Show that $\mathbb{Z}[x]/(x)$ is isomorphic to $\mathbb{Z}$ Let $\mathbb{Z}[x]$ be a ring of polynomials with integer coefficients, $(x)$ be an ideal generated by $x$.
Show that $\mathbb{Z}[x]/_{(x)}$  is isomorphic to $\mathbb{Z}$.
My attempt:
For each polynomial $$z(x) = a_nx^n + \ ...\ +a_1x + a_0$$ 
We have
$$z(x) = x\cdot(a_nx^{n-1}+ \ ... \ +a_1)\ + a_0$$.
Hence if we define $\Phi : \mathbb{Z}[x]/_{(x)} \rightarrow \mathbb{Z}$ as:
$$\Phi(a_nx^n +\ ... \ +a_1x + a_0 +(x)) =\Phi((a_nx^{n-1} +\ ... \ +a_1)x + a_0 +(x)) = \Phi(a_0 + (x)) = a_0$$.
My first question is about notation and equivalence class. How do I write that two polynomials $w(x), z(x)$ are equivalent if their $a_0, b_0$ coefficients are equal?
I will try to show that such $\Phi$ is homomorphism.
For two polynomials $z(x), w(x)$ with coefficients $a_i, b_i$ respectively we have:
$\Phi((z+(x)) + (w+(x))) = \Phi((z+w)+(x)) = \Phi((a_0+b_0)+(x)) = a_0 + b_0 = \Phi(z+(x)) + \Phi(w+(x))$.
And from now I have trouble to show that there is unique element in $\mathbb{Z}[x]/_{(x)}$ that corresponds to an element of $\mathbb{Z}$.
Can it be fixed? My main concern is how to properly write the equivalence relation and how do I proceed with my proof further. Would be grateful for any hints.
Edit:
I will try the other way around.
Define $$\psi : \mathbb{Z} \rightarrow \mathbb{Z}[x]/_{(x)}$$ as follows
$$\psi(a) = a + (x)$$
Showing that $\psi$ is homomorphism
$$\psi(a) + \psi(b) = (a+(x)) + (b+(x)) = (a+b) + (x) = \psi(a+b)$$
Suppose that $\psi(a) = \psi(b)$
We have
$$a + (x) = b + (x)$$
Which leads to $(a-b) \in (x)$ how do I conclude that $a-b = 0$? Since I am missing something.
Moreover, I am struggling with showing that for every element in quotient space there is some number in $\mathbb{Z}$.
Any hints?
 A: Comment on your attempt:
You may find it easier (see "alternate method" below) to try to define the homomorphism the other way around -- from $\mathbb{Z}$ to $\mathbb{Z}[x] / (x)$. The problem is that the way you are doing, it is kind of confusing to work with $\Phi$ because it really is a function on equivalence classes. It can be done, but requires that you show $\Phi$ is first well-defined. Normally you would do this the following way:


*

*First define $\Phi$ on a particular element, not on equivalence classes (i.e. define $\Phi(a_0 + a_1 x + \cdots + a_n x^n) := a_0$)

*Next show this is a well-defined operation on equivalence classes -- to do this you should consider two equivalent polynomials in $\mathbb{Z}[x] / (x)$, and show that $\Phi$ is the same on those two polynomials.
Once you have that $\Phi$ is well defined, the rest of the proof is easier: any time you want to evaluate $\Phi$, it is enough to evaluate it on some member of the equivalence class, so you don't have to worry about having "$+ (x)$" everywhere in your proof.
Alternate method
Alternatively instead of $\Phi$ you can try defining
$$
\psi: \mathbb{Z} \to \mathbb{Z} [x] / (x).
$$
Then you don't have to show $\psi$ is well-defined. Instead, you just have to show that


*

*It is a homomorphism

*It is one-to-one: this amounts to showing that if $\psi(a)$ and $\psi(b)$ are equivalent (differ by something in $(x)$), then $a = b$.

*It is onto: this amounts to fixing some polynomial $a_0 + a_1 x + \cdots + a_n x^n$, and showing it is equivalent to something that $\psi$ produces as output.
