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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.

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    $\begingroup$ That's perfectly correct. $\endgroup$ – Bernard Jul 29 '18 at 10:08
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First $\mathbf Q[α]=\{p(α)\mid \deg p<n\}$.

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.

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