# Is a polynomial solvable by roots iff every irreducible factor is?

Let $F$ be a field, I asked myself if $p(x)\in F[x]$ is solvable by radicals iff every irreducible factor is solvable by radicals.

My thoughts: If every irreducible factor is solvable by roots then it imply that there are field extensions that are solvable by roots, I would like to use the simple fact that the composition field of solvable extensions is also solvable, but in my case the extensions I have are not subfields of one field.

I don't know about the other direction, and I have a feeling it is not true but due to lack of examples that I know I can't think of a counter example.

Is this statement correct ? If so, how can we prove it ?

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Each individual root is by definition so expressible. Is anything more wanted? –  André Nicolas Jul 27 '12 at 21:52
it is more clear in this way, but I was thinking of how to do this in terms of the definition (and not of existance of formula). i.e how can we show that the galois group is solvable ? (and I still don't know about the other direction...) –  Belgi Jul 27 '12 at 21:56

## 2 Answers

Let $L$ be a splitting field for $fg$, let $K$ be a subfield of $L$ and a splitting field for $f$. Assume $G_{fg}$, the Galois group of $L$ over $F$, is solvable. Galois Theory says that the Galois group of $K$ over $F$ is the quotient of $G_{fg}$ by the Galois group of $L$ over $K$, and group theory says a quotient group of a solvable group is solvable, so the group of $K$ over $F$ is solvable.

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For a polynomial $f \in \mathbf{Q}[x]$, we say $f$ is "solvable by radicals" if the Galois group $G_f$ of the splitting field of $f$ is a solvable group.

Now $G_{fg}$ is a subgroup of $G_f \times G_g$, by one of the corollaries to the Fundamental Theorem of Galois Theory. We also know that:

1. The direct product of solvable groups is solvable;
2. Subgroups of solvable groups are solvable.

Thus if $f,g$ are solvable by radicals, so is $fg$.

What about the converse?

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Can you please explain why $G_{fg}$ is a subgroup of $G_f \times G_g$ ? I didn't learn this in class. –  Belgi Jul 27 '12 at 21:59
The splitting field of $fg$ is the compositum of the splitting fields of $f$ and $g$. Do you know how to compute the Galois group of a compositum? –  Bruno Joyal Jul 27 '12 at 22:05
No, I didn't learn that either (Although I do know why the composition is solvable) –  Belgi Jul 27 '12 at 22:10
Well, if you know that the compositum of solvable extensions is solvable, then you're set! It's what I showed above. –  Bruno Joyal Jul 27 '12 at 22:12
At first I didn't understand why we can take the composition, but than I figured that the both splitting fields ar subfields of the splitting fields of the product of the polynomials, so I this direction is ok. It is still interesting that $G_{fg}$ is a subgroup of $G_f \times G_g$, is it easy to prove ? do you have any idea about the other direction ? thank you for your time and help (+1) –  Belgi Jul 27 '12 at 22:15