Exercise 15.4.11

Proof. Let $d=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(A)$, and $f=\det <!--\; FUNCTION\; APPLICATION\; -->(B)$, so that

Then the hypothesis $\frac{B(1∕u)}{A(1∕u)}=\frac{A(u)}{B(u)}$ gives

where

therefore

Note that $a_0\ne 0$ or $b_0\ne 0$, otherwise $A(u)$ and $B(u)$ have $u$ as common factor.

Suppose that $a_0\ne 0$ (since the problem is symmetric about $A,B$, the other case $b_0\ne 0$ is similar). If we develop the preceding equality by decreasing powers, we obtain

where $a_0{a}_d\ne 0$ (but perhaps $b_0{b}_f=0$).

Then $a_0{a}_d{u}^{2d+f}$ is a nonzero term of the right member, of degree less or equal to $d+2f$, thus $2d+f\le d+2f$. Moreover $a_0{a}_d{u}^f$ is a nonzero term of the right member, with valuation greater or equal to $d$, thus $f\ge d$. The inequalities $2d+f\ge d+2f,f\ge d$ imply $d\ge f,f\ge d$, so that $d=f$:

We can rewrite, after simplification by $u^d$, the above polynomial equality under the form

where

If we define $Ã(u)=u^{d}A(1∕u)=a_d+\cdots +a_0{u}^d,\tilde{B}(u)=u^{d}B(1∕u)=b_d+\cdots +b_0{u}^f\in K[u]$, then

Therefore $B(u)$ divides $A(u)Ã(u)$ in $K[u]$, where $B(u),A(u)$ are relatively prime, therefore $B(u)$ divides $Ã(u)$.

Moreover $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(Ã(u))\le d=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(B(u))$, and $Ã(u)\ne 0$, otherwise $A(u)=0$, thus $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(Ã(u))=d$. Therefore there is some $\lambda \in K^{\text{∗}}$ such that $Ã(u)=\mathit{\lambda B}(u)$, that is

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