# Find the kernel of $\mathbb{Z}[x] \to \mathbb{Q}$ given by $f \mapsto f(1/3)$.

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.

Thanks!

• looks good to me! Commented Jul 5, 2022 at 17:53
• 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 $\mathbb{Z}[x]$. Commented Jul 5, 2022 at 18:22