TITLE: Reducibility of $P(X^2)$
QUESTION [10 upvotes]: This question is inspired by a comment discussion in If $K=K^2$ then every automorphism of $\mbox{Aut}_K V$, where $\dim V< \infty$, is the square of some endomorphism..
Let $k$ be a field of caracteristic different from $2$, all of whose elements are squares. Let $P$ be an irreducible polynomial over $k$. Is it true that $Q(X)=P(X^2)$ cannot be irreducible?
I believe I have a proof for polynomials of odd degree (and possibly caracteristic $0$) by contradiction. The idea is to use the automorphism of $\sigma\in G=Gal(\frac{k[X]}{Q(X)}\supset k)$ defined by $X\mapsto -X$ and showing that it maps to $-1$ under $G\rightarrow \{ -1,+1\}, \theta \mapsto det(\theta)^{deg(P)}$ thus giving a normal subgroup of index $2$ of $G$ and via the Galois correspondance a degree $2$ extension of $k$ of which there can be none.
I even wonder wether there may be irreducible polynomials off even degree at all...
EDIT Since I put a bounty on this question, let me clear about what it is I would like to know. First of all, I would like to know wether my proof by contradiction is valid, and whether the restriction to caracteristic $0$ is necessary. Secondly, and most importantly, I ask if irreducibility of $P$ implies $Q$ cannot be irreducible when all elements in the ground field are squares (and $car(k)\neq 2$), thirdly, and this is optional (but would prove the second point if true), I wonder wether there can be any irreducible polynomials of even degree under said assumption on the ground field $k$. Thank you very much for your time!
Also, if the second question turns out to be right, I will be able to answer the question (linked above) that started all this. Bounties will be given to whoever answers the second or third question.
FURTHER EDIT It seems my proof in the odd case is partly and possibly completely invalid : the extension is not generally Galois and so the correspondance fails. One could try to replace it with the splitting field of $Q$ which has a similar automorphism sending each root of $Q$ to its opposite, calculating its matrix and taking determinants and taking some power in order to fall into the square roots of $1$. But I am no longer sure the result maps to $-1$...

REPLY [3 votes]: This is a complement to Lierre's answer. It is well known that if we take a unitary polynomial $P$ of degree $n$, $P=X^n+a_{n-1}X^{n-1} + \ldots +a_1X+a_0$, in a "sufficiently random" way, then there will be no special algebraic relation satisfied by the roots $\alpha_1, \ldots, \alpha_n$, other than those coming from the formulas equating the coefficients to symmetric polynomials in the roots. For those "generic" choices, $P$ will be irreducible and its Galois group will be the full group ${\mathfrak S}_n$ (the key notion here is that of a thin set in the sense of Serre : the set of polynomials not satisfying what I say form a thin set in ${\mathbb Q}^n$).
Similarly, there will be no special relation involving the roots of $Q(X)=P(X^2)$, so the Galois group of $Q$ (over $\mathbb Q$) will be the isomorphic to the group ${\mathfrak S}'_n$ of signed permutations on $\lbrace 1,2, \ldots , n \rbrace$. 
Denote by $\mathbb S$ the square-root-closure of $\mathbb Q$ in $\mathbb C$(in Hagen's terminology), also known as the smallest quadratically closed subfield of $\mathbb C$  (in Lierre's terminology).
Thanks to the Galois correspondence, your question : "Is $Q$ reducible over $\mathbb S$ ?"translates as a purely group-theoretic question, as follows : 
For any group $G$, it is easy to show that any sequence $G=N_0 >N_1>N_2> \ldots N_r$
where $N_{i+1}$ is a subgroup of index $2$ in $N_i$, must always eventually stop at a special subgroup of $G$, that's independent from the sequence. I do not know if that subgroup has a name in the literature already ; let's call it the Begassat subgroup of $G$ for short.
Then, for a generic polynomial $P$, the Galois group of $Q=P(X^2)$ over $\mathbb S$ will be $B_n$, the Begassat subgroup of ${\mathfrak S}'_n$. So $Q$ irreducible over $\mathbb S$ iff $B_n$ acts transitively on $\lbrace 1,2, \ldots ,n \rbrace$.
So it all reduces to computing the Begassat subgroup of ${\mathfrak S}'_n$. Now
${\mathfrak S}'_n$ can be written as a semi-direct product ${\mathfrak S}_n \rtimes \lbrace \pm 1\rbrace^n$ (the left part in the semi-product corresponds to the signed permutations without the signs). Put $$U_n=\lbrace (x_1,\ldots ,x_n) \in \lbrace \pm 1\rbrace^n | x_1x_2x_3 \ldots x_n=1\rbrace$$. Then $U_n$ has index $2$ in $\lbrace \pm 1\rbrace^n$ and ${\cal A}_n$ has index $2$ in ${\mathfrak S}_n$, so if we put $H_n={\mathfrak S}_n \rtimes U_n$ and 
$G_n={\mathcal A}_n \rtimes U_n$, in the sequence ${\mathfrak S}'_n >H_n >G_n$
each term has index $2$ in the preceding term.
Now $D(G_n)=G_n$ for $n\geq 5$, and $D(G_n)$ has index $3$ in $G_n$ for $n=4$. This shows that $G_n$ cannot have an index $2$ subgroup for $n\geq 4$, so $G_n$ is the Begassat subgroup of ${\mathfrak S}'_n$ and hence the Galois group of $Q$ over $\mathfrak S$ is $G_n$, so $Q$ is irreducible over $\mathfrak S$.
To summarize : for "most" polynomials $P$ of degree $n\geq 4$, $Q=P(X^2)$ will stay irreducible over $\mathbb S$, the smallest quadratically closed subfield of $\mathbb C$.
The main difference with Lierre's answer here is that he starts with a "semi-generic" polynomial instead of a fully generic one (with Galois group ${\mathcal A}_n$ instead
of ${\mathfrak S}_n$).