1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Algabraic field extensions (true false questions)

  1. Feb 21, 2010 #1
    1. The problem statement, all variables and given/known data
    Can you please check my answers and help me develop explanation for the ones I can't explain.

    a) every field has nontrivial extensions.
    b) every field has nontrivial algebraic extensions.
    c) every simple extension is algebraic.
    d) every extension is simple.
    e) all simple algebraic extensions of a given subfield of C are isomorphic.
    f) all simple transcendental extensions of a given subfield of C are isomorphic.
    g) every minimal polynomial is monic.
    h) monic polynomails are always irreducible.
    i) every polynomial is a constant multiple of an irreducible polynomial.

    2. The attempt at a solution
    a) true I am thinking of C and adding something that's not in it. like a cube rotation or something call it J. a in C. a+J is a*90degree rotation of the cube. a*J is number of flips. and J+J is a 180 rotation. J*J is again a flip.
    b) false. a flip wouldn't be a solution to anything in C. so it has to be a transcendental extension.
    c) false Q(pi) is simple and transcendetal.
    d) false. only true for when the field has characteristic 0.
    e) false? need explanation
    f) true? need explanation
    g) true. definition of minimal polynomial was that it's monic.
    h) false. x^2+4x+2
    i) false. need explanation
  2. jcsd
  3. Feb 23, 2010 #2
    a) Any field can be the base field for a polynomial ring, so true.

    b) An algebraically closed field has no nontrivial algebraic extensions.

    c) Like you said.

    d) Like you said, except only finite extensions in characteristic zero are necessarily simple.

    e) Q is a subfield of C, and [tex]Q(\sqrt{2})[/tex] is not isomorphic to [tex]Q(\sqrt{3})[/tex].

    f) Transcendentals don't solve any polynomial, so one is as good as another.

    g) Right.

    h) Right.

    i) It doesn't say nonzero constant, and zero is an irreducible polynomial...but that's pretty cheesy.
  4. Feb 24, 2010 #3


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Au contraire, I bet I can write zero as a product of two non-units....
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook