# Prove that$f(\varphi ( \overline{x}))=0$

Let $f(x)$ be a polynomial in $\mathbb{Z}[x]$, and let $\langle f(x)\rangle$ denote the ideal it generates. Let $R$ be a ring. Define a ring homomorphism $\varphi:\mathbb{Z}[x]/\langle f(x)\rangle\rightarrow R$.

Prove that $f(\varphi(\overline{x}))=0$, where $\overline{x}$ is just an abbreviation for the coset $x+\langle f(x)\rangle$.

Any hints will be appreciated. Thanks.

• I have an answer. Do you want that, or a hint? – feralin Feb 25 '17 at 3:58
• @feralin Thanks so much. It would be good if you could just provide the answer. – PropositionX Feb 25 '17 at 4:01

Write $f(x) = a_0 + a_1x + a_2 x^2 + \cdots + a_nx^n$. Now let $\pi : \mathbb Z[x] \to \mathbb Z[x] / \langle f\rangle$ be the projection (ring) homomorphism $\pi(p) = p + \langle f\rangle$, so that we want to prove $$f(\varphi(\pi(x))) = 0.$$ This is quite simple, by using the ring homomorphism properties $$\varphi(\pi(p + q)) = \varphi(\pi(p)) + \varphi(\pi(q))$$ $$\varphi(\pi(pq)) = \varphi(\pi(p))\varphi(\pi(q))$$ $$\varphi(\pi(1_{\mathbb Z[x]})) = 1_R$$ for any $p, q \in \mathbb Z[x]$. Note that this last property, along with the additivity property, gives us (by induction) that $$\varphi(\pi(n)) = \varphi(\pi(n \cdot 1_{\mathbb Z[x]})) = n \cdot \varphi(\pi(1_{\mathbb Z[x]})) = n \cdot 1_R = n,$$ where in any ring $R$ we implicitly write $1 = 1_R$, $2 = 1_R + 1_R$, $3 = 1_R + 1_R + 1_R$, $-1 = -1_R$, etc.
Now we have \begin{align*} f(\varphi(\pi(x)) &= a_0 + a_1 \varphi(\pi(x)) + a_2 \varphi(\pi(x))^2 + \cdots + a_n \varphi(\pi(x))^n \\ &= \varphi(\pi(a_0)) + \varphi(\pi(a_1))\varphi(\pi(x)) + \varphi(\pi(a_2))\varphi(\pi(x^2)) + \cdots + \varphi(\pi(a_n))\varphi(\pi(x^n)) \\ &= \varphi(\pi(a_0 + a_1 x + a_2 x^2 + \cdots + a_nx^n)) \\ &= \varphi(\pi(f)) \\ &= \varphi(0_{\mathbb Z[x]/\langle f\rangle}) \\ &= 0_R. \end{align*}
• Why is it the case that $a_i = φ(π(a_i))$ – PropositionX Feb 25 '17 at 4:12