If $\alpha,\beta\in L$ algebraic over $K$ with degrees $m,n$, then $\alpha\pm\beta$ is algebraic with degree $\leq mn$

Question

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

Answer 1

Hint: Since $k[x] \subseteq k(\alpha)[x],$ it follows that $[k(\alpha,\beta) : k(\alpha)] \leq [k(\beta) : k]$

Answer 2

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 $mn$.