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Let $K$ be a field, $f \in K[X]$ separable and irreducible with $\text{deg}(f)=n$; $x_1,...,x_n$ are the roots of $f$ in a splitting field of $f$ over $K$. Let $g \in K[X]$ be any polynomial with $\text{deg}(g)\leq n-1$. I want to prove the following statement: $$ K[g(x_1)]=K[x_1] \Longleftrightarrow g(x_1),...,g(x_n) \ \text{are pairwise distinct}.$$

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  • $\begingroup$ Can you get one of the inclusions between the two fields? What do you know about the number of conjugates? $\endgroup$ – Jyrki Lahtonen Jun 1 '14 at 9:45
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    $\begingroup$ And welcome to Math.SE! If you followed the general advice of "lurking" for a while before starting, you should have noticed that here we don't always want to simply spoonfeed solutions (sometimes we do - it varies on the question and the person). Our goal is to make the askers understand, and to give answers/hints that help you get there. To that end it is helpful that you describe your own thoughts and give a bit more context: what has been covered in the book/course up to this point et cetera. $\endgroup$ – Jyrki Lahtonen Jun 1 '14 at 9:48
  • $\begingroup$ That inclusion is correct. Getting the other inclusion needs a bit work. But from $K[g(x_1)]=K[x_1]$ you cannot conclude that the two numbers would have the same minimal polynomial. For example $\Bbb{Q}(\sqrt2)=\Bbb{Q}(1+\sqrt2)$, but the minimal polynomials of $\sqrt2$ and $1+\sqrt2$ do not coincide. $\endgroup$ – Jyrki Lahtonen Jun 1 '14 at 13:03
  • $\begingroup$ You can for example consider the questions: What are the conjugates of $g(x_1)$? What does the number of conjugates of $\alpha$ (with a separable minimal polynomial) tell you about the degree of the extension $K[\alpha]/K$? $\endgroup$ – Jyrki Lahtonen Jun 1 '14 at 13:21
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Hint 1:

$K(g(x_1)) \subset K(x_1)$ is always true, so the two are equal iff $[K(x_1):K]=[K(g(x_1)):K]$.

Hint 2:

In any finite Galois extension $L/K$, if $G = Gal(L/K)$, the minimal polynomial of $y$ over $K$ is $f(X)=\Pi_{i=1}^r(X-y_i)$ where $\{y_1,\dots,y_r\}$ is the orbit of $G$ acting on $y$, and the $y_i$ are distinct.* Since $f(X)$ is the minimal polynomial of $x_1$, if $L$ is the splitting field for $f(X)/K$, $G=Gal(L/K)$, then $G$ acts transitively on the roots of $f$ by what I said above. What does this tell us about the minimal polynomial of $g(x_1)$?

*This is because $f(X)$ is fixed by $G$ so it's coefficients lie in $K$, and any polynomial in $K[X]$ which has $y$ as a root has all of the $y_i$ as roots as well.

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  • $\begingroup$ @user144921 if $\sigma \in G$ sends $x_1$ to $x_i$, where does it send $g(x_1)$? It may help to write out $g(X)$ explicitly as a polynomial with coefficients on $K$. $\endgroup$ – Tom Oldfield Jun 1 '14 at 20:37
  • $\begingroup$ @user144921 Given this, and given hint $2$, can you work out the minimal polynomial of $g(x_1)$ over $K$? $\endgroup$ – Tom Oldfield Jun 1 '14 at 20:47
  • $\begingroup$ Not quite. The orbit of $g(x_1)$ under $G$ is $\{g(x_1),\dots,g(x_n)\}$. The minimal polynomial is the polynomial which has each of these as roots exactly once. In other words, if the orbit is $\{y_1,\dots,y_r\}$ where the $y_i$ are all distinct, $f_{g(x_1)}(X)=\Pi(X-y_i)$. Do you see the difference? Can you now give a condition for the degree of this polynomial to be $n$? $\endgroup$ – Tom Oldfield Jun 1 '14 at 20:56
  • $\begingroup$ @user144921 Yes, and that is exactly what we wanted to prove. Does it all make sense? $\endgroup$ – Tom Oldfield Jun 1 '14 at 21:01
  • $\begingroup$ @user144921 No problem, happy to help! $\endgroup$ – Tom Oldfield Jun 1 '14 at 21:07

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