Automorphisms of Q(sqrt2)

1. Nov 25, 2006

MathematicalPhysicist

i need to find how many automorphisms are there for the field Q(sqrt2).
i am given the hint to use the fact that a is a root of a polynomial on Q(sqrt2).
what i did is:
if f:Q(sqrt2)->Q(sqrt2) is an automorphism then if P(x) is a polynomial on this field then:
if a is a root of P(x), then P(a)=0 but then also f(P(a))=f(0)=0
so $$P(x)=b_0+b_1x+...+b_nx^n$$
and $$f(P(x))=f(b_0)+f(b_1)f(x)+....+f(b_n)f(x)^n$$
they have the same degree, so when f(P(x))=0=P(x)
we have: $$f(b_0)+f(b_1)f(x)+...+f(b_n)f(x)^n=b_0+...+b_nx^n$$
from here iv'e concluded that it has only the identity function as an automorphism, am i right?

Last edited: Nov 25, 2006
2. Nov 25, 2006

matt grime

Why have you concluded that from there?

3. Nov 25, 2006

MathematicalPhysicist

cause f(b_i)f(x)^i=b_ix^i
am i wrong here?

4. Nov 25, 2006

MathematicalPhysicist

ok, iv'e looked in mathworld, i think i got my answer.

5. Nov 25, 2006

MathematicalPhysicist

i have another question.
i need to prove that f:R->R has only the indentity automorphism.
well bacause f preserves order, by the fixed point theorem we can find a point a in R suhc that f(a)=a, now if reuse the automorphism on the field R\{a} we can reuse this theorem and thus by reiterating we can get that f is the identity function, am i right in this way?

6. Nov 25, 2006

matt grime

Yes. f is an isomorphism that maps Q to Q. It does not (necessarily) send x to x for all x, merely for x in Q.

7. Nov 25, 2006

matt grime

R\{a} is not a field. R also does not contain a finite number (or even a countable number) of points, and I don't believe you're supposed to use transfinite induction...

8. Nov 25, 2006

MathematicalPhysicist

so how to prove that its only automorphism is the identity function?

9. Nov 30, 2006

mathwonk

automorphisms are symmetries.

consider the case of the roots of x^2+1 = 0.

do you see any symmetries among the roots of this one? or of the root field C?

the symmetry you find should leave the real numbers fixed.