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Field/Subspace Proof

  1. Sep 8, 2010 #1
    1. The problem statement, all variables and given/known data
    Prove whether or not a field F has just two subspaces: {0} and F itself.

    2. The attempt at a solution
    I'm not exactly sure where to start. I believe it's true but I don't know how to prove that F has only two subspaces. I tried doing proof by contradiction (ie. supposing there was a third subspace for F), but I don't know where I can go with that.
  2. jcsd
  3. Sep 8, 2010 #2
    Not sure on this one, but... Can you show that the zero subspace, and the field itself has: (a) the "zero object" of F exists in both subspaces. (b) Each subspace is closed under "object" addition. (c) Each subspace is closed under scalar multiplication. ? Just a thought.
  4. Sep 8, 2010 #3


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    If F is a vector space over itself, what dimension is it?

    Then remember any proper subspace has to have a smaller dimension
  5. Sep 8, 2010 #4
    Well, yes you can, but that would just prove that the empty set and the field itself are subspaces. It wouldn't prove that those are the only two subspaces of F.

    We haven't gotten over dimensions in my class so I don't know that to be honest. Do you know a link that can direct me to know more about this, or do you have the time to explain it? It would seem pretty easy once I know this piece of information.
  6. Sep 8, 2010 #5
  7. Sep 8, 2010 #6
    Suppose that U is a non-zero subspace of F. Then U is closed under scalar multiplication. Let x be any non-zero element of U and let y be any element of F. Can you find a scalar multiple of x that equals y? What does that tell you?

  8. Sep 8, 2010 #7


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    Do you mean subfield rather than subspace? I don't know what it means to say that a field has a subspace.
  9. Sep 8, 2010 #8
    Is it supposed to show that any non-zero subspace of F must be F itself? I can understand the rest of the proof from then on, but how do you show that U must be = F? What scalar multiple of x equals y?
    Last edited: Sep 8, 2010
  10. Sep 8, 2010 #9
    A subspace of a vector space is, by definition, closed under scalar multiplication. I'm trying to get you to show that every element of F is a scalar multiple of a non-zero element of U. Thus, F [itex]\subseteq[/itex] U and so F = U. I'm sure that with some thought you can figure out what scalar (that is, what element of F) times x equals y.

  11. Sep 8, 2010 #10
    I'm taking this to mean that the field is being regarded as a vector space over itself and then subspace means a subset of F that also is a vector space over F.
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