# Degree of a field extension of a minimal polynomial

Suppose a polynomial $f(x)$ of degree $n$ over $\mathbb{Q}$ is the minimal polynomial of an element $\alpha$ in an extension field of $\mathbb{Q}$. Is $[\mathbb{Q}(\alpha):\mathbb{Q}]=n$? Please give reason(s) for your guess.

• That's perfectly correct. – Bernard Jul 29 '18 at 10:08

First $$\mathbf Q[α]=\{p(α)\mid \deg p.
Indeed,we can divide any polynomial $$p(x)$$ by $$f(x)$$ by Euclidean division: \begin{align} p(x)&=q(x)f(x)+r(x),\quad& r&=0\;\text{or}\;\deg r<\deg f=n. \end{align}
Thus $$p(α)=r(α)$$ and $$\mathbf Q[α]$$ is a finite-dimensional $$\bf Q$$-vectorspace, of dimension $$n$$ since $$f$$ is the minimal polynomial of $$α$$.
Next, note that $$\mathbf Q(α)=\mathbf Q[α]$$.
Indeed, since $$\mathbf Q[α]$$ is a finite dimensional $$\mathbf Q$$-vector space, so that multiplication by a non-zero element of $$\mathbf Q[α]$$, which is injective, is also surjective, i.e. $$1$$ is attained – in other words this non-zero element has an inverse in $$\mathbf Q[α]$$, which is therefore a field.