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A polynomial $f(X)$ is said to be reciprocal if whenever $alpha$ is a root, then $1/alpha$ is also a root. We suppose that $f$ has coefficients in a real subfield $k$ of the complex numbers. If $f$ is irreducible over $k$, and has a nonreal root of absolute value $1$, show that $f$ is reciprocal of even degree.

Proof:
Let $f(X)in k[X]$ with $textdegf(X)=ngeq 1$. Since $beta in mathbbC-mathbbR$ and $f(beta)=0$ then $fleft(overlinebetaright)=0$ and $betaneq overlinebeta$. It means that $(X-beta)left(X-overlinebetaright)mid f(X)$, but $(X-beta)left(X-overlinebetaright)=X^2-left(beta+overlinebetaright)X+1$ because $|beta|=1$.

The following cases are possible:

1. If $textdeg f(X)=2$ then $f(X)=cleft(X^2-left(beta+overlinebetaright)X+1right)$ for some $cin k$. Then we see that $f$ has even degree and is reciprocal because its roots are $beta$ and $overlinebeta$.

2. If $textdeg f(X)>2$ then $beta+overlinebetanotin k$ because $f(X)$ is irreducible over $k$. Consider an extension $kleft(beta+overlinebetaright)$ then we see that $kleq kleft(beta+overlinebetaright)leq k(beta)$. Using multiplicativity of towers and taking into account that $[k(beta):k]=n$ then $$n=left[k(beta):kleft(beta+overlinebetaright)right]left[kleft(beta+overlinebetaright):kright].$$

We can show that $k(beta)=kleft(beta+overlinebetaright)(beta)$. Consider the polynomial $X^2-left(beta+overlinebetaright)X+1in kleft(beta+overlinebetaright)[X]$ and we see that $beta$ its root. Hence $left[k(beta):kleft(beta+overlinebetaright)right]leq 2$.

If this value was equal to $1$ then $k(beta)=kleft(beta+overlinebetaright)$ i.e. $beta in kleft(beta+overlinebetaright)$ but $beta$ is nonreal complex and $kleft(beta+overlinebetaright)$ is real subfield. Hence $left[k(beta):kleft(beta+overlinebetaright)right]=2$.

Therefore, we have shown that $n$ is divisible by $2$, i.e. $n=2m$.

Let's show that $f(X)$ is reciprocal polynomial.

Since $beta$ has absolute value $1$ then $k(beta)=kleft(beta^-1right)$ and $[k(beta):k]=left[kleft(beta^-1right):kright]=textdeg f(X)=2m$.

Define the new polynomial $hatf(X)=X^2nf(1/X)in k[X].$ Then I can show that $hatf$ is irreducible over $k$.

Also note that $hatf(beta)=beta^2nf(1/beta)=beta^2nfleft(overlinebetaright)=beta^2noverlinef(beta)=0.$

Thus $f,hatfin k[X]$ are both irreducible polynomials over $k$ with common root $beta$. Since minimal polynomial is unique then $f(X)=hatf(X)$, i.e. $$f(X)=X^2nf(1/X).$$
And from here its easy to deduce that if $alpha$ root of $f$ then $1/alpha$ is also root of $f$.

Remark: When I read this problem this seemed to me very interesting and I solved it but spent couple hours for its solution.

Let me ask you two questions:

1) Is the solution correct?

2) When I was proving that $f(X)$ is reciprocal where did use that $f(X)$ has even degree? I was not able to figure it out by myself.

Your solution looks just fine to me.

[Edit: In hindsight, I missed a critical error! For my part, I should've thought a bit more carefully about what $hat f$ actually looked like, and not taken you at your word that you were able to prove it to be irreducible. However, even if $hat f$ had been irreducible, there would still be a minor issue, which I missed.

Thus $f,hatfin k[X]$ are both irreducible polynomials over $k$ with common root $beta$. Since minimal polynomial is unique then $f(X)=hatf(X)$, i.e. $$f(X)=X^2nf(1/X).$$

Rather, we would only be able to conclude from this that $hat f$ is some non-zero constant multiple of $f.$ That would've still been enough to prove that $f$ is reciprocal, but since $hat f$ isn't irreducible, we need to work a bit harder.
More below.]

You didn't use the fact that $f$ has even degree to show that $f$ is reciprocal; that was just another thing you had to prove.

In general, reciprocal polynomials need not have even degree--after all, $pm 1$ can be roots, too, so it will have odd degree if only one of those two is a root of odd multiplicity. In this case, though, $f$ was irreducible, so neither of these was a root.

Added: As it turns out, $hat f$ is not equal to $f.$ To see this, let's consider what we know about $f.$

Since $fin k[X]$ is irreducible, of degree $n,$ and has $beta$ as a root, then there is a non-zero $ain k$ such that $f(X)=acdot g(X),$ where $g$ is the minimal polynomial of $beta$ over $k.$ Without loss of generality, we can assume that $f=g,$ so that $a=1$. (Do you see why we can assume this?) Note in particular that $f$ is monic.

Now, we know that $$f(X)=sum_j=0^na_jX^j,tag$star$$$ where each $a_jin k,$ and $a_n=1,$ in particular. Moreover, since $f$ is irreducible, we must also have $a_0ne 0.$ (Do you see why?)

Thus, $$fleft(X^-1right)=sum_j=0^na_jX^-j$$ $$fleft(X^-1right)=X^-nsum_j=0^na_jX^n-j$$ $$fleft(X^-1right)=X^-nsum_j=0a_n-jX^j,tag1$$ so since $a_0ne 0,$ then $hat f$ would have degree $2n,$ and so can't be equal to $f$! Fortunately, this isn't a problem.

By $(1)$, we can see that $$hat f(X)=X^nsum_j=0a_n-jX^j,tag2$$ and you're quite right that $hat f(beta)=0.$ Since $hat f$ is a polynomial with $k$-coefficients having $beta$ as a root, then the minimal polynomial of $beta$ over $k$ divides $hat f$ in $k[X].$ But $f$ is that minimal polynomial! Thus, there is a polynomial $h$ with $k$-coefficients such that $$hat f(X)=f(X)h(X),$$ so by $(2)$ we have $$X^nsum_j=0a_n-jX^j=f(X)h(X).tag3$$

Now, given any root $alpha$ of $f,$ we have that $alphane 0$ since $f$ is irreducible over $k,$ and so $alpha^nne 0$ since $k$ is a real subfield. Since $$alpha^nsum_j=0a_n-jalpha^j=f(alpha)h(alpha)=0cdot h(alpha)=0,$$ then $$sum_j=0a_n-jalpha^j=0,$$ so $$fleft(alpha^-1right)=alpha^-nsum_j=0a_n-jalpha^j=0,$$ meaning that $alpha^-1$ is a root of $f.$ Since $alpha$ was an arbitrary root of $f,$ then $f$ is reciprocal, as was to be shown.

Bonus Content: It turns out that we can go a little further, still! Note that $0$ is a root of multiplicity at least $n$ of the left-hand side of $(3),$ so since $0$ isn't a root of $f,$ then $0$ is a root of $h$ of the same multiplicity. However, the left-hand side of $(3)$ has degree $2n,$ so since $f$ has degree $n,$ then $h$ has degree $n,$ as well, whence $h=bX^n$ for some non-zero $bin k.$ Thus, it follows by $(3)$ that $$sum_j=0a_n-jX^j=bcdot f(X),$$ so by $(star),$ we see that $$sum_j=0a_n-jX^j=sum_j=0^nba_jX^j,$$ so that $a_n-j=ba_j$ for each $j.$ In particular, $$a_n=a_n-0=ba_0=ba_n-n=b^2a_n,$$ so $b^2=1,$ and so either $b=1$ or $b=-1.$

I claim that $b=1.$ We prove this by way of contradiction, supposing instead that $b=-1.$ By this assumption, $a_n-j=-a_j$ for each $j.$ In particular, recalling that $n=2m,$ we have $$a_m=a_n-m=-a_m,$$ whence $a_m=0.$ Thus, we have by $(star)$ that begineqnarrayf(X) &=& sum_j=0^na_jX^j\ &=& sum_j=0^m-1a_jX^j+sum_j=m+1^na_jX^j\ &=& sum_j=0^m-1a_jX^j+sum_j=0^m-1a_n-jX^n-j\ &=& sum_j=0^m-1a_jX^j-sum_j=0^m-1a_jX^n-j\ &=& sum_j=0^m-1a_jleft(X^j-X^n-jright)\ &=& sum_j=0^m-1a_jleft(X^j-x^2mX^-jright).endeqnarray Consequently, we have that $$f(-1)=sum_j=0^m-1a_jleft((-1)^j-(-1)^2m(-1)^-jright)=sum_j=0^m-1a_jleft((-1)^j-(-1)^-jright)=0.$$ Since $-1in k$ and $f$ is monic and irreducible over $k,$ then $f(X)=x-1,$ which is impossible, since $f$ has even degree.

Thus, $b=1,$ meaning that $a_n-j=a_j$ for each $j,$ so that $f$ is a palindromic polynomial, and $$f(X)=sum_j=0a_n-jX^j.$$ Equivalently, by $(1),$ we have $F(X)=X^nfleft(X^-1right).$$

The only other adjustment I can think to make to your proof, for simplicity's sake, is that instead of $hat f,$ you consider the function $$F(X):=X^nfleft(X^-1right)=sum_j=0^na_n-jX^j.$$ We can show that this is an element of $k[X]$ of degree $n,$ and that $$F(beta)=beta^noverlinef(beta)=0,$$ whence there is a non-zero scalar $cin k$ such that $F(X)=cf(X).$ Then for any root $alpha$ of $f,$ we have $$sum_j=0^na_n-jalpha^j=F(alpha)=cf(alpha)=0,$$ whence $fleft(alpha^-1right)=0$ as above.

This also lets us prove that $F(X)=f(X)$ even more directly.