I had a question regarding fields.
So I know that when a vector space is defined over a field, that field tells you what kind of coefficients you're allowed to take i.e. scalar multiplication needs to use scalars from field F.
But when I considered factorisation, and specifically the fact that:
$t^2 + 1 = (t + i)(t - i)$ over field C
$t^2 + 1 = t^2 + 1$ over field R
I understand that this is true because the coefficients in the first line is in C (i.e. i), while in the second line, the coefficients are in R.
But my question is:
if you had a vector space of polynomials with degree two over field R, then why can't you just construct the basis
{$t^2, t^2i, t, ti, 1, i$},
and this way you can take linear combinations of the basis with real coefficients to obtain any complex polynomial degree 2 in the vector space
i.e. $(a + bi)t^2 + (c + di)t + (e + fi) = a(t^2) + b(t^2i) + c(t) + d(ti) + e(1) + f(i)$
This way even if you factorise $t^2 + 1$ over field R, you can still get $t^2 + 1 = (t + i)(t - i)$
because $t + i = 1(t) + 1(i)$ and $t - i = 1(t) - 1(i)$. (i.e. you're taking linear combinations of basis elements with real coefficients in R)
However, I feel like this isn't a valid way to think about fields, but I'm not exactly sure where my reasoning breaks down.
These polynomials are not over $\mathbb R$, so it is not a basis of your space of polynomials