The question: Let $K = \mathbb{Q}$ and $f = X^4 + 1$. Show that $f$ is irreducible in $\mathbb{Q} [X]$. Explain why $\mathbb{Q} [X]/⟨X^4 + 1⟩ ≃ \mathbb{Q} [(1 + i)/\sqrt{2}]$.
I have shown that $f$ is irreducible in $\mathbb{Q} [X]$ and I have found that $(1 + i)/\sqrt2$ is a complex root of $f$ but I am not sure how to apply the theorem to prove the isomorphism above. Why does the theorem imply the extension is isomorphic to the smallest set which contains $\mathbb{Q} $ and $(1 + i)/\sqrt2$?
It is a theorem that, in general, if $f \in F[x]$ is irreducible, then $F[x]/\langle f(x) \rangle \cong F[\alpha]$ where $\alpha$ is a root of $f$.
We can prove this with the isomorphism theorems. Consider the evaluation homomorphism $\text{ev}_{\alpha}:F[x] \rightarrow F[\alpha]$ defined as follows: $g(x) \mapsto g(\alpha)$. You'll want to prove that this is indeed a homomorphism.
Now, because $F[x]$ is a principal ideal domain, and because $f$ is the minimal polynomial over $F$ with $\alpha$ as a root, $\ker(\phi) = \langle f(x) \rangle$. Finally, we know that $\text{ev}_\alpha$ maps surjectively to $F[\alpha]$.
Hence, by the isomorphism theorems, we have $F[x]/\ker(\phi) \cong \text{Im}(\phi) \Longrightarrow F[x]/\langle f(x) \rangle \cong F[\alpha]$.