The problem is to find the kernel of the homomorphism $\mathbb{Z}[x] \to \mathbb{Q}$ given by $f \mapsto f(1/3)$. I think I've solved this using elementary techniques but I want to make sure the solution works.

Claim: The kernel is the principal ideal $(3x - 1)$.

Proof: Clearly $(3x - 1)$ is in the kernel. Conversely, suppose $f(1/3) = 0$. Then $x - 1/3$ divides $f$ in $\mathbb{Q}[x]$ so we may write $f = (x - 1/3)(C_0 + C_1x + \cdots C_n x^n)$ for $C_n \in \mathbb{Q}$. To conclude, we show that all the $C_i \in 3\mathbb{Z}$. Writing out the product, we see that $$f(x) = -\frac{1}{3}C_0 + \left(C_0 - \frac{1}{3}C_1 \right)x + \cdots + \left(C_{n - 1} - \frac{1}{3} C_n\right) x^n + C_nx^{n + 1}.$$ Since $f \in \mathbb{Z}[x]$, the coefficients are all integers so that $C_0 \in 3\mathbb{Z}$ and $$C_i - \frac{1}{3} C_{i + 1} \in \mathbb{Z}$$ for all $0 \leq i \leq n - 1$. Now, if $C_i \in 3\mathbb{Z}$ then $C_{i + 1} \in 3\mathbb{Z}$ by the above identity. We can then apply induction since $C_0 \in 3\mathbb{Z}$ to see that $C_i \in 3\mathbb{Z}$ for all $0 \leq i \leq n$ as required.


  • 1
    $\begingroup$ looks good to me! $\endgroup$ Commented Jul 5, 2022 at 17:53
  • 3
    $\begingroup$ Nicely done. I'd explicitly mention the final step that $$ f(x) = (3x-1)\left(\frac{C_0}{3} + \frac{C_1}{3} x + \cdots + \frac{C_n}{3} x^n\right) $$ where the fact $C_i \in 3 \mathbb{Z}$ means that the second polynomial is an element of $\mathbb{Z}[x]$. $\endgroup$
    – aschepler
    Commented Jul 5, 2022 at 18:22


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