Proof: Galois Group Maps Roots to Roots

so we are going to prove that field

extension automorphisms map roots to

Roots a special case of this is that the

galwa group will map roots of polinomial

to other Roots let's look at the setup

we start with some field F and an

extension field e we're going to

consider a function f which has

coefficients in the base field F now we

will consider an element Fe that is in

the automorphism group of e over F now

what does this mean it means we're

taking an automorphism of the field e

that fixes F so Fe is in the

automorphism group of

e and Fe restricted to the base field f

is the identity function so it leaves

elements of the base field F

unchanged now we have this function f

which is in this base field polinomial

ring what we're going to do is consider

some element R which is in the extended

field e and we're going to ask what is f

of f of

R well in order to do that let's first

make this polinomial a little bit more

concrete of course we can write any

polinomial in terms of powers of X so

let's say the first term is a subn x x^

of N and then after that we'll have a

n-1 x

nus1 and so on all the way down to a Z

being the constant term so we want to

ask what is Fe of f of R well F of R is

just this except that we plug in this

value R being an element of e into the

function so we're looking at F of a subn

r the N plus a sub nus 1 R to the

nus1 plus and so on A1 R and then

a0 now we said at the beginning that Fe

is an automorphism so in particular it's

a homomorphism and what that means is

first of all that f of x + y equal F of

X+ F of Y so we can either add these

elements inside of the fee and then

apply fee or we can apply fee first and

then add the results and those will

always be equal that's one of the

properties of a homomorphism so in this

case we are adding a bunch of terms we

can split this up and apply fee to each

term

individually so here I've split up this

expression by separating each term

individually and applying fee to those

all separately from here we can do

another step which is that in each of

these cases we are multiplying two

things inside of the fee we have a subn

* R the N for examp example and one of

the other properties of homomorphisms of

rings is that f of x * Y is f of x * V

of

Y so in this case if a subn is X and R

to the N is y then we can split this up

as F of a subn * F of R to the n and we

can do that for all of the other terms

as well so I've gone ahead and written

out the separated version of each of

these factors and now we have just a

couple more steps to take first of all

remember that one of these stipulations

for fee being an automorphism of this

field extension is that fee restricted

to the base field f is the identity

function now our polinomial f is a

polinomial with coefficients in the base

field F so if F restricted to F is the

identity then Fe applied to any of the

coefficients is going to be the identity

as well in particular a subn a subn

minus1 a sub 1 a Sub 0 those are all

coefficients in the base field so Fe of

a sub I is just equal to a sub I so we

can go ahead and remove the application

of Fe to the coefficients because of

applying Fe to an element of the base

field doesn't do anything and finally if

we look at F of R to the N well R to the

N is just R * R * r a bunch of times and

we know that f of x y equals f of x f of

Y

so in this case we can bring the power

of n to the outside because F of R * R *

R is the same as F of R * F of R * F of

R so F of R to the N is the same as F of

R and then we raise this result to the

power of n so this is our final

simplification we have that F applied to

F of R is equal to a subn * F of R the N

plus a subn -1 * F of R nus1 and so on

all the way down to A1 F of r +

a0 now if we take a look at this last

line and we compare it to the definition

of f ofx using these powers of X we can

see that it's actually the same we have

a subn * something to the power of n a

sub nus1 * something to the power of n-1

and so on the only difference is in this

case the input to the polinomial is X X

and in this case the input to the

polinomial is f of R in all of those

cases so this is actually the polinomial

F applied to the input F of R and what

we found now is that F of f of R equals

F of f of R and now we get to the point

of the video which is we want to show

that Fe Maps roots of this polinomial F

to other roots of f so to see whether

that's true let's suppose that R is a

root in other words F of R equals zero

this is the definition of R being a root

of f given that R is a root our goal is

to prove that F of R is also a root well

what that means is we want to prove that

F of f of R equals

z and we just proved that F of f of R is

equal to F of f of R so this statement

is true if and only if F of f of R

equals zero but we know what F of R is

because we assumed that R is a root of f

so this inside is just zero and now

we're asking whether F of 0 equals 0 but

because f is a homomorphism that is

always going to be true and therefore we

have found that F of f of R equals z

which which means that F of R is a root

of f so we have proved all in all that

if e is a field extension of f and f is

a polom with coefficients in that base

field f is an automorphism of the field

extension e over F and R is a root of

this polinomial then F of R is also a

root of the same polinomial in other

words if we're considering a polinomial

with coefficients in the base field then

an automorphism of a field extension

will map roots to roots

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