Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Automorphisms of Q(sqrt2)

  1. Nov 25, 2006 #1
    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 [tex]P(x)=b_0+b_1x+...+b_nx^n[/tex]
    and [tex]f(P(x))=f(b_0)+f(b_1)f(x)+....+f(b_n)f(x)^n[/tex]
    they have the same degree, so when f(P(x))=0=P(x)
    we have: [tex]f(b_0)+f(b_1)f(x)+...+f(b_n)f(x)^n=b_0+...+b_nx^n[/tex]
    from here iv'e concluded that it has only the identity function as an automorphism, am i right?
    Last edited: Nov 25, 2006
  2. jcsd
  3. Nov 25, 2006 #2

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    Why have you concluded that from there?
  4. Nov 25, 2006 #3
    cause f(b_i)f(x)^i=b_ix^i
    am i wrong here?
  5. Nov 25, 2006 #4
    ok, iv'e looked in mathworld, i think i got my answer.
  6. Nov 25, 2006 #5
    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?
  7. Nov 25, 2006 #6

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    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.
  8. Nov 25, 2006 #7

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    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...
  9. Nov 25, 2006 #8
    so how to prove that its only automorphism is the identity function?
  10. Nov 30, 2006 #9


    User Avatar
    Science Advisor
    Homework Helper

    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Automorphisms of Q(sqrt2)
  1. Automorphism Group (Replies: 20)

  2. Automorphisms of Z_3 (Replies: 3)

  3. Automorphism groups (Replies: 7)

  4. Inner Automorphism (Replies: 1)