I am having some trouble with the following problem:

Let $K\subseteq L$ be fields. Suppose $\alpha,\beta\in L$ are algebraic elements over $K$ of degrees $p,q$ respectively, where $p$ and $q$ are distinct primes. Show that $\alpha+\beta$ is algebraic over $K$ with degree $pq$.

I have shown that $\alpha+\beta$ is algebraic over $K$, but I am having trouble showing that the degree is $pq$. I have previously shown that $[K(\alpha,\beta):K]=pq$, so I have been trying to use this result. I thought that if I could show that $K(\alpha,\beta)=K(\alpha+\beta)$ then I'd be done. To show this I first noted that clearly $K(\alpha+\beta)\subseteq K(\alpha,\beta)$. Then we have $$pq=[K(\alpha,\beta):K]=[K(\alpha,\beta):K(\alpha+\beta)][K(\alpha+\beta):K]$$ so that $[K(\alpha+\beta):K]=1,p,q,pq$. I tried looking at the cases when it is equal to $1,p,q$ and deriving a contradiction, but have been unsuccessful.

Another way I thought of solving this is to show that $\alpha,\beta\in K(\alpha+\beta)$, which would allow me to conclude that $K(\alpha+\beta)=K(\alpha,\beta)$, but I am not sure how to complete this either.

I am looking for some assistance to show that $[K(\alpha+\beta):K]\neq 1,p,q$, or that $\alpha,\beta\in K(\alpha,\beta)$. If you have any other solutions to this problem I would like to see those too.

  • 2
    $\begingroup$ This does not hold without future restrictions on $K, L$. Let $K = \mathbb{F}_3 (s,t)$, where $s,t$ are indeterminates. Let $\alpha, \beta$ satisfy $$\alpha^2 - s = 0 \qquad \beta^3 - \beta s - t = 0$$ both equations are irreducible. However, $\alpha + \beta$ satisfy an equation of degree 3: $$(\alpha + \beta)^3 - s (\alpha + \beta ) - t = 0$$ $\endgroup$ – pisco Jan 29 '18 at 5:59

As pointed out in my comment above, this does not hold in general. I will provide a solution for characteristic $0$ case. Assume $p<q$.

Let $K_1$ be the Galois closure of $K(\alpha)/K$, $K_2$ be the Galois closure of $K(\beta)/K$. Denote $\beta:= \beta_0, \beta_1, \cdots, \beta_{q-1}$ be all the conjugates of $\beta$ over $K$. Becuase $|\text{Gal}(K_2/ K)|$ is divisible by $q$, there exists $\sigma\in \text{Gal}(K_2/ K)$ which is a cyclic permutation. Let $\sigma(\beta_0) = \beta_1 , \sigma(\beta_1) = \beta_2 , \cdots, \sigma(\beta_{q-1}) = \beta_0$. Since the extension $K_1K_2/K_2$ is normal, we can extend $\sigma$ to an element in $G:= \text{Gal}(K_1K_2/ K)$. The number $[K(\alpha+\beta):K]$ is the orbit size of $\alpha+\beta_0$ under $G$.

Now, consider the numbers $$\alpha,\sigma(\alpha), \cdots, \sigma^{q-1}(\alpha)$$ they're roots of minimal polynomial of $\alpha$ over $K$, since $p<q$, two of them must coincide, so we have an integer $1\leq k \leq q-1$ such that $\alpha = \sigma^k (\alpha)$. Now we apply $\sigma^k$ to $\alpha + \beta_0$ sucessively: $$\alpha + \beta_0 \mapsto \alpha + \beta_k \mapsto \alpha + \beta_{2k} \mapsto \cdots \mapsto \alpha + \beta_{(q-1)k} $$ where the subscripts have to be interpreted modulo $q$. Since $(k,q) = 1$, we see that all $\alpha + \beta_0 , \cdots ,\alpha + \beta_{q-1}$ are in the orbit, and they're all distinct. Hence $[K(\alpha+\beta):K] \geq q$

If $[K(\alpha+\beta):K] = q$, then $\alpha + \beta_0 , \cdots ,\alpha + \beta_{q-1}$ are all the roots of the minimal polynomial of $\alpha + \beta$ over $K$, hence $$(\alpha + \beta_0) + (\alpha + \beta_1) + \cdots + (\alpha + \beta_{q-1}) \in K \implies q\alpha \in K$$ where we have used the fact that $\beta_0 + \cdots + \beta_{q-1} \in K$. Since $K$ has characteristic $0$, we have $\alpha \in K$, contradiction. Therefore $[K(\alpha+\beta):K] = pq$, the proof is completed.

  • $\begingroup$ In the third case you, if $f(x),g(x)$ are the polynomials with $\alpha$ as a root of degree $q,p$, respectively, then you say the division algorithm gives $$f(x)=g(x)h(x)+r(x)$$ for some $h(x),r(x)\in K(\beta)[x]$ with $\deg(r)<\deg(p)$ or $r\equiv 0$. What happens if $r\equiv 0$? $\endgroup$ – Dave Jan 28 '18 at 20:56
  • $\begingroup$ In fact, doesn't $g(x)$ necessarily divide $f(x)$ since $g(x)$ is an associate of the minimal polynomial of $\alpha$ over $K(\beta)$? $\endgroup$ – Dave Jan 28 '18 at 21:53
  • $\begingroup$ How do you know that $[K(\alpha)(\beta):K(\alpha)] = q$? I seems quite likely true to me, but my residual skepticism wants proof. This is exactly where I got stuck on the problem. Any ideas? Anyone? $\endgroup$ – Robert Lewis Jan 28 '18 at 22:22
  • 1
    $\begingroup$ @RobertLewis $$[K(\alpha)(\beta):K(\alpha)]=\frac{[K(\alpha)(\beta):K]}{[K(\alpha):K]}=\frac{pq}{p}=q$$since $K(\alpha)(\beta)=K(\alpha,\beta)$. $\endgroup$ – Dave Jan 28 '18 at 22:24
  • $\begingroup$ I think this solution is a bit beyond where we are in our course right now. However, I really appreciate your time on this question. I think I'll appeal to your counter example in your comment. $\endgroup$ – Dave Jan 31 '18 at 17:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.