Dimension of quotient construction

Question

If I have an irreducible polynomial, $f$ with $deg(f) = n$ and I look at the quotient:

$$R = \frac{\mathbb{Q}[x]}{(f)}$$

How can we show that the dimension of $R$ as a $\mathbb{Q}$ vector space is equal to the degree of $f$? I guess this should be pretty standard, but I am struggling to understand it.

I want to show that we have a basis

$$1 + (f), x + (f), x^2 + (f), ..., x^{n-1} + (f)$$

Now I can show that this is spanning by using the division algorithm with elements of $R$, however how do we go about showing this is linearly independent?

Thanks

Answer 1

Hint: what does it mean for $g \in R$ to be equal to 0? If $deg( g) < deg (f)$, can $g + (f) = 0$? Can you therefore show that $$\sum_{n=0}^{n-1}a_ix^i + (f) \ne 0\text{ } \forall a_i \in \mathbb Q$$

Answer 2

One way to do this is to prove that, if $f \in F[x]$ is irreducible, then $F[x]/\langle f(x) \rangle \cong F[\alpha]$.

To prove this, consider the evaluation homomorphism $ev_\alpha : F[x] \rightarrow F[\alpha]$ defined such that $g(x) \mapsto g(\alpha)$. Of course, if you've never seen this before, you'll want to prove that this is indeed a homomorphism. :)

Now apply the isomorphism theorem.