No ring homomorphisms $f: R[x]/ \langle p(x) \rangle \to K$ if $p(x)=0$ has no solutions in K. I encountered this claim here. In general it would read:
Let $R$ and $K$ be commutative rings with unity. There are no homomorphisms $f: R[x]/ \langle p(x) \rangle \to K$ if $p(x)=0$ has no solutions in $K$.
This is probably very basic question but I'm still not sure if I understand correctly why the above works. Here's my thought process:
Let's denote $p(x) = \underset{k=0}{\overset{n}{\sum}} a_kx^k$. It is clear to me that since $f$ is a ring homomorphism we have $f(0_{{}_R} + \langle p(x) \rangle) = 0_{{}_K}$. I can also see that $f(p(x)) = \underset{k=0}{\overset{n}{\sum}} f(a_kx^k) = 0$ yields the desired result if $f$ is a function from $R[x]$.
But $f$ is a function from $R[x]/ \langle p(x) \rangle$, so I'm not sure why it would be legal to 'invade' the coset $(0_{{}_R} + \langle p(x) \rangle)$ with this function.
 A: I think you over-generalized. The statement "$p(x)=0$ has no solutions in $K$" makes no sense without further assumptions, since $p(x)$ has coefficients in $R$ and there is no multiplication between elements from $R$ and $K$.
If $K$ however is an $R$-algebra, which means that $R$ comes equipped with a ring homomorphism $\varphi:R\to K$ then the statement makes sense, since you can use $\varphi$ to transform the coefficients of $p(x)$ from $R$ to $K$.
The correct statement would then be:

There are no $R$-algebra homomorphisms $ R[x]/ \langle p(x) \rangle \to K$ if $p(x)=0$ has no solutions in $K$.

This is because if $f: R[x]/ \langle p(x) \rangle \to K$ is an $R$-algebra homomorphism, then for $s=f([x])$
$$
0=f([0])
=f([p(x)])
=f([\underset{k=0}{\overset{n}{\sum}} a_kx^k])
=\underset{k=0}{\overset{n}{\sum}} f([a_k])[x]^k])
=\underset{k=0}{\overset{n}{\sum}} \varphi (a_k)s^k=:p(s)
$$
The stronger requirement that $f$ is a homomorphism of $R$-algebras ensures that $f([r])=\varphi(r)$ for all $r\in R$.
In the linked question $R=\mathbb Z$, $K=\mathbb Z_n$ and in this case any ring homomorphism $\mathbb Z[x]/\langle p(x)\rangle\to \mathbb Z_n$ is automatically a $\mathbb Z$-algebra homomorphism.
