EnjoysMath

Theorem.  Let $f \in F[X]$ be irred of degree $n$ & let $K = F[X]/(f)$.  Let $\theta = x + (f) \in K = F[X]/(f)$.  Then the elements $\{1, \theta, \theta^2, \dots, \theta^{n-1}\}$ are a basis for $K$ as a vector space over $F$ so that $[K:F] = n$ and $K = \{a_0 + a_1 \theta + \dots + a_{n-1}\theta^{n-1} : a_i \in F\}$. Proof.  $F[X]$ is a Euclidean domain so divide $g\in F[X]$ by $f$ using the Euclidean algorithm.  The remainder is a coset representative for $g + (f)$. Examples: $\Bbb{R}[X]/(X^2 + 1) \simeq \Bbb{C}$ $\Bbb{Q}[X]/(X^2 + 1) \simeq \Bbb{Q}(i)$ so $[\Bbb{Q}(i) : \Bbb{Q}] = 2$

Theorem.   Let $F$ be a field & $f \in F[X]$ an irreducible polynomial.  Then there exists a field ext $K$ containing an isomorphic copy of $F$ in which $f$ has a root.  Proof.  Consider the ideal $(f) \subset F[X]$.  Since $F[X]$ is a PID $(f)$ is maximal since $f$ is irred.  Let $K = F[X]/(f)$ which is a field since $(f)$ is maximal.  Let $\pi : F[X] \to K$ be the natural projection $\pi(g) = g + (f) \equiv \overline{g}$.  Then $f(\overline{x}) = \overline{f(x)} = 0_K \pmod{(f)}$.  Further, $\pi |_F$ embeds $F$ into $K$ as an isomorphic copy.  Identifying $F$ with the isomorphic copy, there essentially exists $K/F$ in which $f(x) = 0$ for some $x \in K$.