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in my Algebra class, it was given as an exercise to find all possible Galois groups of reducible polynomials of degree 5 over $\mathbb{Q}$ without repeated roots.

Where, for a field $F$, we call the Galois group of a polynomial $f(x)\in F[x]$ the Galois group of one of its splitting extensions $E/F$.

Now, I've already classified all possible Galois groups of irreducible polynomials over $\mathbb{Q}$ of degree 1 (the trivial group $\{1\}$), 2 ($S_2$), 3 ($A_3$ and $S_3$) and 4 ($V_4$, $C_4$, $D_4$, $A_4$ and $S_4$). My idea is to recover the Galois group of a reducible $f(x)\in \mathbb{Q}[x]$ of degree 5 without repeated roots starting from these groups as follows.

Every map in the Galois group $G$ of $f(x)$ over $\mathbb{Q}$ can only shuffle separately the roots of the irreducible factors of $f(x)$. So, say that $f(x)=g_1(x)\cdot\ldots\cdot g_k(x)$ is a decomposition in irreducible factors in $\mathbb{Q}[x]$, $G\leq G_1\times\ldots\times G_k$, with $G_i$ Galois group of $f_i(x)$ over $\mathbb{Q}$. We distinguish then the cases:

i- $deg(g_i)=1$ for all $i\leq k=5$, then $G=\{1\}$;

ii- $deg(g_1)=2$ and $deg(g_i)=1$ for all $i\neq 1$, then $G=G_1=S_2$;

iii- $deg(g_1)=3$ and $deg(g_i)=1$ for all $i\neq 1$, then $G=G_1$, which can be only $A_3$ or $S_3$;

iv- $deg(g_1)=4$ and $deg(g_i)=1$ for all $i\neq 1$, then $G=G_1$, which as before can be only $V_4$, $C_4$, $D_4$, $A_4$ or $S_4$;

v- $deg(g_1)=deg(g_2)=2$ and the remaining factor of order 1, then we consider the splitting field $E_j\subseteq\bar{\mathbb{Q}}$ of $g_j(x)$ over $\mathbb{Q}$, for $j\leq k$. The intersection $E_1\cap E_2$ is a $\mathbb{Q}$-vector subspace of $E_1$ and $E_2$ containing $\mathbb{Q}$, so it has at least dimension 1. If $dim(E_1\cap E_2)=1$, then $E_1\cap E_2=\mathbb{Q}$ and then $G=G_1\times G_2=S_2\times S_2$. Conversely, if $dim(E_1\cap E_2)=2$, then $E_1=E_2$ and $G=G_1=G_2=S_2$. These seem to be the only possibilities according to Linear Algebra;

vi- $deg(g_1)=3$, $deg(g_2)=2$ and $k=2$. Then, arguing as above, $E_1\cap E_2$ is a $\mathbb{Q}$-vector subspace of $E_1$ and $E_2$ of dimension at least 1 and can only have:

(a)- $dim(E_1\cap E_2)=1$, then $E_1\cap E_2=\mathbb{Q}$ and then $G=G_1\times G_2=G_1\times S_2$, with $G_1\in \{A_3, S_3\}$;

(b)- $dim(E_1\cap E_2)=2$, then $\mathbb{Q}\subseteq E_1\cap E_2=E_2\subseteq E_1$. From here by using the Fundamental Theorem of Galois Theory we can get that $Gal(E_1/E_2)$ is a normal subgroup of $G_1$ of index $|G_2|=|S_2|=2$. However, except for this, I don't see how to settle this last case, is there anything I'm missing?

Moreover, is the argument above correct? Also, do you know any examples of polynomials for the cases v and vi-(a)?

I would appreciate any help, thanks.

F. Salviati
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    It's not clear to me, what do you think is missing from (vi-b)? When $Gal(E_1/\Bbb{Q})$ has an index two subgroup, it has to be all of $S_3$. This happens for example, when $$f(x)=(x^3-2)(x^2+3).$$ Anyway, looks like you have all the cases. – Jyrki Lahtonen Jun 23 '24 at 10:17
  • Well, the goal is to compute $G$ also in this case. My argument above is incomplete since it does not provide $G$ – F. Salviati Jun 23 '24 at 10:18
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    As $E_2\subseteq E_1$, you see that $E_1$ is the splitting field of $f$. So the Galois group is $G=S_3$. – Jyrki Lahtonen Jun 23 '24 at 10:21
  • Correct. Well done! Feel free to relocate that comment into an answer box. That way you can mark this question as answered, and also get feedback from others. – Jyrki Lahtonen Jun 23 '24 at 15:45
  • Done, thank you again! I've just also edited the question because I'd like to see some examples of such polynomials for the cases (v) and (vi)-(a). I'll close the post as soon as I'll get them – F. Salviati Jun 23 '24 at 15:53

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I'm just resuming here @JyrkiLahtonen nice suggestion.

Since we have that $f(x)=g_1(x)\cdot g_2(x) \cdot (x-\alpha)$ for some $\alpha\in \mathbb{Q}$, $E_i$ is the splitting field of $g_i(x)$, for $i=1,2$ (i.e. the extension of $\mathbb{Q}$ obtained by adding just the roots of $g_i(x)$) and $E_2\subseteq E_1$, we have that $E_1$ is the splitting field of $f(x)$ over $\mathbb{Q}$. Hence, $G=G_1$, which can only be $S_3$ since this is the only subgroup of $S_3$ having a proper subgroup of index 2.

F. Salviati
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