Irreducibility and finite fields

Trying to do i) and iii) on this past exam paper

For part i) I'm pretty stumped

I've said that the possible roots of the polynomial are +- all the factors of T

In particular rt(T) needs to be a factor of T but this can't be possible?

Doesn't sound too good but its the best I've got.

Part iii) I know this means that every element in K is seperable over k, i.e that the minimal polynomials of elements in K are seperable, where they have no repeated roots.

Not sure how to apply this though..

Maybe K = k(rt(T))

so the minimal polynomial of K is X^2 - T which has repeated root rt(T) so it is inseperable?

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For part i:

[STRIKE]Because this is a finite field, you must show that the polynomial g(X) has no roots in k. Because k is a finite field of 2 elements, you can just try plugging in 0 and 1 because k = {0,1}.

So g(X) is irreducible if and only if 0^2-T is not equal to zero and 1^2-T is not equal to zero. Since T is an extended element onto k and is therefore not equal to either 0 or 1, which is what we need to make the above two equations equal to zero, I think that we can safely draw our conclusion. What do you think?[/STRIKE]

edit: i'm dumb. i'll rethink this.

For part i:

Because this is a finite field, you must show that the polynomial g(X) has no roots in k. Because k is a finite field of 2 elements, you can just try plugging in 0 and 1 because k = {0,1}.

So g(X) is irreducible if and only if 0^2-T is not equal to zero and 1^2-T is not equal to zero. Since T is an extended element onto k, I think we can safely draw our conclusion. What do you think?

isn't k = {0, 1, T, 1+T} since k = {a + bT | a, b in {0,1}}

Otherwise I see what you mean. Thanks for the reply

yeah you're completely right and i noticed my mistake right after i posted that. i overlooked the whole extended element thing. my bad.

bump for confirmation