# Galois Group of an reducible polynomial

If $$f(x)$$ is a reducible polynomial of degree $$n$$ over a field $$F$$, then its Galois Group is proper subgroup of $$S_n$$. Am I correct?

I feel this is true because, if it is $$S_n$$, then each root is being permuted with other. But we know that root permutations happen amongst an irreducible polynomial, so I think $$f(x)$$ is irreducible.

More precisely, if If $$f(x)$$ is a reducible polynomial of degree $$n$$ over a field $$F$$, further $$[E:F]=n$$, where $$E$$ is splitting field of $$F$$. then its Galois Group is proper subgroup of $$S_n$$. Am I correct?

Are these both statements correct? In not, can you help me formulate a precise statement?

• If $f$ is reducible, then you do not know whether $[E:F]=n$. For example, the splitting field of $(x^2+1)^2$ over $\mathbb{Q}$ has degree $2$, not degree $4$. In fact, the splitting field will almost never have degree $n$, but I don't think that matters here. Regardless of the degree of $E$ over $F$, you get a natural realization of $\mathrm{Gal}(f)$ as a subgroup of $S_n$ by looking at the action on the roots of $f(x)$. Commented Aug 23, 2022 at 17:28

If $$n=r+s$$, $$r,s\gt 0$$, then $$S_n$$ has (proper) subgroups isomorphic to $$S_r\times S_s$$, obtained by requiring the permutation to restrict to a permutation of $$\{1,\ldots,r\}$$ and of $$\{r+1,\ldots,r+s=n\}$$ (or of any partition of $$\{1,\ldots,n\}$$ into a set with $$r$$ alements and a set with $$s$$ elements).
If $$f(x)=g(x)h(x)$$ and $$g(x)$$ and $$h(x)$$ have no roots in common in $$E$$, with $$\deg(f)=n$$, $$\deg(g)=r$$, and $$\deg(h)=s$$, then any automorphism of $$E$$ over $$F$$ will permute the roots of $$g$$ among themselves, and the roots of $$h$$ among themselves, since no root of $$g$$ is a root of $$h$$ and vice versa. This means that the Galois group of $$f$$ lies in the proper subgroup $$S_r\times S_s$$ of $$S_n$$, and in particular is a proper subgroup.
The only case in which a reducible polynomial cannot be factored this way is if $$f(x) = (p(x))^m$$ for some irreducible polynomial $$p$$ and some integer $$m\gt 1$$. In this case, $$\deg{p}\lt n$$, and the Galois group of $$f$$ just permutes the roots of $$p$$ amongst themselves, which means its image in $$S_n$$ lies in the diagonal subgroup of a (proper) subgroup of $$S_n$$ of the form $$S_k\times\cdots\times S_k$$ (where $$\deg(p)=k$$ and $$km=n$$).
Either way you get a proper subgroup of $$S_n$$.
(Note, as I mention in my comment, that you will almost never get that $$[E:F]=\deg(f)$$ when $$E$$ is the splitting field of $$f$$ and $$f$$ is reducible. You could; for example, the splitting field of $$x^3-2$$ has degree $$6$$, so the splitting field of $$f(x) = (x^3-2)^2$$ has degree $$\deg(f)$$; but in general, you will not get degree $$n$$, yet that does not matter here.)