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I was just thinking about this

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?

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  • $\begingroup$ 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 $f(x)$. $\endgroup$ Commented Aug 23, 2022 at 17:28

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Yes, this is correct, though you failed to consider the possibility that the polynomial is a perfect power.

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.)

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