Finding the minimal polynomial in this field extension of $\mathbb Q$?

Question

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?

Answer 1

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 $[\mathbb{Q}(a):\mathbb{Q}]=2$.

Answer 2

Your analysis is exactly correct; in fact $a = \bar{x}$ in $K$, as you said. What element does $a$ correspond to in $\mathbb{Q}(\sqrt{5})$? If you can determine the minimal polynomial of this element over $\mathbb{Q}$, that will tell you what the minimal polynomial of $a$ is over $\mathbb{Q}$ as well.