1. ## Question

F is a field, K is a sub field of F , There is a subspace over the field K (and therefore F), and there is A=(a1,a2,....,an) vectors that belongs to the subspace V
Prove or disprove the next to claims

1-)If A is linearly independent over the field K , then A is linearly independent over the field F too.
2-)If A is linearly independent over the field F, then A is linearly independent over the field K too.

So, the second claim is wrong, and I gave a counter-example, the finite fields Z2 and Z3.
And I think the first claim is right because every vector that is in K, it is in F too. That's why if it is linearly independent in K then it should be in F too, F wouldn't make a difference for those vectors, however, this is not a proof, so How should I prove the first claim? Maybe it's not even correct, but I couldn't find a counter-example for that.
Appreciate if someone can help me proving this.

By the way, how should I do the prove\disprove question? Because there are a lot of claims where you think it's right , and you ''prove'' it and you find out it's wrong, so every time I do these kind of questions I should try to disprove it first right?

2. ## Re: Question

To disprove the second claim is easy:

Let A = {1,√2} in R. Then this set is linearly independent over Q, and linearly dependent over R.

Your counter-example doesn't work because Z2 is not a sub-field of Z3.

To prove 1, I suggest proving the contrapositve: If A is linearly dependent over F, it is linearly dependent over K.

3. ## Re: Question

Why isn't Z2 a sub field of Z3?

4. ## Re: Question

What are the conditions in order to be a subfield?

5. ## Re: Question

Well, first and foremost a subfield is a subset of the larger field.

It may seem like Z2 = {0,1} is a subset of Z3 = {0,1,2} (sometimes written {-1,0,1}) but the "1" in each field is a different kind of animal.

In Z2, we have 1 + 1 = 0.

In Z3, we have 1 + 1 + 1 = 0, and 1 + 1 = 2 = -1 ≠ 0.

In general, if F is a subfield of K, then F and K will have the same CHARACTERISTIC, and share a common PRIME FIELD (the additive subgroup generated by the multiplicative identity of all subfields).

So if we have a field, like Z3, with characteristic 3, any subfield would ALSO have to have characteristic 3. Z3 is its own prime field, it doesn't have any subfields.

To get a finite field with another finite field as a subfield, you would have to go to the galois field of order 4. One representation of this is as:

F = {0,1,u,u+1} where:

0+x = x, for all x in F
1+1 = 0
1+u = u+1
1+(u+1) = u
u+u = 0
u+(u+1) = 1
(u+1)+(u+1) = 0

0*x = 0, for all x in F
1*x = x, for all x in F
u*u = u+1
u*(u+1) = 1
(u+1)*(u+1) = u

This field contains {0,1} = Z2 as a subfield.