I have the following here:
Let $R=\text{Map}(\mathbb{R},\mathbb{R})$ with addition and multiplication defined, as usual, by $(f+g)(x)=f(x)+g(x)$ and $(f \cdot g)(x)=f(x)g(x)$. Let $S$ be the ring of sequences $(a_n)_{n\geq0}$ with entries $a_n \in \mathbb{R}$, and define:
$$\varphi: R \rightarrow S$$ $$f \mapsto (f(n))_n$$
a) Show that $\varphi$ is a ring homomorphism.
b) Use the First Isomorphism Theorem to show that there is a well-defined ring isomorphism $\overline{\varphi}:R/I \rightarrow S$ satisfying $\overline{\varphi}(f+i)=(f(n))_n$ for all $f\in R$, where $I=\{f\in R|f(n)=0 \text{ for all } n\in N\}$.
c)Let $1\in R$ be the constant function $\mathbf{1}:x \mapsto 1.$ Find a non-constant function $f\in R$ such that $f+I=\mathbf{1}+I$.
a) I got this. It was fine.
b) I know the first isomorphism theorem implies that:
$\frac{R}{\text{ker}(\varphi)}\rightarrow\text{Image}(\varphi)$
$a+\text{ker}(\varphi) \mapsto\varphi(a)$
I'm not really sure how to use this though. It seems almost too simple. I know here the image is $S$ and the kernel is $I$. So does that mean I just need to write:
$\frac{R}{I} \rightarrow S$
$f+I \mapsto (f(n))_n$ ?
I'm not 100% sure about this part. It seems too simple.
I am not sure if I am understanding how to actually "write" the first isomorphism theorem correctly.
c) I know that $f+I=\mathbf{1}+I$ implies $f-\mathbf{1} \in I$ so could I not just pick $f=f(n)-1$ as my function? Again, I'm not 100% sure about this part.