# 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.