My task is to show following:

Let $ f \in K[X]$ irreducible with odd degree.

If $\alpha , \beta \in \bar K$ are distinct roots of $f$, where $ \bar K$ is an algebraic closure of $K$, neither $\alpha + \beta \in K$ nor $ \alpha \beta \in K$.

I have no idea, how to go on. Some hints or (partial) solutions would be really nice.

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    $\begingroup$ Are $\alpha,\beta\in\overline{K}\setminus K$? How are $\alpha,\beta$ related to $f$? Perhaps also $f$ has degree at least $3$? $\endgroup$ – Michael Burr Jan 21 '18 at 16:35
  • $\begingroup$ It isn't said in the task, but I think yes, they should be. Otherwise the statement would be false, since K is a fiel, so addition and multiplication is closed. $\endgroup$ – Myrkuls JayKay Jan 21 '18 at 16:36
  • $\begingroup$ Sorry, I forgot to say that they are distinct roots of $f$. Otherwise $f$ wouldn't have a role. Just edited it. $\endgroup$ – Myrkuls JayKay Jan 21 '18 at 16:38
  • $\begingroup$ The case where $\deg(f)=3$. In this case, $f(x)=ax^3+bx^2+cx+d$ where $a,b,c,d\in K$. Observe that $\frac{b}{a}$ is the negative sum of the roots of $f$ and $\frac{d}{a}$ is the product of the roots of $f$. If $\alpha+\beta$ were in $K$, then $\alpha+\beta-\frac{d}{a}$ would be the third root of $f$ and would be in $K$, so $f$ is not irreducible. If $\alpha\beta$ were in $K$, then $\frac{d}{a(\alpha\beta)}$ would be the third root of $K$, so $f$ is not irreducible. $\endgroup$ – Michael Burr Jan 21 '18 at 16:52
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    $\begingroup$ Observe that if the roots of $f$ are $\alpha$, $\beta$, and $\gamma$ (in the fixed algebraic closure), then $f(x)=a(x-\alpha)(x-\beta)(x-\gamma)$. $\endgroup$ – Michael Burr Jan 21 '18 at 17:16

Let us first assume that $f(x)$ is separable.

Let $L$ be the splitting field of $f(x)$. Then $L/K$ is Galois, and the Galois group $G=Gal(L/K)$ acts transitively on the set of zeros of $f(x)$.

Assume that $\alpha+\beta=z\in K$ for some roots $\alpha,\beta$. In other words, $\beta=z-\alpha$ is also a root. For each $\sigma\in G$ we get that $$z-\sigma(\alpha)=\sigma(z)-\sigma(\alpha)=\sigma(\beta).$$ Here $\sigma(\alpha)$ and $\sigma(\beta)$ are also roots of $f(x)$, and by transitivity of $G$ any root of $f(x)$ will appear as $\sigma(\alpha)$ for some $\sigma\in G$. We have proven that:

For any root $\gamma$ of $f(x)$, the element $z-\gamma$ is also a root of $f(x)$.

So the mapping $\delta:\alpha\mapsto z-\alpha$ is a permutation of the roots. If $\delta(\alpha)=\alpha$, then $2\alpha=z$. This is a contradiction because either we have $\alpha=z/2\in K$, or (in characteristic two) $z=0$ and $\alpha=-\alpha$ is a double root. So $\delta$ cannot have fixed points.

As $\delta^2(\alpha)=\alpha$ it follows that $\delta$ permutes the roots in 2-cycles. But there are an odd number of them so this is a contradiction. An alternative end game would be to use the fact that by Vieta relations the sum of all the roots of $f(x)$ is an element of $K$ when pairing up all the roots, save for a single exception, would imply that the unpaired root would also be an element of $K$.

As you asked for hints only I am leaving it to you to figure out what modifications to this argument are needed to show that $\alpha\beta\notin K$.

If $f(x)$ is non-separable and irreducible then, $K$ has characteristic $p$, $f(x)=g(x^{p^t})$ for some $t$, and $g(x)$ is an irreducible separable polynomial of odd degree over $K$. The roots of $g$ are gotten by raising the roots of $f$ to the power $p^t$. So if $\alpha$ and $\beta$ are distinct root of $f$ then $\alpha^{p^t}$ and $\beta^{p^t}$ are roots of $g$ (you prove that they are distinct, it's not hard!). Therefore the above result tells us that $$\alpha^{p^t}+\beta^{p^t}=(\alpha+\beta)^{p^t}$$ is not an element of $K$. Therefore neither is $\alpha+\beta$. Ditto for the product.

  • $\begingroup$ Thank you a lot for this answer ! That the powers of $\alpha$ and $\beta$ are distinct, if $\alpha$ and $\beta$ are distinct, should follow by injectivity of the Frobenius-Homomorphism. $\endgroup$ – Myrkuls JayKay Jan 21 '18 at 18:08
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    $\begingroup$ Indeed. Good job @Myrkuls! $\endgroup$ – Jyrki Lahtonen Jan 21 '18 at 18:08

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