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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.

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    $\begingroup$ 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
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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.

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  • $\begingroup$ 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
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    $\begingroup$ @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

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