Let $\alpha$ be a root of $(x^2-a)$ and $\beta$ be a root of $(x^2-b)$. Provide conditions over $a$ and $b$ to have $F=K(\alpha+\beta)$. QUESTION: Let $K$ be a field of characteristic different of $2$. Let $F$ be a splitting field for $(x^2-a)(x^2-b)\in K[x]$. Let $\alpha$ be a root of $x^2-a$ and $\beta$ be a root of $x^2-b$. Provide conditions over $a$ and $b$ to have $F=K(\alpha+\beta)$.
MY ATTEMPT:
Let $\alpha=\sqrt{a}$, $\beta=\sqrt{b}$ and $\gamma=\alpha+\beta$.
First of all, we have $F=K(\alpha, \beta)$ due to the definition of splitting field.
Defining $K(\alpha+\beta)=K(\gamma)$.
Let's show that $K(\alpha, \beta)\subset K(\gamma)$:

*

*From $\gamma=\alpha+\beta$ follows that
\begin{align*}
\gamma^2&=(\alpha+\beta)^2\\
&=\alpha^2+2\alpha\beta+\beta^2\\
&=(\sqrt{a})^2+2\sqrt{a}\sqrt{b} +(\sqrt{b})^2\\
&=(a+b)+2\sqrt{a}\sqrt{b}\qquad (*)\\
\end{align*}

*Now we are going to show that $\sqrt{b}\in K(\gamma)$
Indeed, multiplying both sides in $(*)$ by $\sqrt{b}$ we have:
$\gamma^2\sqrt{b}=(a+b)\sqrt{b}+2\sqrt{a}(\sqrt{b})^2$. Then $$\sqrt{b}=\frac{2b\sqrt{a}+(a+b)\sqrt{b}}{\gamma^2}\in K(\gamma)$$

*

*Similarly, $\sqrt{a}\in K(\gamma)$, this is

$\gamma^2\sqrt{a}=(a+b)\sqrt{a}+2(\sqrt{a})^2\sqrt{b}$, then
$$\sqrt{a}=\frac{(a+b)\sqrt{a}+2a\sqrt{b}}{\gamma^2}\in K(\gamma)$$
MY DOUBT:
I guess there is no conditions over $a$ and $b$ such that $\alpha=\sqrt{a}$ and $\beta=\sqrt{b}$, however I'm not sure. And I don't know how to connect this with the hypothesis that $K$ has characteristic different of two. Would you help me please?
 A: Once you know that $[ K( \alpha,\beta ) : K ]=4$, with basis $\{1,\alpha,\beta,\alpha\beta\}$, you can proceed as follows:
$$
\begin{pmatrix} 1 \\ \gamma \\ \gamma^2 \\ \gamma^3 \end{pmatrix}
=
\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ a+b & 0 & 0 & 2 \\ 0 & a+3b & 3a+b & 0
 \end{pmatrix}
\begin{pmatrix} 1 \\ \alpha \\ \beta \\ \alpha\beta \end{pmatrix}
$$
The matrix has determinant $4(b-a)$ and so is invertible iff $a\ne b$ since the characteristic of $K$ is not $2$. Therefore, $\{1,\gamma,\gamma^2,\gamma^3\}$ is also a basis and so generates the same space, that is, $K( \alpha,\beta ) = K(\gamma)=K( \alpha + \beta )$.
Bottom line: The main condition is that $[ K( \alpha,\beta ) : K ]=4$, or equivalently, $\beta \not\in K( \alpha)$.
This approach does not work in characteristic $2$ because $[K(\gamma):K]\le 2$ since $\gamma^2 = a+b \in K$.
A: We assume that $x^2-a,x^2-b$ are irreducible over $K$ and $b\ne a$, since otherwise the problem is trivial.

*

*If $\sqrt{b}\not \in K(\sqrt{a})$ then show that $\sqrt{a}+\sqrt{b}$ has 4 distinct conjugates (that's where we use $char(K)\ne 2$) which implies that $[K(\sqrt{a}+\sqrt{b}):K] = 4$.


*If $\sqrt{b}=u+v\sqrt{a} \in K(\sqrt{a})$ then $v\ne 0$ so $(u+v\sqrt{a})^2\in K$ implies $u=0$. Since $b\ne a$ then $v\ne \pm 1$ and hence $K(\sqrt{a}+\sqrt{b})= K((v+1)\sqrt{a})=K(\sqrt{a})$.
