I've recently been learning about field extensions, but I've ran in to a few problems.
Say we're given a field $F$ and an irreducible polynomial $f \in F[x]$, then we can construct a field $E = \frac{F[x]}{<f>}$.
Now, if we let $\alpha = x + <f>$, then it is clear that $\alpha$ is a root of $f$ in $E$.
It is then stated that $(E:F) = (F(\alpha):F) = deg(f)$ ; where $(E:F)$ denotes the degree of $E$ over $F$.
Now, firstly, I don't understand how we can view $E$ as an extension of $F$? How is $F$ a subfield of $E$?
Secondly, when computing the degree of $E$ over $F$, how is $(E:F) = (F(\alpha):F)?$ Is this saying that $E$ is equal to the field generated by $\alpha$ over $F$ ; $F(\alpha)$? If so, how? These seem like different objects to me?
Thank you for an help!
We have the ring homomorphism $F\to F[x]$ that sends to the constant polynomials. Compose with the canonical projection $F[x]\to F[x]/\langle f\rangle$. This gives us a ring homomorphism $F\to E$, which must be an embedding because we are dealing with fields. Via this canonical embedding (which is of course $a\mapsto a+\langle f\rangle$) we can view $F$ as a subfield of $F$.
Btw, you also have to be careful when saying that $f$ has a root in $E$. To this end, we should first make $f$ a polynomial with coefficients in $E$. In order to avoid confusingly repeated variable names, assume that $f=a_0+a_1x+\ldots+a_nx^n\in F[x]$ and consider $a_0+a_1y+\ldots+a_ny^n\in E[y]$ (where strictly speaking, each $a_j$ should be written as $a_j+\langle f\rangle$), then we find that plugging in $x+\langle f\rangle$ for $y$, we obtain the zero element of $E$. It is in this sense that $f$ obtains a root in the extension $E$.
We have $F\subseteq E$ via our canonical embedding. We have $\alpha\in E$, of course. And the smallest subfield of $E$ that contains both $F$ and $\alpha$ (vulgo $F(\alpha)$) is $E$ itself. This follows because $F(\alpha)$ contains $g(\alpha)$ for any $g\in F[x]$, and $g+\langle f\rangle$ with $g\in F[x]$ is precisely the general element of $E$. Therefore, $E=F(\alpha)$