Showing that $\mathbb{Z}[x]/_{(x)}$ is isomorphic to $\mathbb{Z}$, where $(x)$ is an ideal generated by $x$.
My attempt:
I will try to show that $\psi : \mathbb{Z} \rightarrow \mathbb{Z}[x]/_{(x)}$ defined below is indeed isomorphism:
$$\psi(a) := a + (x)$$
1) Being homomorphism
$$\psi(a) + \psi(b) = (a + (x)) + (b+(x)) = (a+b) + (x) = \psi(a+b)$$
2) Being injecitve
Suppose that $\psi(a) = \psi(b) \Rightarrow (a-b) \in (x)$, but the only constant in $(x)$ is $0$ hence $a=b$
Is it correct up to this point?
3) Being surjective
I have trouble with this point.
Any hints would be great.
Also I am posting possible duplicate, a post of mine but with different approach:
Different method
Best Answer
Your proof is fine up to this point. To prove it is surjective, let $$p(x)=\sum_{i=0}^n a_ix^i$$ Then this is equal to $$a_0+x\sum_{i=1}^na_ix^{i-1}$$ But $x\displaystyle\sum_{i=1}^na_ix^{i-1}\in (x)$. Since this polynomial was arbitrary, all polynomials are of the form $a+xf(x)$ for some polynomial $f(x)$, hence all elements in $\mathbb Z[x]/(x)$ are of the form $a+(x)$ for $a\in\mathbb Z$.