# 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?

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.

• Thank you I will try your method. But the question is how do I find a form of $\psi : \mathbb{Z} \rightarrow \mathbb{Z}[x]/_{(x)}$? I mean does it have an exact form? May 13 '20 at 16:09
• @janusz Basically, you just define $\psi: \mathbb{Z} \to \mathbb{Z}[x]$; you don't need to worry about the $/ (x)$ until you are talking about the one-to-one and onto parts. This is because the "up to equivalence mod $x$" doesn't matter for defining the function
– 6005
May 13 '20 at 16:56
• To be honest I don't how to proceed right know. Is there an exact form of such $\psi$? Without it I'm not able to operate. May 13 '20 at 17:03
• I was thinking about defining $\psi(a) := a + (x)$, showing that this is indeed homomorphism is fine, showing that this is injective I think wouldn't be a problem since only consant element of $(x)$ is $0$, so it will result in $a=b$. But what about being surjective? May 13 '20 at 17:13
• @janusz It is clearer to think of polynomials "up to equivalence mod $x$" rather than equivalence classes like $a + (x)$. That avoids dealing with set arithmetic everywhere like $(x) + (x)$. So I would just define $\psi(a) := a$. To show surjectiveness, you need to show that for any polynomial in $\mathbb{Z}[x]$, there is an equivalent polynomial that $\psi$ maps to.
– 6005
May 13 '20 at 17:33