subjective ring homomorphism, then the image is isomorphic to what the source divided by the kernel. So in particular, l is a subfield t a. And we have to basically show however that it is equal to k a because our statement is that this. So the first point is that it contains k as the image is cons of the constant polynomials. And l also contains a because if we take the polynomial x it is mapped to a a is equal to the evaluation at a of the polynomial x. So we have that l is a subfield of k a which contains a and k. But this means it is equal to k because k a is the smallest field which contains k and k a. So thus, so this describes this thing. Now the second statement is that we want to show to compute the degree and so we have so two in one. We have in particular shown that the evaluation map at a from k x to k a is surjective because after all l was just the image. So we have a subfield of k x to k a because after all l was just the image and l turns out to be k so it's a surjective map. So therefore we can write k a just as the image of the evaluation map that means it's the set of all g of a where g is a polynomial in k x. That's what it is. Now we want to show but maybe we don't need all polynomials we want to show that first show that so if m is the degree of this minimal polynomial that these elements here spend k a as a k vector space. So one a until a m minus one generate k a is k vector space. So let's take any element in k a we can write it like this. Let's k of g of a in element in k of a. So any element in k of a looks like this. So it's a sum g in k x. So we want to show that you can find another polynomial of lower degree which gives us the same element in k a and obviously we do this by division with rest. Maybe I write it differently. So just to make it logically clear that b being element in k of a then but by what we have seen here then it exists a polynomial g in k x. So we want to show that we can find another polynomial in k x such that b is equal to g of a. So if the degree of g is bigger equal to m the degree of the minimal polynomial we can make division with rest. So right do we have by division with rest the g is equal to q times f a plus r with q and r in k x and r is equal to zero or the degree of r is smaller than m. Okay and now the point is obviously that r of a is equal to g of a. So then we have because f a of a is equal to zero we have that g of a is equal to q times f a of a which is equal to zero plus r of a so q of a which is equal to r of a but that means that we can write this is equal to r of a and so if you write r as sum i equal to zero to m minus one because the degree is at most m b i x to the i we find that b is equal to sum i equal to zero m minus one b i a to the i that means it is in the linear span of a of one up to a. Okay so this shows that these elements generate k a as a k vector space and now we also want to show that they are linear dependent so that they are basis. So assume one a m minus one are linear dependent so then this kind of the obvious thing then there exist say elements say b one b zero to b m b m minus one in k such that you know this linear combination is zero so sum i equal to zero to m minus one b i a to the i is equal to zero but obviously this means we have a polynomial of this degree which vanishes at a so thus for g equal to sum i equal to zero to m minus one b i x to the i we have g of a is equal to zero so we know that the so that means that g lies in the kernel of this evaluation morphism of a so it is a multiple of f a is in the kernel of the evaluation at a which is the ideal generated by f a so we have the g is equal to so it means that g is in f a but you know the degree of g is at most m minus one the degree of f is m so this is only possible if g is equal to zero so that means that g is equal to zero so these coefficients were all zero zero is equal to b one m minus one is equal to zero that means these elements were linearly independent so we have not found a non-trivial linear combination which is zero so one a until a to the m minus one are linearly independent so we have seen it somehow I think this proof is quite straightforward maybe most of you would have found it not so exciting from a certain point because one only does the tries the obvious things and they work so now in particular we can if you think of it this gives you a completely explicit description of this field k a as this portion k x model then s k x model of the f a you can also maybe make it slightly more ever so slightly more explicit as follows so if k a over k is a simple algebraic extension and c of degree m and we have f is the minimal polynomial or maybe I still write f a of a then we can explicitly say describe this as follows can describe k a as follows so as a set or as a k vector space k a is equal to the set of all kind of polynomials evaluated a sum I equals zero to m minus one say bi a to the I where the bi okay bi elements of k so as I said it's just this vector space of linear combinations of the ai's and the addition and the multiplication are essentially the obvious ones as polynomials so addition and multiplication defined in the obvious way as if these were polynomials no you just s for polynomials in a so remember the rules you add them component wise you multiply by this product formula with the additional rule so so with the extra relation so if you write f a equal to sum i equals zero to m say c i x to the i then we want if we multiply two polynomials we can get we get the polynomial in a whose degree might be bigger than m bigger equal to m and so we have to see how to get lower and we do this by applying this namely with the extra relation that a to the m is equal so this is a monic polynomial so we have that c m is equal to one equal to minus sum i equals zero to m minus one c i a to the m to the i and this is used to eliminate all polynomials in a of degree bigger equal to m or whenever we have a power of a which is bigger than m we can replace the m by this until we get only terms of lower degree so it's a very simple description of this ring of this field so just as an example for instance you can view the construction of the complex numbers out of the real numbers joining i as a as a simple algebraic extension so so you have that c is equal to r i is a simple algebraic extension where i is the square root of minus one in the complex numbers and the minimal polynomial of i is x squared plus one which means i squared is equal to minus one and so then if we just apply this rule we find that c so thus according to this remark we have that c can be written as a set of all a plus where a and b are real numbers as it should be and the addition is the obvious one so a one plus b one a i plus a two plus b two i is just component wise so it's a one plus a two plus b one plus two times i and the multiplication is given according to this rule so if I take say a one plus b one i times a two plus b two i according to this we first just multiply them out as polynomials in i so this is a one a two plus a one b two plus a two b one times i plus i squared times say b one times i squared and then we apply this thing that the i squared is replaced by minus what we have here so i squared is replaced by minus one so this is equal to a one a two plus a one b two plus a two b one so i squared is minus one minus b one b two plus this term here a one b two plus a two b one times i the rule of computing and the complex numbers that you know can be viewed as a special case of this okay so much for the moment about these simple extensions so now we want to characterize finite field extensions you remember that the finite field extension is one where the degree of the field extension is finite and so the claim is that the field extension is finite if and only if it is algebraic and is obtained by adjoining by you know and is generated by finitely many elements so it's a following theorem say let L over K be a field extension then L over K is a finite field extension if and only if first it is algebraic and second there exist finitely many elements so a one whatever a n in L such that L is the field extension generated by these finitely many elements so if I alright you know it was not clear that a finite extension has to be algebraic because for extension to be algebraic it means that every element of L is algebraic over K if we take for instance the extension generated by just one element contains many other elements it could be some of them aren't but we now easily see that this is not the case the proof is not particular difficult but so we assume we have a finite field extension we want to show it's algebraic and it's generated by finitely many elements so let M be the degree of the field extension which is some finite number so then if I take any element a it's kind of the same trick we have used before any element a in L we find that the elements one a until a to the M so these are M plus one elements in L they must be linear dependent over K in the K vector space so linearly dependent over K so in the K vector space L you know we have one more than this dimension so it follows there exist P1 P0 to Pm in K such that that's obviously the same thing as before that sum I equals 0 to M A to the I is equal to 0 and the Bi are not all 0 so that means if I take the polynomial which has these as coefficients then this vanishes at A G equal to sum I equals 0 M Bi X to the I which is the polynomial with coefficients in X and it's not the 0 polynomial satisfies G of A is equal to 0 so A is algebraic A is algebraic over K so we even find that we have a bound on the degree of any polynomial which vanishes and on the other and we can also simply how do we find these finding many elements let us just take any basis of L as K vector space so if A1 until A in this case the degree is actually M is a basis of L as a K vector space then we can write any element in L as a linear combination of of these AIs in particular the field generated by these elements is the whole field so this is one direction and now the other one so we prove this so assume that L is algebraic and we have this with this number N generated by N elements we make induction on N in fact we don't even so if the case N equals 0 is trivial because then we just have our field L is equal to K and the degree is 1 we make the induction step so we assume that our L can be written as K A1 to An plus 1 for some elements AI in L and we put L prime generated by the first N of them then obviously there is some degree so we have that L prime over K is some finite number M by induction so it's finite which I can call M by induction and so I have to see by the degree theorem we have to say see only what L prime what L over L prime is so we know that An N plus 1 is algebraic over K and thus over L prime because we know that it's algebraic over K and so every element in L is algebraic over K particular also this thing so if we take G the minimal polynomial of An plus 1 over L prime and then we know that the degree of L over K is equal by the degree theorem to the degree of L over L prime times the degree of L prime over K which this one was M and this degree is equal to the degree of the minimal polynomial because L is equal to L prime from L L prime adjoint we just have adjoint one more element so this is also finite so we see that every finite algebraic every finite field extension is algebraic and is kind of obtained by joining finitely many algebraic elements and vice versa so we have a few we have a small corollary so if you have any field extension we can look at a bigger field which are algebraic over the smaller field and then the claim is these themselves form a field and they form an algebraic field extension of the smaller field so let K over K be a field extension and at L be the set of all elements say A in K such that A is algebraic over the smaller field K then L over K is an algebraic field extension L is a subfield of K so we know obviously K is contained in L and so to see that this is an algebraic obviously because every M K is algebraic over K now we want to see that L we have to see that L is a subfield of K because obviously every element by definition every element of L is algebraic over K so the field extension will automatically be algebraic so to show L is a subfield of large K well that's actually quite simple let us just take two elements in L and these are two we have to see that the sum, the product whatever the difference or the quotient is contained in L I mean unless one of them is 0 so maybe we can assume that they are not 0 so to show that to say A plus B A minus A A times B and B A to the minus 1 are all elements in L then it is a field the point is we can look so if we look at KAB the field extension generated by A and B this is a field extension generated by two algebraic elements so this is a finite according to the theorem this is a finite extension so it is algebraic thus algebraic so therefore KAB is contained in L because all elements in KAB are algebraic and therefore obviously the sum and the product and so on are also in KAB ok so that's quite simple and as I said that this is an algebraic extension so as an example we use this kind of abstract but we can for instance look at Q bar this is supposed to be the set of all elements all complex numbers which are algebraic over Q so we can look at this set and this is an algebraic field extension of Q so Q bar over Q is an algebraic field extension but obviously it's not a finite extension you can take nth roots numbers for any n so of infinity ok that's simple we can also see that if we take a composition of algebraic field extensions it will also be algebraic so if we have a smallest field we make an algebraic field extension of it and then we make an algebraic field extension of the bigger one then the biggest field or the smallest field is also algebraic and in fact yeah so it is independent of whether these field extensions are finite or not so let say k over L and L over small k the algebraic field extensions then the extension of the largest field over the smallest is also extension is also algebraic so let's see so I don't make the assumption that these field extensions are finite but I still want to use the fact that how you do that finite field extensions are how they are related to algebraic extension that finite extensions are always algebraic so how do we do that so proof so let U any element of k we have to show that k is that this element U is algebraic over small k first we know it's algebraic over L so then there exists a polynomial say in L of x which is not the 0 polynomial such that f of U is equal to 0 so we can write f sum i equal to 0 n whatever n is a i x to the i where the a i are some elements of f now we want to so we want to know whether the thing is algebraic over k so see where the i so we have that this a i are elements in L and therefore the algebraic over k so we take m make another field m which is the smallest field k and we join the element the coefficients a 0 to a n the a i are elements in L so they are algebraic over k and they are finally many of them in this case n plus 1 so this m over k is a finite extension a i thus m over k is a finite extension and we also have if I take m of U over m so if U is an algebraic element so it's algebraic over L but it's also it's also algebraic over m because the coefficients of the polynomial are all out of m so this extension m U over m is also an algebraic extension and as we have only joined one element it's a finite algebraic extension and so we find that the degree of m of U over k is again by the degree theorem the degree of m of U over m times the degree of m over k and these numbers are both finite so therefore this is finite so this extension here is a finite extension so as it's a finite extension we know it's an algebraic extension every element here is algebraic over k U is algebraic over k it's finite thus algebraic therefore U is algebraic over k and so we indeed find that U was an arbitrary element in k so every element in k is algebraic over small k so now we will talk about extensions of field homomorphisms so basically field homomorphisms so we will later do Galois theory which means to study field extension via their Galois group and the Galois group is the group so if L over k is a field extension the Galois group will be the group of k isomorphisms k automorphisms of L so somehow therefore as a something one needs to therefore one somehow needs to study how field homomorphisms from a from a smaller field extend to a field extension and so we want to do this here so so the thing basically show at this point is we want to show that if so I have again say k over k field extension we want to show that if say we have two elements A and B in the larger field k which have the same minimal polynomial over the smaller field field small k then there is an automorphism then there is a small k automorphism of large k which sends A to B and there is a unique one then there exists a unique k automorphism not of from k sends so phi maybe with phi of A is equal to B so if two polynomials if two elements have the same minimal polynomial then there is an isomorphism between the fields automorphism then there is an isomorphism from the field the field extension generated by this element the field extension generated by that which sends this element to that and there is actually a unique one there is only one such thing so we prove however and this will be important when we want to do this Galois theory as I try to hint and we will however need a slightly more general result where the difference is that we even here we don't just start with one field k but we have two different k's we have an isomorphism between them and we want to extend it so let me formulate it so definition so let say phi from k to k prime we have field isomorphism maybe I and let say L over k and L prime over k prime B field extensions so an isomorphism say large phi from L to L prime is called an extension of small phi if it's restriction to k is this given map it's in particular if we restrict phi to the small k it maps to k prime okay so let's now we want to study for this simple algebraic extension the existence and uniqueness of such things so maybe I for need this trivial observation if I have a field isomorphism so if phi from k to k prime is a field isomorphism then this gives me also an isomorphism between polynomial rings in the obvious way then I could call this phi star from kx to k prime x which sends any polynomial some i 0 n ai x to the i to the polynomial where we apply phi to the coefficients so this is a ring isomorphism so this is essentially completely obvious you can see that it sends the sum to the sum and it's easy to check it sends the product to the product and obviously as this is an isomorphism if I take the inverse isomorphism here and apply it in this way this gives us me the inverse map between the polynomial rings so now we want to make the statement about the extension of polynomial rings so let again phi from k to k prime be a field isomorphism and we have again our field extension l over k l prime over k prime field extensions and now we take a in l which is algebraic over k and we take so with the specified polynomial polynomial f a and now we take the element a prime in l prime to be a 0 of the image of the minimal polynomial so then there exists a unique extension of this field isomorphism so maybe I will a unique extension of say phi from k a to k prime of a prime of our isomorphism phi with so this means that the extension means that phi restricted to k is equal to k prime and the thing which makes it unique is that phi of a is equal to a prime so there exists a unique extension of this original map phi which sends a to a prime as usual if something exists and is unique the normal approach is always that you first prove the uniqueness you just see that the conditions determine it uniquely and this will actually tell you what it is and then you check for that thing that it satisfies what you want so that occurs all the time okay so if we start with the uniqueness so let phi from k a to k prime of a prime be such an extension so with phi of a is equal to a prime now we have to see whether we can and am I so note we have seen that k of a can be written as the set of all g of a with g is an element in kx in the same way k prime of a prime is equal to the set of all whatever g prime of a prime h of a prime where h is an element in k prime of x so every element in k a so it can be written as a polynomial evaluated on a and in the same way well and so and then the ring structure is in particular involves that you additional modification of polynomials we know if so so if b is an element of k a then you can write b equal to g of a for a some polynomial for g in kx so if you write g for instance equal to i equals 0 so n bi x to the i bi elements in k well then what is phi of our element b the same as phi of g of a that is phi of some i equals 0 to n and 8 to the i now this is a ring homomorphism so it so it we know that phi is supposed to send a to a prime and you know and on the bi which elements of k it's supposed to be small phi so this is equal to the sum i equals 0 to n like first write it again phi of bi phi of a to the i because it's a ring homomorphism and phi of bi is equal to phi so this is equal to phi small phi of bi and phi of this is equal to a prime so this thing is equal to some which in other words is nothing else than g prime or was it phi prime of a prime so we find that we have you know we have actually computed the element so that it's uniquely determined because just from the definition we were able to compute it so now we have to check that this is a ring a field isomorphism maybe we should have checked it but anyway I don't existence so we define phi k a to k prime of a prime by phi of g of a and k of equal to g of a is equal to phi star of g of a prime so by this so yeah so in principle I should have checked that this is well defined so it's easy this is well defined so if two polynomials g and h give me the same element in k a then the difference is a multiple of the minimal polynomial and then you can easily see that the difference maps to zero and so gives me the same so we get the same element here you can easily check that same as before and but on the other hand it's clear from the definition that this is a homomorphism I mean once it's well defined it's because phi is a homomorphism no because the multiplication is precisely given by and it is clearly and phi is also clearly subjective because we know that this map phi star is an isomorphism from kx to k prime so obviously I get if I apply this if I have all polynomials applied to a by applying phi star I get all polynomials applied to a prime so the map is subjective and now we have a subjective morphism between two fields then it is an isomorphism is an isomorphism I mean in the notes there's something more complicated but because you know the kernel of a field isomorphism is an ideal so it must be either 0 or the whole of the field but it cannot be 0 because this map is subjective it cannot be the whole of the kernel cannot be the whole of the field so the map cannot be the 0 map because it's subjective so it is an injective map so p is injective as a non-trivial homomorphism of fields so we find that this is an isomorphism of fields and finally it is clear by definition where do I have a thing so for an element b in k we have that b is a constant polynomial so we apply this you know if you apply to a constant polynomial this map it's just the map phi the constant polynomial is just if I have the constant polynomial b then it's b times a to the 0 and the map is by sending it to phi of b is equal to phi star of the constant polynomial b which is just phi of b so that means that phi restricted to k is the small phi and by definition we also have that if we take phi of a you know remember how do I get a by as evaluating a polynomial in kx at a we do this by taking the polynomial x so this is phi of the polynomial x evaluated at a so according to our definition we are supposed to apply phi star to this but you know the coefficient here is 1 so the phi star maps x to itself so this is you know the map here is obtained by evaluating it at a prime so this is x evaluated at a prime which is a prime and so we have precisely the statement that we have a unique extension of this map from k of a to k prime of a prime which sends which is the map yeah this is obviously a misprint phi restricted to k the original map phi and phi of a is equal to a prime okay and the thing which we are most interested in is the thing which I mentioned in the beginning namely if we send set k equal to k prime and let the map and let the map phi that we have here to be the identity that we get in particular the following corollary so let say L over k here be field extension and we take two elements in L which have the same minimal polynomial so algebraic with the same minimal polynomial so the algebraic with the same minimal polynomial then there exists a unique k isomorphism phi from k of a to k of a prime which sends a to a prime so somehow you if you have a field extension you have several zeros of the same polynomial then you find that there is an extension there is a field isomorphism which sends one and the unique one so we will later see that this is an important tool for us this is up and you know this is a special case where we have to prove nothing so before going on I wanted to just make a few remarks about the algebraic closure of a field we are not really going to use it in the future it's just you know for completeness so that you have seen what the algebraic closure is I also will not prove things I will just state stated and state the theorem about the existence of an algebraic closure so so we want to introduce the algebraic closure of a field so this will be so this will be a field extension L over K an algebraic field extension such that every polynomial with coefficients in the smaller field K has a zero in L so all polynomial equations in one variable with coefficients in K can be solved in L but the field extension is algebraic for instance it's certainly not true that C is the algebraic closure of Q because it's not an algebraic extension so let me write the definition once again completing precisely although it's already written so let L be a field we say L we say L is algebraically closed here I actually wanted well whatever what I said is also correct but it's it's maybe not what I will say later it's maybe it would be an exercise for you to check that it's equivalent to what I would say now so I take a field we say that L is algebraically closed if every polynomial F with coefficients in L has a zero in L if well I wanted to actually say differently if the following equivalent statement holds so the first one is that every polynomial F which coefficients in L has a zero in L and the second statement is that every polynomial L with coefficients in L can be written as a product of linear factors so it splits into linear factors so every F in Lx splits over L into linear factors by this I mean the following I can write I mean I mean I'm obviously supposed to mean that I should say it so I mean that F can be written as say B times X minus A1 times say X minus An so there exist B and A1 to An elements in our field L such that F looks like that so F is a product of linear factors I need to B because it might not be money now obviously these two statements are equivalent 1 and 2 equivalent 2 implies 1 is completely trivial because you can see the zeros F has zeros A1 to An in this case which are elements in L and the other direction is not much is by trivial induction so 1 2 so so let A be a zero so A and L be a zero of an element F in Lx so then we can write F is equal to X minus A times G where G is an element in Lx and the degree of F of G is equal to the degree of F minus 1 and so then easily you can see that you can deduce the statement of 2 by induction on the degree of F on the degree of F I mean that's clear because in C that Q is not algebraically closed for instance X squared minus 3 has no zero in Q and R is not algebraically closed because the typical polynomial X squared plus 1 has no zero in C in R obviously it has zero in C but I claim that this Q bar which I have introduced before is algebraically closed this does not I mean completely directly follow from the definition but you can easily check it because by definition it consists of all the elements which are algebraic over Q so every polynomial with coefficient has a zero but I claim it follows from that that also polynomial with coefficient Q bar has a zero in Q bar that's okay and then I want to introduce this algebraic closure and I think with that I will stop so definition a field over K is called an algebraic closure so a field extension is an algebraic closure of K by closure if L is algebraically closed and the field extension is algebraic if L over K is algebraic is algebraically closed and you might check as an exercise that this is equivalent to what I wrote here so one can prove that every field has an algebraic closure it's a statement of very much of the same kind that every vector space has a basis in fact it's proven in the same way with the lemma of so on so with the action of choice it's a kind of standard application of the action of choice so I will just try it out because we don't really care so much so theorem so the first statement is every field K has an algebraic closure and it's also say and it's also essentially unique it's unique up to isomorphism so second if K and L are two just K then K and L are small K isomorphic so there's an isomorphism between them which fixes K in particular they are isomorphic and K are K isomorphic so then I should write it out then there exists K isomorphism phi from K to L as I said it's quite simple and so one can check it's kind of clear that so as an example you could easily check that for instance Q bar is an algebraic closure isomorphism is a unique algebraic closure of Q and obviously it's also and Mark and you know that C is algebraically closed I mean I don't know whether you have had it yet this is proven in complex analysis I don't know whether you have already covered it but that's the way and it's therefore C is the algebraic closure is an algebraic closure unique after isomorphism of the real numbers maybe as my time is essentially up I will stop here so we see each other the day after tomorrow are there questions comments okay so thank you