# Field extension and irreducibility

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 $K$ ?

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Relevant? wiki/Gauss's_lemma_(polynomial) – user2468 Mar 8 '12 at 18:30

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 $K$.

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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 $k\subset K\subset K(x)$]

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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....

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Dear Hurkyl, if you don't like fixing an algebraic closure of $k$, take a splitting field $\Omega$ of $P(X)$ seen as a polynomial $P(X)\in K[X]$. The field $\Omega$ will contain everything you want, in particular $K$ and $\alpha$ . – Georges Elencwajg Mar 8 '12 at 18:55
I guess I didn't really say, but my irritation is more about fixing an embedding into an ambient field, and viewing an extension as being a subfield of the ambient field formed by adjoining a particular element of the ambient field. e.g. I found algebraic number theory far more tractable when I started viewing number fields as quotients of polynomial rings, rather than subfields of $\mathbb{C}$. – Hurkyl Mar 8 '12 at 19:49