Let $R$ be a ring. Show that the map $\theta : R[x] \rightarrow R$ given by $\theta(a_0+a_1 x+\cdots+a_nx^n) = a_0$ is a ring homomorphism. Show that $ker (\theta)$ is the principal ideal $(x)$. Here $R[x] = \{a_0+a_1 x+\cdots+a_nx^n \mid a_i \in R\}$, the ring of polynomials, and $(x) = \{ax\mid a\in R\}= xR[x]$, the ideal generated by $x$.
I already showed that this was a ring homomorphism, but I don't know how to show that $\ker(\theta) = (x)$. Obviously
$\ker(\theta) = \{a \in R[x] \mid \theta(a) = 0\} =\{a_0+a_1 x + \cdots + a_nx^n \mid a_0 = 0, a_i \in R, i\neq0\}$
and
$$(x) = \{ax\mid a\in R[x]\}= xR[x].$$
One thing that I was confused about was if $R[x]$ are polynomials that are upto degree $n$ wouldn't $(x)$ contain polynomials upto degree $n+1$? I could see that the first term in each elements of these sets are have the same deree since for any $a = a_0+a_1 x+\cdots+a_nx^n$, $x(a_0+\cdots+a_nx^n) = a_0x+\cdots+a_nx^{n+1}$. If the degrees were the same, I was just going to say that since they contain the exact elements $\ker(\theta) = (x)$, but I'm not sure that'd be enough. What would be a good strategy to show this?
It's cleary that $$(x) \subset ker(\theta)$$, conversely if $$f \in ker(\theta)$$ write $$f=q.x+a_0$$ with $$a_0 \in R$$ thus $$\theta(f)=a_0=0$$ thus $$f=qx$$.