Let $\mathbb F$ be an extension of a field $\mathbb K$ such that $[\mathbb F: \mathbb K] = 2$. Show that $\mathbb F$ is a field of roots over $\mathbb K$.
If $f(x)$ is a root field then $f(x)$ decomposes into $K$, that is, $f(x)=c(x-\alpha_1)…(x-\alpha_r)$ for certain $\alpha_1,…,\alpha_r \in K.$ E $K =f(\alpha_1,…,\alpha_r).$
And if $[\mathbb F: \mathbb K] = 2$, then we have $1, \alpha, \alpha^2$. So I need to get to
$f(x) = c(x-\alpha_1)(x-\alpha_2) = a_0 + a_1x +a_2x^2.$ Correct?
Assuming it's correct, I tried to do the following.
$(cx -c\alpha_1)(x-\alpha_2) = cx^2 – cx\alpha_2 – cx\alpha_1 + c\alpha_1\alpha_2 = a_0 + a_1x +a_2x^2 \to a_0 = c\alpha_1\alpha_2,\ a_1x=(c\alpha_1-c\alpha_2)x,\ a_2x^2=cx^2$.
I don't think it's right but I can't think of anything else.
Thank's.
Best Answer
I am new and I can't comment until I have 50 points of reputation. I want to say that to apply the primitive element theorem you need that your extension is separable.
How do you know that $$\mathbb{K} \hookrightarrow \mathbb{F}$$ is separable?
I think that this is a posible alternative without using primitive element theorem:
Take $\alpha \in \mathbb{F} | \alpha \not\in \mathbb{K}$ (how $[\mathbb{F} : \mathbb{K}]=2$ note that $\alpha$ exists), then we have $\mathbb{K} \hookrightarrow \mathbb{K}(\alpha) \hookrightarrow \mathbb{F}$ and for the degree theorem we have that $[\mathbb{F}:\mathbb{K}(\alpha)]=1$ so $\mathbb{F}=\mathbb{K}(\alpha)$.
Now we can continue with the same procediment that use the other user.