Prove Field F Has Only 2 Subspaces: {0} & F Itself

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Homework Help Overview

The discussion revolves around proving whether a field F has only two subspaces: the zero subspace {0} and the field itself F. Participants explore the properties of subspaces in the context of vector spaces and question the definitions involved.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Some participants attempt to start with proof by contradiction, questioning the existence of a third subspace. Others suggest examining the properties of the zero subspace and the field itself, including closure under addition and scalar multiplication.
  • Questions arise about the dimension of F as a vector space over itself and how this relates to the existence of proper subspaces.
  • There is a discussion about whether the term "subspace" is appropriate in this context, with some suggesting "subfield" instead.
  • Participants inquire about the implications of scalar multiplication in relation to non-zero subspaces and how this leads to the conclusion that any non-zero subspace must be F itself.

Discussion Status

The discussion is active, with various lines of reasoning being explored. Some participants are providing insights into the properties of subspaces, while others are seeking clarification on definitions and dimensions. There is no explicit consensus yet, but the conversation is progressing towards understanding the relationship between subspaces and fields.

Contextual Notes

Participants note that they have not yet covered dimensions in their coursework, which may affect their ability to engage fully with the problem. The definitions of subspaces and their properties are under scrutiny, particularly in relation to fields viewed as vector spaces over themselves.

Anonymous217
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Homework Statement


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

Then remember any proper subspace has to have a smaller dimension
 
Samuelb88 said:
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.
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.

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

Then remember any proper subspace has to have a smaller dimension
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.
 
Bumppp.
 
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?

Petek
 
Do you mean subfield rather than subspace? I don't know what it means to say that a field has a subspace.
 
Petek said:
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?

Petek

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:
Anonymous217 said:
Is it supposed to show that any non-zero subspace of F must be F itself?
Yes.
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?

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.

Petek
 
  • #10
Anonymous217 said:
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?

HallsofIvy said:
Do you mean subfield rather than subspace? I don't know what it means to say that a field has a subspace.

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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