Let $k$ a field, $P \in k[X]$ irreducible of degree $n \geq 2$, $K$ an extension field of $k$ with degree $m$ such as $\gcd(m,n) =1$. How can I show that $P$ stays irreducible over $K$ ?
Thank for answers.
Let $k$ a field, $P \in k[X]$ irreducible of degree $n \geq 2$, $K$ an extension field of $k$ with degree $m$ such as $\gcd(m,n) =1$. How can I show that $P$ stays irreducible over $K$ ?
Thank for answers.
Let's draw a diagram of When working with many field extensions, it is often very useful to draw a chart showing how they're all related. e.g.
$ \begin{array}{ccccc} K & & & & k[X] / P \\ &\nwarrow & & \nearrow \\ & & k \end{array} $
Actually, let's fix an algebraic closure of $k$, and that $X$ maps to a specific root $\alpha$ of $P$. (I don't often like doing this, but it's useful for this sort of problem)
$ \begin{array}{ccccc} K & & & & k(\alpha) \\ &\nwarrow & & \nearrow \\ & & k \end{array} $
It helps to annotate these diagrams with the degree of the extension:
$ \begin{array}{ccccc} K & & & & k(\alpha) \\ &\nwarrow m & & \nearrow n \\ & & k \end{array} $
Now, can we translate your question into one about field extensions? I claim yes -- if $f$ is reducible, then what can you say about the minimal polynomial of $\alpha$ over $K$? Anyways, let's fill in the missing interesting field:
$ \begin{array}{ccccc} & & K(\alpha) \\ &\nearrow & & \nwarrow \\ K & & & & k(\alpha) \\ &\nwarrow m & & \nearrow n \\ & & k \end{array} $
Now, try to fill in the missing annotations....
Let $\Omega$ be a field containing $K$ and a root $x$ of $P$.
Show first that $[K(x):k]=mn$ and then that $[K(x):K]=n$.
[Use the towers $k\subset k(x)\subset K(x)$ and $k\subset K\subset K(x)$]
This is basically the same as the other answers.
Let $x$ be a root of $P$ in some extension of $K$. Then $P$ is the minimal polynomial of $x$ over $k$. Let $d$ be the degree of $K(x)$ over $k$.
Contemplating the tower $k\subset K\subset K(x)$, we see that $m$ divides $d$.
Contemplating the tower $k\subset k(x)\subset K(x)$, we see that $n$ divides $d$.
Contemplating any of the above the towers, we see that $d$ doesn't exceed $mn$.
As $m$ and $n$ are coprime, we have $d=mn$, and the first tower shows that $K(x)$ has degree $n$ over $K$. This implies that $P$ is irreducible over $K$.