Let $F$ be a field, and $p(x)\in \mathbb{Z}[x]$ an irreducible polynomial. We want to show that $$\phi: F \longrightarrow F[x]/ \langle p(x) \rangle,$$ $$c\mapsto \phi (c):=c+\langle p(x) \rangle$$ is a field monomorphism.
Proof:
- We have that $\phi(c_1+c_2)=(c_1+c_2)+\langle p(x) \rangle = (c_1+\langle p(x) \rangle) + (c_2+\langle p(x) \rangle)=\phi(c_1)+\phi(c_2)$. Working by this way, $\phi(c_1\cdot c_2)=\cdots =\phi(c_1)\cdot \phi (c_2)$.
- If $\phi(c_1)=\phi(c_2) \iff c_1+\langle p(x) \rangle=c_2+\langle p(x) \rangle \iff (c_1-c_2)+\langle p(x) \rangle=\langle p(x) \rangle \iff c_1-c_2 \in \langle p(x) \rangle \iff \exists\ q(x)\in F[x] : c_1-c_2 =q(x)\cdot p(x)$. But now, if we take degrees from both sides, we take that $$\deg (c_1-c_2)=\deg q(x)+\deg p(x),$$ where $\deg (c_1-c_2)=0$, if $c_1\neq c_2$ or $\deg (c_1-c_2)=-\infty$, if $c_1=c_2$. From the other side, $\deg q(x)\in \mathbb{N}\cup\{-\infty \}$.
My question, now, is: Can from this conclude that obligatory $\deg (c_1-c_2)=-\infty \implies c_1=c_2$, so $\phi$ is injective, and finally monomorphism?
Thank you in advance.