If $\alpha,\beta\in L$ algebraic over $K$ with degrees $m,n$, then $\alpha\pm\beta$ is algebraic with degree $\leq mn$ Sorry if this is a duplicate, I couldn't find anything on here with $m,n$ not being coprime. 
My attempt thus far: first observe that $[k(\alpha):k]=m$, $[k(\beta):k]=n$.
Since $(k(\alpha,\beta):k(\alpha))$,  $(k(\alpha,\beta):k(\beta))$ are finite extensions they are algebraic and we have $[k(\alpha,\beta):k(\alpha)]= k_1$ and $[k(\alpha,\beta):k(\beta)]= k_2$ for some $k_1,k_2$
Therefore $[k(\alpha,\beta):k]=mk_1=nk_2$ and $n\vert mk_1$ and  $m\vert nk_2$. 
But I don't see how to conclude though that its less than $nm$, Any hints would be appreciated
 A: Hint: Since $k[x] \subseteq k(\alpha)[x],$ it follows that $[k(\alpha,\beta) : k(\alpha)] \leq [k(\beta) : k]$
A: We are given that
$[K(\alpha):K] = m, \; [K(\beta):K] = n; \tag 1$
we observe that
$\alpha \pm \beta \in K(\alpha, \beta) = K(\alpha)(\beta); \tag 2$
using (1), by the tower law we have
$[ K(\alpha, \beta): K]$
$= [ K(\alpha, \beta):K(\alpha)][K(\alpha):K] =  [ K(\alpha, \beta):K(\alpha)]m. \tag 3$
Now 
$ [ K(\alpha, \beta):K(\alpha)] = [K(\alpha)(\beta): K(\alpha)]$
$= \deg m_\alpha(x) \in K(\alpha)[x], \tag 4$
where $m_\alpha(x)$ is the minimal polynomial of $\beta$ over $K(\alpha)$;  denoting by
$m(x) \in K[x] \tag 5$
the minimal polynomial of $\beta$ over $K$, it is easily seen we also have
$m(x) \in K(\alpha)[x], \tag 6$
since 
$K \subset K(\alpha) \Longrightarrow K[x] \subset K(\alpha)[x]; \tag 7$
it follows then from the minimality if $m_\alpha(x)$ over $K(\alpha)$ that 
$\deg m_\alpha(x) \le \deg m(x); \tag 8$
but 
$\deg m(x) = [K(\beta):K] = n; \tag 9$
we may then transform (4) into  
$ [ K(\alpha, \beta):K(\alpha)] = \deg m_\alpha(x) \le \deg m(x) = n ,  \tag{10}$
and so (3) becomes
$[ K(\alpha, \beta): K] \le  nm; \tag{11}$
we thus conclude in light of (2) that $\alpha \pm \beta$ is algebraic over $K$ with degree at most $mn$.
