Upload Number Field with airSlate SignNow

Decline number field

welcome to field theory prerequisite is basically ring theory so that's the ring theory of fine and first abstract algebra course for this first part we'll review the main results for fields from basic ring theory it will give examples to motivate the rest of the course we begin with definitions we call commutative ring K is called a field all the nonzero elements are units so all nonzero elements have multiplicative inverses with a definition we split fields into two classes by characteristic so recall the characteristic of a field so we've been smallest non-negative integer n such that we had any element in the field to itself n times we get zero to find the characteristic we form the subfield generated by one and then two things can happen if one generates copy of the rational numbers can we the characteristic of K is equal to 0 otherwise one generates a prime field Z my P then would P a prime then the characteristic of K is equal to P if we put all these fields okay if type two together we call these the fields positive characteristic we're mostly interested in here fields with characteristic zero now for examples fields of characteristic zero we have number fields so the subfields of the complex numbers okay our main interest here we have function fields over the rationals so here we're going to take for instance rational functions with coefficients in the rational numbers its polynomials over polynomials we also have versions with coefficients over the reals and the complex numbers for another type of field okay so this type is advanced so we're not going to say much about it here we have P attic fields now p-adic fields are worth mentioning okay we'll take the rational sitting aside of real numbers we have a notion of distance just given by measurement today you take the usual notion of length for the P attic fields start with the rationals then we can measure length by considering divisibility by a prime that's gonna make for a very different kind of field and it requires a different intuition now a main theme in the course is going to be field autumn orphism x' so we want to review some results on ring homomorphisms now if K is a field the only ideals and K are going to be 0 and K itself so if I have a ring homomorphism say by 10 K to any other ring then 5 will have to be one to one so recall and I take kernel of a homomorphism must be an ideal because K is a field the only options are 0 and can't sell now our definition of ring homomorphism I had that v 1 goes to 1 so it can't be can't sell the kernel must be equal to 0 and that forces Phi to be 1 to 1 as promised with that any homomorphism between fields is essentially an inclusion that leads us to ask questions like when it's K a subfield eval when are K and L isomorphic and what are the field automorphisms on k ok field automorphism it's just going to be a ring isomorphism thye carrying k back into itself now if we're working with characteristic of k equal to 0 so the rational numbers live inside of k if I have a ring homomorphism v 1 is equal to 1 then we'll have that Phi is gonna fix the rational numbers pulling wise now this idea of having fixed field field automorphism and subfields we put all these together that's Galois theory and that's what we're gonna work up to for now we just will look at an example to get an idea of field automorphism so let's consider the real number square root of two so there's gonna be a word of polynomial x squared minus two over the rational numbers we could form the field Q adjoin squared - this is all real numbers of the form a plus B squared - 2 where a and B irrational if I want to formalize things okay we note I have a map going from polynomials over Q to Q 1 square root of 2 given by evaluation so I'll send the polynomial f of X to F evaluated square root of 2 this is a ring homomorphism and it's on 2 so we can invoke the first isomorphism theorem for rings to do that we need to identify the kernel now kernel of e 1 is just going to be principle ideal generated by x squared minus 2 because x squared minus 2 is irreducible over the rationals it's going to be a maximum ideal and Q a join ax no that means we have a ring isomorphism or in this case field isomorphism between Q join square root of 2 and cue a joint square root of X might or maximal ideal we could have developed this in a different direction so instead of evaluating at square root of 2 it could have evaluated at minus square root of 2 okay we'll call this e 2 so everything would run the same we'd wind up with our field isomorphism now I can consider okay II 1 inverse compose with E 2 so what's happening here well we're run this slice of morphism backwards okay here then we're going to run this one forwards so it's going to carry Q 1 square root of 2 2 Q 1 square root of 2 we follow what happens we're carrying a plus B square to 2 2 a minus B square root of 2 now because this is a composition of isomorphisms okay this is going to be a field autumn morphism to clarify the isomorphisms for instance e1 we're gonna send the ideal 4x square root of 2 so for the inverse where sin squared of 2 2 the ideal 4x and likewise for e2 now do we have any other field autumn morphisms here ok the answer is only the identity so let's see that if we apply 5 to a plus B square root of 2 we could separate everything then we notice it's a and B or rational numbers they get sent in themselves so everything is decided by where we send 5 square root of 2 now 2 equals 5 2 that's equal to 5 square root of 2 times square root of 2 we've left by 2 each term so I have to have 5 square root of 2 squared equals 2 the only possibilities are plus or minus square root of 2 so that gives everything now to bring in group theory we note the autumn morphisms on a field we're going to form a group under composition so in this case we only have a two element group so the autumn morphisms of Q to join square root of 2 as a group or isomorphic to Z mod 2 ok it's a simple check that by squared is equal to the identity now how about subfields of Q a joint square root of 2 by observation we have the rationals and we have Q a join square root of 2 itself so how about anything else well if I look for a K inside of Q join square root of 2 ok 1 is in there so it's going to contain the rationals but that's gonna mean this K is gonna be a vector space over the rationals and now we just check dimensions so the rationals themselves are gonna take up one dimension Q joint square root of 2 takes up 2 that means there's no room in between for a K so we're only going to have two subfields in this case for somewhat different example let's consider the cube root of 2 in the reals this is a root of the polynomial X cubed minus 2 it's irreducible over the rationals this has roots using the mobster M given as cube root of 2 Omega cube root of 2 and Omega squared cube root of 2 here Omega is the cube root of unity given as minus 1/2 plus square root of 3 over 2i this satisfies Omega cube equals 1 Omega squared equals the complex conjugate of Omega and Omega and Omega squared or roots of the polynomial x squared plus 2x plus 1 two things to notice we take Q a join cube root of 2 okay this is going to be a subfield of the real numbers so it's not going to contain all the roots of X cubed minus 2 if we consider automorphisms in this field okay we'll note using our procedures before ok the cube root of 2 is gonna have to go to one of the roots of a polynomial but since we're in a subfield of the reals it can only go to the cube root of 2 so that means the only autum morphism is the identity now doesn't seem like there's a lot going on here if we consider our set up using polynomials ok well if we work that out we're just gonna have like some morphisms between Q a join roots of the polynomials okay so these are all gonna be isomorphic as fields so it's a little bit better to really get something useful instead want to consider Q adjoin cube root of 2 and Omega now in here we want automorphisms okay well if we look at our polynomials so X cubed minus 2 and x squared plus X plus 1 possibilities are okay I could send the cube root of 2/2 cube root 2 Omega cube root of 2 and Omega squared cube root of 2 and for Omega we could send it to Omega or Omega squared okay and Omega squared let's just go to correspond a complex conjugation now it turns out to consider any combination of these we will get a filled autumn morphism for this field that's gonna mean group automorphisms 4q a join cube root of 2 and omega will have six elements so this group it's gonna be isomorphic either to an s3 or cyclic group of order 6 let's treat the automorphisms as group elements under composition for instance by 3 kalkaryn cube root of 2 2 Omega cube root of 2 Omega 2 Omega now I want to find the order of 5/3 so how many times we have to apply 5/3 before we get back to the identity now on Omega it's just going to keep giving back Omega so I have to track out the cube root of 2 we apply 5/3 once we get Omega cube root of 2 we plaid again Omega goes to itself cube root of 2 goes to Omega cube root of 2 we're gonna make a square cube root of 2 we apply 5/3 one more time Omega squared goes to 1 mega squared cube root of 2 goes to Omega cube root of 2 Omega cubed is 1 so get cube root of 2 that means Phi 3 has order 3 on the other hand let's take C giving us complex conjugation so this fixes the cube root of 2 sends Omega 2 Omega squared cube root of 2 just gets sent to itself as we apply C so we track out Omega so Omega goes to Omega squared goes to Omega to the fourth power which is Omega so C has order to now straight forward to C by 3 and C do not commute so the automorphism group has six elements and it's non abelian that means it's isomorphic to s 3 to finish let's consider subfields now by joining elements we can get the following six subfields note if we join any two roots of X cube minus 2 wind up getting qeh join cube root of 2 Omega captain mansions ok we'll have 6 2 and then 3 with 3 we'll see you later on the recipe we're pulling out your intermediate field what we do we're take our automorphism group we pick a subgroup and then we take all points fixed under the Autumn morphisms of that subgroup that's going to be fixed field for instance if we take the subgroup generated by K complex conjugation so that's two elements your fixed under complex conjugation that means you're a real field in this case that's going to be Q adjoin the cube root of 2 on the other hand if we take subgroup generated by 5/3 ok you'll note and Phi 3 what are we doing we keep changing up the cube roots of 2 but we're gonna leave Omega alone so in this case the fixed field is gonna be cute a joint Omega I'll leave it to you to figure out the other subgroups and their fixed fields