I have a field extension
\begin{equation*}
K = \mathbb Q[x]/(x^2 – 5)
\end{equation*}
of $\mathbb Q$, and an element $a = \bar x \in K$. I need to find the minimal polynomial of $a$ over $\mathbb Q$.
I have worked out that $[K:\mathbb Q]=2$ and $K \simeq Q(\sqrt 5)$. My main problem is that I'm not sure how to interpret a polynomial involving $a$, as $a$ is basically $x$ if I'm interpreting this correctly. Can anyone give me any hints?
Best Answer
An element of $K$ can be written as $a=p+q\sqrt{5}$, so $$ a-p=q\sqrt{5} $$ and $$a^2-2ap+p^2=5q^2.$$ Therefore $a$ is a root of $$ X^2-2pX+p^2-5q^2. $$ This is the minimal polynomial if and only if $q\ne0$, because in this case the degree of $a$, as you correctly thought is $2$. Indeed, we have $$ [\mathbb{Q}(\sqrt{5}):\mathbb{Q}(a)][\mathbb{Q}(a):\mathbb{Q}]= [\mathbb{Q}(\sqrt{5}):\mathbb{Q}]=2 $$ and so either $[\mathbb{Q}(a):\mathbb{Q}]=1$ (that is, $a\in\mathbb{Q}$) or $[\mathbb{Q}(a):\mathbb{Q}]=2$.