Reducibility of $P(X^2)$

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 $-1$...

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Definitions and Goal

Definition: A field $K$ with the property $K=K^2$ is called square-root-closed (src).

Definition: A field $K$ is called odd, if every irreducible polynomial $f\in K[X]$ has odd degree.

A field $K$ of characteristic $p\neq 2$ is odd if and only every separable, irreducible polynomial $f\in K[X]$ has odd degree.

Every separably closed field of characteristic $p\neq 2$ is odd.

A field $K$ of characteristic $p\neq 2$ is odd if and only if every finite extension has odd degree.

Every algebraic extension of an odd field of characteristic $p\neq 2$ is odd.

Every odd field is src.

Question: Is every src-field $K$ of characteristic $p\neq 2$ odd?

Partial answer given in this post: the answer is Yes for the following classes of fields:

Class 1: $K$ an algebraic extension of a finite field.

Class 2: $K$ an algebraic extension of the rationals, that carries a valuation $v$ such that the residue field $k(v)$ has characteristic $p\neq 2$ and $(K,v)$ is henselian.

The method provided here to prove the answer is Yes for the fields in class 2 actually can be applied recursively thus creating infinitely many classes of fields for which the answer is Yes. Precisely the following result is proved:

Theorem: let $K$ be an src-field that carries a valuation $v$ such that $(K,v)$ is henselian, the characteristic of $k(v)$ is $\neq 2$ and $k(v)$ ia an odd field. Then $K$ is odd.

Proof of the assertion for fields in class 1

The square-root-closure of a finite field $\mathbb{F}_q$ with $q=p^n$ elements, $p\neq 2$.

(A) In $\mathbb{F}_q$ one has $-1\neq 1$ and $(-1)^2=1$. Thus $\mathbb{F}_q\neq\mathbb{F}_q^2$.

(B) In a fixed algebraic closure of $\mathbb{F}_q$ there is exactly one extension $F/\mathbb{F}_q$ of degree $2$, namely $\mathbb{F}_{q^2}$. Hence every $x\in\mathbb{F}_q\setminus\mathbb{F}_q^2$ has its square roots in $\mathbb{F}_{q^2}$.

This is a general property of finite fields: within a fixed algebraic closure there is exactly one extension of a given degree.

(C) The field $K:=\bigcup\limits_{k\in\mathbb{N}}\mathbb{F}_{q^{2^k}}$ is the smallest square-root-closed field containing $\mathbb{F}_{q}$.

By construction $K$ must be contained in every square-root-closed field containing $\mathbb{F}_{q}$.

(D) Every irreducible polynomial $f\in K[X]$ has odd degree.

Proof: let $x$ be a root of $f$ and choose $k$ such that $K_0:=\mathbb{F}_{q^{2^k}}$ contains all the coefficients of $f$. Then $[K(x):K]=[K_0(x):K_0]$ and $K_0(x).K=K(x)$. Now $K_0(x)/K_0$ is a cyclic extension, because every finite extension of finite fields is cyclic generated by the Frobenius map. Thus if $2$ divides $[K_0(x):K_0]$, there exists an intermediate field $K_0\subset M\subseteq K_0(x)$ such that $[M:K_0]=2$. But then $M\subset K$ and thus $[K(x):K]<[K_0(x):K_0]$.

(E) For every algebraic extension $L$ of $K$ every irreducible polynomial $f\in L[X]$ has odd degree. In particular $L$ is itself square-root-closed.

Proof: let $f\in L[X]$ be irreducible and choose a finite extension $L_0/K$ such that $L_0$ contains the coefficients of $f$. Since $L_0/K$ is separable, its degree by (D) is odd. Let $x$ be a root of $f$. Then the degree of $L_0(x)/K$ is odd too. Consequently the degree of $f$ being equal to the degree of $L_0(x)/L_0$ must be odd.

The following is worth noting: (E) is true (same proof) for every separable extension $L$ of a field $K$ for which every irreducible polynomial has odd degree..

(F) Every square-root-closed algebraic extension of $\mathbb{F}_q$ is an algebraic extension of $K$.

Facts from Valuation Theory

I use the following sources:

[E] O. Endler, Valuation Theory

[Z] P. Samuel, O. Zariski, Commutative Algebra II

For convenience I recall some basic definitions and facts:

(1) A valuation $v$ of a field $K$ is a surjective group homomorphism $v:K^\ast\rightarrow\Gamma$, where $\Gamma$ is a totally ordered abelian group (called the value group of $v$), such that $v(x+y)\geq\min (vx,vy)$.

(2) The valuation ring of $v$ is the subring $O:=\{x\in K^\ast : vx\geq 0\}\cup 0$. It is a local ring with maximal ideal $M:=\{x\in K^\ast : vx> 0\}\cup 0$. The field $k(v)=O/M$ is called the residue field of $v$.

(3) A valuation $v$ of $K$ has at least one prolongation $w$ to an extension field $L$ of $K$. The value group $\Delta$ of $w$ contains $\Gamma$ as a subgroup. The residue field $k(w)$ is an extension field of $k(v)$. If $L/K$ is algebraic, then $k(w)/k(v)$ is algebraic and $\Delta /\Gamma$ is a torsion group. [E], Chapter 2, Paragraph 9 or [Z], Chapter 6, Paragraph 4.

(4) A valuation $v$ (or the pair $(K,v)$) is called henselian, if $v$ has exactly one prolongation to any algebraic extension field of $K$.

(5) For every valuation $v$ of a field $K$ there exists an algebraic extension field $K^h$ of $K$ and a prolongation $v^h$ of $v$ to $K^h$ such that $(K^h,v^h)$ is henselian and the value groups and the residue fields of $v$ and $v^h$ coincide. $(K^h,v^h)$ is called the henselisation of $(K,v)$. [E], Chapter 3, Paragraph 17.

(6) Let $(K,v)$ be henselian. Then for every finite extension $L/K$ the following equation holds: $[L:K]=(\Delta :\Gamma )[k(w):k(v)]d(w/v)$, where $w$ is the unique prolongation of $v$ to $L$, $\Delta$ is the value group of $w$, $(\Delta :\Gamma )$ is the index of $\Gamma$ in $\Delta$ and $d(w/v)$ is a power of the characteristic of $k(v)$ (called the defect of $w$). [K], Chapter 11, Lemmas 11.1 and 11.17

(7) Let $K$ be an src-field and $v$ a valuation on $K$, then $\Gamma$ is $2$-divisible (that is $\Gamma =2\Gamma$) and $k(v)$ is an src-field.

If $(K,v)$ is henselian, $\Gamma$ is $2$-divisible and $k(v)$ is an src-field, then $K$ is an src-field.

[K], Chapter 9, Corollary 9.38

Proof of the assertion for fields in class 2 and Remarks

Proof of the Theorem: let $L/K$ be a finite extension and let $w$ be the unique prolongation of $v$ to $L$. By (7) the value group $\Gamma$ of $v$ is $2$-divisible, hence $(\Delta :\Gamma )$ is odd. Since $k(v)$ is an odd field by assumption, the degree $[k(w):k(v)]$ is odd. The defect $d(w/v)$ is odd since $k(v)$ has characteristic $p\neq 2$. Hence by (6) the degree $[L:K]$ is odd.

Proof of the assertion about class 2: by (3) the residue field $k(v)$ is an algebraic extension of a finite field. By (7) $k(v)$ is an src-field. Thus $k(v)$ is in class 1 and therefore an odd field. The theorem now yields the assertion.

Remarks: let $v_0$ be a valuation on a field $K_0$ such that $k(v_0)$ has characteristic $\neq 2$. Given an algebraic extension $k/k(v_0)$ Valuation Theory shows that there exists an algebraic extension $K/K_0$ and a prolongation $v$ of $v_0$ to $K$ such that $k\subseteq k(v)$ and the value group of $v$ is $2$-divisible.

If $K_0$ is not algebraically closed, the field $K$ can be chosen to be rather small compared to the algebraic closure of $K_0$.

Combining these facts with (5) one can prove that class 2 contains more fields than just the algebraic closure of $\mathbb{Q}$: take $K_0$ to be the rationals, $v_0$ a $p$-adic valuation with $p\neq 2$, $k$ an algebraic src-extension field of the finite field $\mathbb{F}_p$. Let $K_1$ be the field obtained by adjoining all elements of the form $p^{1/2^k}$, $k\in\mathbb{N}$ to $K_0$ and let $v_1$ be a prolongation of $v_0$ to $K_1$. Let $K_2$ be an algebraic extension of $K_1$ such that for some prolongation $v_2$ of $v_1$ to $K_2$ the inclusion $k\subseteq k(v_2)$ holds. Finally let $(K,v)$ be the henselisation (5) of $(K_1,v_1)$. Then by construction the value group of $v_1$ and thus of $v$ is $2$-divisible. Moreover $k\subseteq k(v)$, hence $k(v)$ is an src-field. Therefore by (7) $K$ is an src-field.

Another example: let $K$ be the field obtained by adjoining to the Laurent series field $\mathbb{C}((t))$ all elements of the form $t^{1/2^k}$, $k\in\mathbb{N}$. The valuetion $v$ is the unique extension of the natural discrete valuation of $\mathbb{C}((t))$ to $K$.

Lierre's contribution

The proof of the complement indeed shows that the fields in class 2 of my post are odd: let $L/K$ be a finite extension and let $M/K$ be its Galois hull. Then one has the normal series $V\subseteq T\subseteq G$ of the Galois group $G:=\mathrm{Gal}(M/K)$, where $T$ is the inertia group of the prolongation $w$ of $v$ to $M$, and $V$ is its ramification group. Ramification theory shows that $G/T$ is isomorphic to the abelian group $\mathrm{Gal}(k(w)/k(v))$, $T/V$ is abelian and $V$ is a $p$-group. Thus $G$ is solvable. Using Lierre's proof of the complement shows that $[M:K]$ and thus $[L:K]$ are odd.

This way of proving the assertion for the fields of class 2 is much more elegant than mine! ${}$

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Thank you for your post. Give me some time to look this stuff up, because I don't know the things you are writing about and I quite honestly don't understand a word of it ^^ – Olivier Bégassat May 9 '12 at 14:19
Hello Olivier. Combining the stuff collected so far, I am pretty sure that one can prove your assertion about the degree of irreducible polynomials also for a large class of algebraic, square-root-closed extensions of the rational numbers. – Hagen Knaf May 10 '12 at 7:51
An'y proof is welcome! Could you provide a little bit of extra explanation about what $\mathcal{O}_v$ is and its maximal ideal? I think I understand your example, but why is the extension $K_0(x)/K_0$ cyclic? Is it because of the Frobenius automorphism? This seems to ring a bell with me... – Olivier Bégassat May 10 '12 at 10:06
I added some clarification to my post in the part about finite fields. I will come back to the part about valuation theory, but don't know exactly how to do that, since some material has to be added. The post then might become to lengthy ... – Hagen Knaf May 10 '12 at 12:57
I reorganized my post and provided proofs and references. Still have to check Lierre's contribution, which fits with my own ideas. Your problem is really inspiring. – Hagen Knaf May 12 '12 at 0:57

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 ${\mathfrak S}_n$).

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So many good answers and now you propose the notion of a "Begassat" subgroup ^^, this is a great example of something being misattributed in maths. I need to read those answers to try to understand them. – Olivier Bégassat May 13 '12 at 11:01

I think I have a counter example. To make a long story short, a finite extension of a quadratically closed field need not be quadratically closed. However, a solvable extension of a quadratically closed field is quadratically closed, as it is proved at the end of the answer.

The counter example is $P = x^5 + 20 x - 16$, over $K$ the quadratic closure of $\Bbb Q$. The polynomial $P(x^2)$ is irreducible over $K$.

Let $L$ be the decomposition field of $P$ over $\Bbb Q$ and $M$ be the decomposition field of $Q = P(x^2)$ over $\Bbb Q$.

I claim that $KM$ is a proper quadratic extension of $KL$, and that this implies that $Q$ is irreducible over $K$.

Note that $K \mid \Bbb Q$ is a Galois extension, and so $L\cap K \mid \Bbb Q$ and $M\cap K \mid \Bbb Q$ are Galois extensions as well.

1. Description of some Galois groups over $\Bbb Q$ — The Galois group over $\Bbb Q$ of the Galois extension $L$ is $A_5$. I took an example from the Wikipedia page. The Galois group over $\Bbb Q$ of $M$ is a group of order 960, isomorphic to $C_2^4 \rtimes A_5$. This group admits only one proper normal subgroup, which has order 16. I made these computations with Sage and GAP, but I guess it should not be too hard to do it by hand.

2. $P$ is irreducible over $K$ — The Galois group $\operatorname{Gal}_K KL$ is isomorphic to $\operatorname{Gal}_{K\cap L} L$ since $L$ is a Galois extension [by Bourbaki, A V, §10, no. 8, thm. 5]. Since $L\cap K \mid \Bbb Q$ is Galois, $\operatorname{Gal}_{K\cap L} L$ is a normal subgroup of $\operatorname{Gal}_{\Bbb Q} L \simeq A_5$. Since $A_5$ is simple then $\operatorname{Gal}_K KL$ is $1$ or $A_5$. It cannot be $1$ since $L$ is not included in $K$, so it is $A_5$. In particular $\operatorname{Gal}_K KL$ acts transitively on the roots of $P$, thus $P$ is irreducible over $K$.

3. Expressing the reducibility of $Q$ over $K$ — I claim that the polynomial $Q$ is reducible over $K$ if and only if $KL = KM$. Only the left-to-right implication is needed, but I will prove both. I don't know if it is a general fact, but in this case, I have a specific proof. ❧ Assume that $KL = KM$. Let $\alpha$ be a root of $Q$. Then $K(\alpha^2)$ has dimension 5 over $K$ since $P$ is irreducible, and $[K(\alpha):K(\alpha^2)]$ is 1 or 2, thus $[K(\alpha):K]$ is 5 or 10. If it is 5 then $Q$ is not irreducible, because the minimal polynomial of $\alpha$ would be a degree 5 factor. Assume that $[K(\alpha):K]$ is 10. We have a tower of extensions $K < K(\alpha^2) < K(\alpha) < KL$ which gives a chain of subgroups $A_5 > H_1 > H_2 > 1$, with $[A_5:H_1] = 5$ and $[H_2 : H_1] = 2$. But every subgroups of $A_5$ of index 5 is isomorphic to $A_4$ which does not contain any subgroup of index $2$. ❧ Assume now that $Q$ is reducible. It splits into two factors of degree $5$. Let $\alpha$ be a root of $Q$. Since $Q$ splits we know $[K(\alpha) : K] = 5$, but since $P$ is irreducible, we have $[K(\alpha^2):K]$ is 5 as well, so $\alpha \in K(\alpha^2) \subset KL$, hence the claim.

4. $KM$ is a proper extension of $KL$ — The Galois group $\operatorname{Gal}_K KM$ is isomorphic to $\operatorname{Gal}_{K\cap M} M$, thus we have an exact sequence $$1 \longrightarrow \operatorname{Gal}_K KM \longrightarrow \operatorname{Gal}_{\Bbb Q} M \longrightarrow \operatorname{Gal}_{\Bbb Q} K\cap M \longrightarrow 1.$$ In particular $\operatorname{Gal}_K KM$ is a normal subgroup of $\operatorname{Gal}_{\Bbb Q} M$, containing $A_5$ as a subgroup. Studying the structure of $\operatorname{Gal}_{\Bbb Q} M$ shows that this normal suggroup is the whole group $\operatorname{Gal}_{\Bbb Q} M$ which has order 960. In particular $KM$ is a proper extension of $KL$, and hence $Q$ is irreducible over $K$. Since $M$ is spanned by the roots of $Q$, we have $KM = KL$

Complement : A solvable extension of $K$ is quadratically closed — Since a solvable extension is splits into a tower of cyclic extension, it is enough to consider a cyclic extension $L$ of $K$. Let $M$ be a quadratic extension of $L$. We have an exact sequence $$1 \longrightarrow \operatorname{Gal}_L M \longrightarrow \operatorname{Gal}_K M \longrightarrow \operatorname{Gal}_K L \longrightarrow 1$$ which gives an isomorphism $\operatorname{Gal}_K M \simeq C_2 \rtimes C_n$. But such a semi-direct product is always a direct product. In particular, $\operatorname{Gal}_K M$ has $C_n$ as a normal subgroup of index 2. The Galois connection then gives a quadratic extension of $K$, which is a contradiction. I think that this result contains the results of Hagen.

Note — This counter-example gives also a counter example for the question linked by Olivier : If $K=K^2$ then every automorphism of $\mbox{Aut}_K V$, where $\dim V< \infty$, is the square of some endomorphism.

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Salut Pierre :D you use this fact twice: why is the subgroup $Gal(L/L\cap K)$ a normal subgroup of $Gal(L/\mathbb{Q})$? I.e. ( I believe) why is $L\cap K/\mathbb{Q}$ Galois? – Olivier Bégassat May 11 '12 at 10:26
Because L and K are. Is it OK ? – Lierre May 11 '12 at 10:56
Non, je ne me sers plus de celle de l'ENS. J'utilise olivier (point) begassat (point) cours (arobase) gmail (point) com – Olivier Bégassat May 13 '12 at 13:06