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How can I prove that a field F is a vector space over itself? Intuitively, it seems obvious because the definition of a field is nearly the same as that of a vector space, just with scalers instead of vectors.

Here's what I'm thinking:

Let V = { (a) | a in F } describe the vector space for F. Then I just show that vector addition is commutative, associative, has an identity and an inverse, and that scalar multiplication is distributary, associative, and has an identity.

Example 1: Commutativity of addition, Here x,y are in V

(x)+(y) = (y)+(x)

(x+y)=(y+x): vector addition

x + (X+y) = (X + x)+y: associative property

Example 2: Additive inverse x,y,0 in V (x)+(y)=(0)

(x+y)=(0) vector addition

Let y=-x in V

(X+-x)=(0) substitute

(0)=(0) simplify

I don't know if I'm going in the right direction with this, although it seems like it should be a pretty simple proof. I think mostly I'm having trouble with the notation.

Any help would be greatly appreciated! Thanks in advance!

ijuneja
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Konstantine
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1 Answers1

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If in the axioms of vector spaces you assume that the vector space is the same as the field, and you identify vector addition and scalar multiplication respectively with addition and multiplication in the field, you will see that all axioms are contained in the set of axioms of a field. There is nothing more to check than this.

Marc van Leeuwen
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    So would the following be acceptable as a proof?

    Assume that the vector space V={aI a in F} so that the vector space is the same as the field F. We see, then, that vector addition is defined the same as addition in F and that scalar multiplication in V is defined the same as multiplication in F.

    If F is a field, addition and multiplication are both commutative in F, and therefore commutative over V.

    If F is a field, addition and multiplication are associative in F, and therefore associative over V. (continued)

     – Konstantine Sep 14 '13 at 17:57
  • Continue with identity elements, inverse elements, and distributivity.

    We see that V is a vector space over F. Because V is the same as F, we conclude F is a vector space over itself, and thus our proof is complete.

     – Konstantine Sep 14 '13 at 17:57
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    Basically that's it, yes. The precise list of vector space axioms to verify may depend on your textbook, but the distributive laws and associativity of multiplication are the most important ones; these become instance of field axioms. I don't think any vector space axiom explicitly ask for inverse elements, though of course nonzero scalars must be invertible because they live in a field. – Marc van Leeuwen Sep 14 '13 at 18:24