I found the following in my textbook: Consider the field extension $\mathbb{Q} \subset \mathbb{Q}(\alpha_1)=:K$ ($\alpha_1$ is a root). Since $f=x^4-2$ is irreducible by Eisenstein's criterion or by reducing $f$ modulo $5$, we find that $[\mathbb{Q}(\alpha_1):\mathbb{Q}]=4$.
I don't understand how the dimension follows from the fact that $f$ is irreducible. Could somebody explain this to me? Thanks!
On
Since , the minimal polynomial is of degree 4, hence the elements of the field can be expressed as linear combination of four basis elements of 1,a,a²,a³ over $\mathbb{Q}$ ,where a is the root(a⁴=2). You can think also it as vector space with basis {1,a,a²,a³} over $\mathbb{Q}$.(since, any field is a vector space over any of it's subfield).
On
Consider a monic irreducible polynomial $f(x)$ of degree $n$ in the ring $k[x].$ Given a root $\alpha$ of $f(x)$ in some field extension of $k,$ observe that the evaluation homomorphism $\operatorname{ev}_\alpha : k[x] \to k(\alpha)$ defined by $\operatorname{ev}_\alpha(p(x)) = p(\alpha)$ is surjective by definition of $k(\alpha).$ Considering that $\ker \operatorname{ev}_\alpha$ is an ideal of the principal ideal domain $k[x],$ it follows that $\ker \operatorname{ev}_\alpha = (p(x))$ for some monic polynomial $p(x).$ Consequently, we have that $p(x) \,|\, f(x).$ By hypothesis that $f(x)$ is a monic irreducible polynomial, we conclude that $p(x) = f(x)$ so that $\ker \operatorname{ev}_\alpha = (f(x)).$
By the First Isomorphism Theorem, we conclude that $k[x] / (f(x)) \cong k(\alpha).$ Considering that $$k[x] / (f(x)) = \{a_0 \overline 1 + a_1 \overline x + \cdots + a_{n - 1} \overline x^{n - 1} \,|\, a_i \in k \}$$ is a $k$-vector space, it follows that its dimension over $k$ is $n,$ hence we have that $[k(\alpha) : k] = n.$
Since the polynomial is of degree 4, a basis of $K$ over $\mathbb{Q}$ is $1, \alpha, \alpha^2, \alpha^3$ (since $\alpha^4=2\in \mathbb{Q}$).