Field Automorphisms and Galois Groups Explained

[00:00:01]welcome to lecture 6.2 field automorphisms recall that an automorphism of a group G was an isomorphism from that group to itself we will Define automorphisms of fields similarly here's the formal definition let F be a field a field automorphism of f or just automorphism of f if it's understood that it's a field is a bje fe from F to F such that for all elements A and B in F or just numbers in f as we often say F of a plus b equals F of a plus F of B and F of a * b equals F of a * F of B in other words Fe must preserve the structure of the field compare this to an automorphism of group which is a structure preserving map from the group to itself now if a group was written additively then Fe had to be a homomorphism so it satisfied this condition F of a plus Bal F of a plus F of B however if F was written multiplicatively then it satisfied this condition F of a b equals F of a * F of B now a field is both an aelan group under addition and an aelon group under multiplication if we take out zero so it must satisfy both or fee must satisfy both of these properties it preserves the structure of addition and preserves the structure of multiplication for example let F be the field Q ajoin root2 on your homework you will verify that the following function Fe from this field to itself defined as follows it sends a + b < tk2 to a minus bk2 that that is an automorphism in other words you have to show explicitly that Fe of the sum of two elements let's say A + B < tk2 + C + dunk2 is equal to so I'm putting dot dot dot here F of a + b < tk2 plus F of C plus dunk2 it's not hard you just need to basically plug in the definition and then rearrange terms and show that this holds well that shows this first part next you need to show that it is a multiplicative homomorphism in other words that F of a + bk2 * C + dunk2 so multiply these things out apply this function in other words swap the sign s i GN of everything in PL in front of aun2 and show that that is equal to F of a plus b < tk2 * F of C plus dunk2 and you also need to show it's a bje but I claim that that's clear that it it's very clear that this function is not only onto everything can on the right hand side gets hit but also that it's one to one here's a question I'd like to pose what other field automorphisms of Q jooin root2 are there well there's clearly the identity automorphism the map that just fixes everything but can you think of something else open-ended question we will answer that later field automorphisms are Central to gawa theory in fact the gawa group of an extension or of a polom is a group of automorphisms of a certain field we'll see more details shortly but first here is a defining property of field automorphisms if Fe is an automorphism of an extension field F of Q then this says that Fe fixes every rational number so F of Q equals Q for all rational numbers Q let's prove this so the automorphism Fe is a bje so it sends one to some rational number Q let's try to figure out what Q can be clearly Q cannot be equal to zero think about why this is well if it were then Fe of two would would be F of 1 + 1 which is f of 1 + f of 1 which is 0o and all of a sudden we don't have a objection because one and two both get sent to zero so Q is not equal to zero observe that Q which is find to be F of one is f of 1 * 1 clearly and then by the property of a multiplicative homomorphism is equal to F of 1 * V of 1 which is equal to q^2 so what rational numbers Q are equal to their Square well there's one minus one and zero and we've already eliminated zero similarly Q is equal to F of 1 by definition but that's also equal to F of 1 * 1 * 1 which is f of 1 * F of 1 * F of 1 which is Q cubed and so on so now what we have a rational number q that is equal to its Square equal to q^2 and equal to Q cubed so the only rational number that does that is zero or one and we've eliminated zero so Q has to be one now if you want to you can say that Q to the N equals Q for every Q greater than or equal to 1 so Q has to be equal to one and that's the end of the proof I forgot my box but I'll just put it right there here's a simple corollary of this proposition root2 is irrational let me say a few words about why well on the previous slide we constructed a field automorphism from Q of < tk2 to Q of < tk2 I think we called it Fe yeah V and it was defined by sending a + b < tk2 to a minus B < tk2 and then I asked you to verify in the homework that this in fact was an automorphism but notice that F of < tk2 equals < tk2 so if we can if we proved that any automorphism of an extension field of Q fixes everything in Q and here is an automorphism of an extension field that does not fix this particular number run2 then this number cannot be in Q This is an alternative proof that Ro root2 is irrational it uses a lot more High powerered Theory than probably what you were used to when you saw your first intive proofs class remember how that proof went so assume < tk2 was rational so you assume that < tk2 was equal to A over B and then you squared it and you got that 2 = a^2 over b^ 2 and then you arrived at a contradiction from there so really simple proof I like it I think it's cute it's a proof using gawa Theory speaking of gawa theory well that's the term that I thrown around a lot but I have yet to Define exactly what it is so I think it's time to do that now so the set of all automorphisms of a field form a group under composition here's the formal definition let's let F be an extension field of Q the gawa group of f is the group of automorphisms of F and we denote this as gal of f or just the gawa group of f so for groups we said we called it the automorphism group of G and here I guess we could Define the automorphism group of f but we we are going to call this the gawa group of F and Define that to be the automorphism group here are some examples without proof first of all the gawa group of Q adjoined root2 is well technically isomorphic to C2 so I asked you to think about what other automorphisms of Q jooin ro2 were there besides the identity and the one that sends a plus bi to a minus bi which if you want to abbreviate you can write it like this because we know it fixes the rational number so it fixes the a part so you going to Define f as the automorphism or the mapping that sends < tk2 to negative < tk2 and in fact these are the only two automorphisms of this field a little bit of a trick question so the gawa group has two automorphisms so it has to be isomorphic to the only group of order two namely C2 next an automorphism of Q adjoined < tk2 and I is completely determined by where it sends < tk2 and I respectively there are four possibilities there's of course the identity map e let's call it but also there's the map that swaps the sign of Ro < tk2 so h of < tk2 = < tk2 and H of IAL I there's the map that fixes root2 and swaps the sign of I so V of < tk22 = < tk2 and V of I = I and there's the map that swaps both of them the V of let's call it r r of < tk2 = < tk2 and R of IAL negative I now I've defined these mappings using the letters h v and R to suggest the hint at that the automorphism group or the gawa group of this extension is isomorphic to the kli 4 group V4 CU remember when we first saw V4 we saw it as containing four elements a horizontal flip a vertical flip and a rotation which is what you get when you do both of these flips independently in other words the gawa group of this field extension is generated by H and V these two automorphisms or really any two of these automorphisms will do and it is isomorphic to V4 let's do another example and as I've done previously I will introduce this with a question what is the smallest extension field F of q that is the rationals that contains all roots of this polinomial G of x = x cubus 3 so this is the first example that we've seen where we're looking at roots of a polinomial of degree more than two let's begin by letting Zeta denote the complex number e 2 pi I over 3 this is in polar coordinates but in cartisian coordinates this is - one2 it's the real part plus imaginary part < tk3 over 2 I so here's a picture of the complex plane over on the right and recall that every every number on the unit circle has the form form e to the I Theta where Theta is the angle um from that number to to one which is right here so e to 2 pi I over3 is at a is on the unit circle at angle 2 pi over 3 radian so that's 1/3 of the way around and Theta squared remember how you multiply complex numbers the angles add and the lengths multiply so you multiply this by itself the length is still one and the angle just doubles so it gets down here so that's e 4 Pi I over 3 and similarly Zeta cubed you just rotate it three times and you go right there or twice I should say to get Zeta squared and then one more time to get back to one zeta is called a third root of unity Unity just meaning one and that just means that it is it is a cube root of of one formally that means it is a root of the polinomial X Cub - 1 which by the way factors as x -1 * x^2 + x + 1 this degree 2 polinomial has two Roots which are Zeta and Zeta squar and then the third root of this polinomial of course is one so all of these are third roots of unity although we call these two primitive third roots of unity because they are not roots of any lower degree polinomial so this is a first let's say a primitive first root of unity these are primitive third roots of unity next I want you to notice that the roots of g ofx g of X is this polinomial up here x Cub minus 2 are the following well first there's the real cube root of two then there's the cube root of two time Zeta and then there's the cube root of two time Zeta squar so to see why this is let me show you both algebraically and geometrically graphically well Z2 cubed is it's hard to draw Zeta I almost forgot Zeta cubed times cube root of 2 cubed and obviously Cub as we said is one and this these cancel so we we get two and similarly for this one zeta 3 cubed is Zeta um that's not quite a Zeta but that's close enough Zeta squar to the 3 which is Zeta to the 6 time the cube root of 2 cubed which is also two because Zeta to the 6 if you go around six time times 1 2 3 4 5 6 so both of these clearly solve x Cub - 2 if you want to see graphically why this is now the cube root of two is approximately 1.26 so if I were to draw it on the unit circle it would be around well make that a little better it would be around there so this this is the cube root of two and then I'm going to exaggerate a little bit just so I can fit it in there's going to be one up there this is that's Zeta time the cube root of two and then down here again I'm going to exaggerate just so I don't write over this that this is going to be Zeta squar time the cube root of two and notice that if I multiply this by itself or if I if I Cube this thing then the angles add so if I If I multiply this by itself it rotates over here and it gets a little longer so I would say it would get to be about there because the the lengths multiply so whatever 1.26 * 1.26 is that's the distance of this and then when I multiply it again I get a rotate let's rotate it one more time the lengths multiply so now I get a length of two this is two and the angles add so I get back to the real axis so graphically multiplying this by itself three times or cubing it gets here and then gets there and similarly If I multiply this guy by itself or if I Cube this thing then well Zeta squ squared angles add it gets up here a little bit longer and then multiplying it by I guess this has a name I can call it Z3 multiplying by Z3 again adds this angle and you get back you get back to two so there's are two different ways to see why these things are the roots clearly the the field that we seek the smallest extension field F of q that contains all the roots can be constructed by just starting with q and throwing in the roots and seeing what that generates but I claim that there's a better way to write this that I don't need to throw in all three roots all I need to do is throw in Zeta and throw in the cube root of two let's try to understand why notice first of all that this field contains Z1 Z2 and Z3 because it contains Z1 there's Z1 it contains Z2 because I can multiply Zeta times cubot of Two And it contains Z3 because I can multiply Zeta by itself get Zeta squar and then multiply that by the the cube root of two to get Z 3 so we just showed that everything in this field is contained in that field so that shows that we have a subset inclusion like that we have containment like this so this thing is at least as big as here let's now argue why everything in here is also in here so let's all we have to do is show that if we can produce Zeta and the cube root of two from this field then we're done well clearly the cube root of two is in this field because it's just Z1 so I claim that I can construct both of these from actually from just Z1 and Z2 using arithmetic and so how would I do that well what what if I take Z2 and I'm allowed to divide and divide by Z1 then I get Zeta and so indeed every generator of this field is in that field and that proves this containment so they have to be equal and that establishes this claim that our field f is equal to Q adjoined Zeta and cuot of two a little bit of algebra can show that this field extension of q that is Q adjoin Zeta and CU root of two can actually be written as this set of complex numbers namely a * 1 + B cuun of 2 plus C cubt of 4 plus d Zeta plus e Zeta cube2 plus F Zeta cubot of 4 where a b c d e and f are all rational numbers in other words the following set is a basis for this field extension over Q let me put colon there so one the cube root of two the cube root of four which is just the cube root of two squared Zeta Zeta time the cube root of two and Zeta time cube root of 4 so that's our basis and it's not hard to see why everything can be written like this let's do it like we did in the previous lecture so let's start with Q adjoined Zeta then adjoined the cube root of two so things in here we can write as well I should say this is a degree 3 extension because this is a degree 3 polinomial so can write things like or this like Alpha * 1 + beta * the cube root of two plus gamma times the cube root of four that's all we need because if we if we uh take this thing multip and Cube it then we get a rational number two now alpha beta and gamma gamma all live in in this field Q adjoins Zeta and so they can actually be written as so Alpha can be written as a plus btimes Zeta and then beta can be written as C plus d Zeta C + D Zeta and then times the cube root of two and then finally gamma can be written as what do we have e+ F time Zeta time the cube root of four and then if we multiply this all out and we will get something that looks like this now I may have permuted my C's D's e and Fs it doesn't matter um I feel like I was going to tell you something else about this oh yeah so I I have a Zeta here but I don't have a Zeta squar and there's a Zeta squar over here well why is that so several reasons first of all Zeta squar can be generated by Zeta so if you take Zeta you take negative Zeta maybe it's easier to look at when you take negative of this so negative Zeta is is over here this is negative Zeta which is positive 12 minus the square < TK of 3 out of three I over 2 and then so so if I start with Zeta and I take negative Zeta and then if I subtract one from it then this this just well then I get Zeta squ so in other words Zeta generates Zeta squ so I don't need both of these and also because Zeta and Zeta squ are roots of this degree 2 polinomial first of all you can see right away that Zeta squar and I've set set this equal to zero you can see right away that Zeta squar is equal to negative Zeta minus one as I claimed and so this is actually a degree 2 extension so we only need two things namely one and Zeta so again to derive this I say a little bit of algebra here it is write an element an arbitrary element of this field a join cube of two like this and then each alpha beta and gamma actually can be written um as as such a plus b Zeta C plus d Zeta and E plus F Theta let me conclude this slide with one more observation about this extension field so since Zeta lies in Q adjoined Zeta and Q of two then so does Zeta plus 12 * 2 and notice what that is if I take Zeta and I add 1/2 to it then this - 1/2 goes away and If I multiply by two I get < tk3 I which is just the square < TK of -3 therefore I can write the extension field Q adjoined Zeta and cuot of two as Q adjoined otk -3 and cubot of two which is just q adjoined < tk3 i and cubot of two and we can do this because these two numbers Zeta and cube root of two clearly generate these two numbers square of -3 that's what we just verified up here and obviously cube of two it's the same thing and vice versa from these two numbers if I start with root3 I can then divide by two and I can subtract 1/2 to get Zeta and then I can generate these two numbers so these are just two different ways to express the same field some people like like writing it like this because well one one advantage to that is if you look at the square root of -3 you can see right away that that is a root of x^2 + three so it's arguably clear that q a join the square < TK of -3 is indeed a extension of degree 2 where it wasn't quite so clear that Q would joined Zeta was an extension of degree 2 at least not until we saw this polinomial here x^2 + x + 1 now I want to turn our attention to the subfields of Q adjoined Zeta and cube root of two so this set is the full field but what if we only take some of these elements like for example you can take q and join the cube cube root of two only that's going to be a subfield and it's clearly not equal to the entire field because it's contained in the real numbers we could also take q and adjoin just Zeta or we could take q and adjoin the cube root of four and so forth or we could adjoin the product of these two things and it's not clear that we're going to get different fields in all of these cases and actually in some cases we don't for example let's compare Q adjoined Zeta Square to just Q ad join Zeta and actually it's it's not hard to see why these are the same field one way to see it is if you take Zeta squ and you square it you get back Zeta so from Zeta squar I can create Zeta or generate Zeta and from Zeta I can create Zeta squ and therefore these field extensions are the same and both of them consists of all numbers of the form a plus b Zeta where A and B are rational and there's other ways to see this too as I think I mentioned before Zeta squ is actually is just equal to Zeta minus 1 here's one that's less obvious I claim that the field you get when you would join the cube root of two and the cube root of four are actually the same thing and here's how to see that so let's take the cube root of four and let's see if we can use it to generate the cube root of two and we can do that so we if we Square it we get two cubot of two and we divide by two so we have used this thing to generate cube root of two and similarly if we start with the cube root of two and we Square it we get the cube root of four so either one of these things can generate the other via operations of arithmetic therefore these two fields are the same set of numbers and both of those are all real numbers of the form a plus B cuun of 2 plus C Cub of 4 where a B and C are rationals it turns out that there are two more subfields and as we did before we can arrange them in a lattice recall we did this for I think it was the field 2 adjoined < tk2 and Ro < tk3 you remember that so we had q adjoined < tk2 q adjoined < TK 6 and Q adjoined < tk3 those are all subfields and they all contained the rational so we got this lattice which look just like the subgroup lattice of the Kleine 4 group V4 well now let's do that with this field here is the subfield lattice does this look familiar and let me mention that I've labeled these edges with the degree of the extension so I start with the rationals to get up to here I have to adjoin a root of a degree 3 polinomial and same thing with these things so it's not completely clear that these are different fields but I encourage you to play around with this on paper and see if you can verify that you cannot create Zeta 2 * cubot of two from just Zeta cubot of two and vice versa and it's also not clear that there's not other fields lurking in here besides these but let me ask you does this look familiar here is the subgroup lattice of D3 notice it is identical in structure to the subfield lattice well assuming I turn this one upside down and even the edge labels are the same for example this three is the degree of this field extension it's because I can create this field by adjoining a root of a degree 3 polinomial namely this root to q and this three comes from the fact that this is a index 3 subgroup in D3 so it has three coets this is not a coincidence I've also highlighted in blue over here the normal subgroups of D3 and the corresponding subfields over here are actually called normal field extensions of Q now we'll study this in more detail later but let me just give you the informal definition with an example so this is a normal field extension because I adjoined a root of a degree 2 polom and I get the other root for free it's it's automatically in there recall that the other root is Zeta is negative Zeta minus one whereas over here I adjoined a root of a degree 3 polinomial cube root of two but I don't get the other two roots in that field for free the other two roots are are not real like Zeta cubot of two is not a real number and this subfield is contained in R so these are not normal field extensions because when I join a root of a polinomial I don't get all of the roots whereas this one is normal and it is not a coincidence that the normal field extensions correspond exactly to the normal subgroups hopefully I've piqued your interest about this amazing connection between subfields and subgroups which is at the heart of gawa theory so let's step back and summarize what we've learned about Fields so far roughly speaking a field is a group under both addition and multiplication or that is the latter if we exclude zero and the distributive law connects these two operations we are mostly interested in the field of rationals den noted q and certain extension Fields f of Q now we can do this whole thing over finite Fields like Z2 and zp and we can get a lot of the same results and sometimes they're a little bit different but this this is the best way to begin and that's all we are going to look look at and it's probably the most relevant because it's directly applicable to things like the quadratic cubic and quartic equations and the non-existence of a quintic equation some of the extension fields that we've encountered include Q joined root2 Q ad joined i q ad joined root2 and i q ad joined root2 and root3 and Q adjoined Zeta and the cube root of two next an automorphism of a field is a structure preserving map and it automatically has to fix the rationals this is a lot like an automorphism of a group which had to fix the identity so here we don't just fix the identity I guess there's two identities the additive and multiplicative but we fix everything that those generate which are the rationals the set of all automorphisms of an extension field F of Q forms a group and it's called the gawa group of F and it's denoted as G of F this is how we write it there is an intriguing but so far mysterious connection between the subfields of F and the subgroups of the gawa group of F and actually until now I haven't actually said that that subgroup lattice that I've showed you like D3 in the example in that previous slide was the gawa group so I guess I'm hinting at that now so there's this connection between the subfields and the subgroups of the gawa groups and this is at the heart of gawa theory this is where it just it all comes together let me finish with a question there's something for you to ponder how the heck does all of this relate to solving pols with radicals and in particular to proving that there is no closed formula for the roots of a general quintic polom so that's something to think about that's what we will address and answer in the next two lectures and then after that we will conclude with two lectures really for fun of where we apply these tools of gawa theory to some very old problems in Geometry that stumped the ancient Greeks and these problems can be solved quite easily with our tools from gawa Theory so stay with us