Degree of $\alpha+\beta$ over $\mathbb{F}_{2011}$

Question

Let $\alpha, \beta$ be the roots of $x^2-5$ and $x^2+5$, respectively, over an algebraic closure of $\mathbb{F}_{2011}$. Show that $\alpha+\beta$ has degree $2$ over $\mathbb{F}_{2011}$.

My proof: Clearly the degree of $\alpha+\beta$ over $\mathbb{F}_{2011}$ is at most $2$. Now, since $2011\equiv 3 \pmod 4$, $-1$ is not a square in $\mathbb{F}_{2011}$ and exactly one between $5$ and $-5$ is not a square in $\mathbb{F}_{2011}$. Therefore $\alpha+\beta \not\in\mathbb{F}_{2011}$ and $\mathbb{F}_{2011}(\alpha+\beta)=\mathbb{F}_{2011^2}$.

My teacher told me that my proof is incomplete: I have also to show that $\mathbb{F}_{2011}(\alpha+\beta)=\mathbb{F}_{2011}(\alpha,\beta)$. Why?

Answer 1

Hints

By QRT:

$$\left(\frac5{2011}\right)=\left(\frac{2011}5\right)=\left(\frac1{2011}\right)=1$$

and thus $\;x^2-5\in\Bbb F_{2011}[x]\;$ is reducible and $\;\alpha=\sqrt5\in\Bbb F_{2011}\;$ .

By a similar calculation as above, $\;x^2+5\in\Bbb F_{2011}[x]\;$ is irreducible , so ...

Answer 2

It is not immediately clear that $\alpha+\beta$ is at most $2$. In general $k(\alpha+\beta)\subset k(\alpha,\beta)$, and if $\alpha$ and $\beta$ are both of degree $2$ but $k(\alpha) \subsetneq k(\beta)$, then the sum will be of degree $4$. Also, while primitive elements for extensions often exist, there is no a priori reason why we cannot in have $k(\alpha+\beta)\subsetneq k(\alpha,\beta)$. It is also possible that the degree of the sum is smaller than the degree of the summands, e.g., $(2+\sqrt{2})+(3-\sqrt{2})$. However, you have done all the actual work necessary to give a full and complete proof.

By your calculation, one of $\alpha$ and $\beta$ is already in the field, while one lies in a degree $2$ extension. The hint by @DonAntonio is that we can calculate which one, but this is unnecessary. By renaming, if necessary, let $\alpha$ be the root of the reducible polynomial, $\beta$ the root of the irreducible one. Then $k(\alpha,\beta)=k(\beta)$ is of degree $2$, and $k(\alpha+\beta)=k(\beta)$ because $\alpha\in k$.