Let $\mathbb F$ be an extension of a field $\mathbb K$ such that $[\mathbb F: \mathbb K] = 2$...

Question

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.

Answer 1

Applying the primitive element theorem there exist some $\alpha\in F$ such as $K(\alpha)=F$. Since the degree of the extension is two then the elements, $1,\alpha,\alpha^2$ are linearly dependant, so there exist some $a,b,c\in K$ such as $a\alpha^2-b\alpha-c=0$ not all equal to zero. We can suppose without loss of generality that $a\not =0$ so dividing by zero we get some $d,e\in K$ such as $\alpha^2-d\alpha-e=0$. By the Tschirnhaus transform (completing squares in these case), we have the following: $$(\alpha-\frac{d}{2})^2=\alpha^2+\frac{d^2}{4}-\alpha d=\frac{d^2}{4}+e\in K$$. Now taking $\beta=\alpha-\frac{d}{2}$, and since $K(\beta)=K(\alpha)$ we have found that $\beta^2\in K$, so $F$ is just the splitting field (the accurate name for field of roots) of the polynomial $f(t):=t^2-\beta^2$

Answer 2

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.

Answer 3

From what I can gather from this MSE post is that a field of roots means a splitting field .

In that case you need to produce a polynomial for which $\Bbb{F}$ is the smallest field over which the polynomial splits into linear factors.

Let $\alpha\in \Bbb{F\setminus K}$ (such an alpha exists other wise $\Bbb{K}=\Bbb{F}$) . Then let $p(x)$ be the minimal polynomial for $\alpha $ over $\Bbb{K}$. Now $\deg p(x)$ cannot be equal to $1$ as $\alpha\notin\Bbb{K}$ . And $\deg(p(x))$ cannot be greater than $2$ as $\Bbb{K}(\alpha)\subseteq \Bbb{F}$ and hence $\deg(p(x))=[\Bbb{K(\alpha)}:\Bbb{K}]\leq [\Bbb{F}:\Bbb{K}]$

Hence $\deg(p(x))=2$ .

But this immediately gives you that $\Bbb{K}(\alpha)=\Bbb{F}$ as $[\Bbb{K(\alpha)}:\Bbb{K}]=[\Bbb{F}:\Bbb{K}]=2$

Now over $\Bbb{K}(\alpha)$ we have $p(x)=(x-\alpha)q(x)$ where $q(x)$ has degree $1$. And hence $p(x)$ splits over $\Bbb{K}(\alpha)$ . And any splitting field of $p(x)$ must contain $\alpha$ and $\Bbb{K}$ and hence contain $\Bbb{K}(\alpha)$ .

$\Bbb{K}(\alpha)$ is the smallest field containing $\Bbb{K}$ and $\alpha$ and since the splitting field must also contain $\alpha$ and $\Bbb{K}$, we have the splitting field is contained in $\Bbb{K}(\alpha)$ . Hence $\Bbb{K}(\alpha)$ is the splitting field of $p(x)$ .

This gives you that $\Bbb{K}(\alpha)=\Bbb{F}$ is the splitting field of $p(x)$.