I’d like to show from a ring theoretic approach that any polynomial of odd integer coefficients and even degree has no rational roots. This exercise came from Joseph Gallian’s 9th edition of Contemporary Abstract Algebra. The textbook solution started by assuming such a rational root $\frac{a}{b}$ exists where $\mathrm{gcd}(a,b)=1$ and then plugging it into the polynomial and working it out from there.
Rather than that, I want to use content presented in the section on polynomial rings, which is also where the exercise came from. The following is my attempt to use homomorphisms, polynomial rings, and a certain ideal to complete the exercise.
Let $f(x)=a_mx^m + a_{m-1}x^{m-1} + \cdots + a_0$, where the $a_i \in 2Z+1$ and $m\in 2Z$. Note that the sum of the coefficients is even, or equivalently, $\sum_i a_i \in 2Z$. This will be relevant in a bit.
I want to consider the homomorphism $\psi:Z[x]\to (Z/2Z)[x]$ such that $c_sx^s+c_{s-1}x^{s-1}+ \cdots + c_0 \mapsto (c_s + 2Z)x^s + (c_{s-1}+2Z)x^{s-1} + \cdots (c_0 + 2Z)$.
I also want to consider the ideal $I$ of $(Z/2Z)[x]$ such that $I=\{f\in (Z/2Z)[x] | \sum_j (\alpha_j + 2Z) = 2Z\}$. I got this from the fact that for any commutative ring $R$, $J=\{a_rx^r+\cdots + a_0 \in R[x] | \sum_k a_k = 0\}$ is an ideal in $R[x]$. Still in $R[x]$, $(x-1)$ is a factor of any element in $I$. So in $(Z/2Z)[x]$, it follows $(x-1+2Z)=(x+1+2Z)$ is a factor of any $\psi(f)$ where $f$ is the original polynomial described above.
Now I’m not sure how to turn it back around in the direction of showing some sort of statement contradicting $f$ having any rational roots. My thought process behind this was that I would turn the problem from one about polynomials in $Z[x]$ into another (ideally easier) about quotient ring polynomials - especially after seeing that the coefficient sum was even at first.
Any hints, suggestions, and/or remarks would be appreciated.