I am reading through Dummit and Foote's Abstract Algebra, and while I'm usually pretty good with my ring and field theory, this theorem is giving me some trouble.
Let $p(x)$ $\in F[x]$ be an irreducible polynomial of degree n over the field $F$, and let K be the field $F[x]/(p(x))$. Let $\theta = x \mod (p(x)) \in K$. Then the elements $1,\theta,\theta^2,...\theta^{n-1}$ are a basis for $K$ as a vector space over $F$.
The proof they give makes perfect sense to me, but the term $\theta= x \mod(p(x))$ $\in K$ doesn't make sense to me. Are they saying that $\theta$ is any element of the residue class of $x$? Is it the entire residue class of $x$ over the quotient ring $F[x]/(p(x))$, which would make $\theta = x + (p(x))$?
If it is the first, then for example, in the ring $\mathbb{Q}[x]/((x^2 +1))$, $x$ would be equal to $x\mod(x^2+1)$, but so would $x+x^2+1$. Either of these bases has to construct $\mathbb{Q}(i)$, so would there be any difference in the construction? Is that the entire point of classifying $\theta$ as a residue class? To say all elements of x mod(p(x)) construct the same field extension?
You can ignore the multiplicative structure on $K$ and simply look at it as a vector space over $F$. When looked at that way, yes, the elements of $K$ (including $x$) are cosets of $(p(x))$, and it's not hard to see that the vector space operations on $K$ are well defined. And when you reimpose the multiplicative structure and look at the elements as field elements, we're saying that $p(x)=0$ in the quotient ring, which is a field, so it makes sense that adding $p(x)$ to a representative of an equivalence class doesn't change anything.
There's an error in your question. I think you want to say (in your last paragraph) that either base has to consruct $\Bbb Q(i)$.