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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?

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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.

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  • $\begingroup$ 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? $\endgroup$
    – janusz
    May 13 '20 at 16:09
  • $\begingroup$ @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 $\endgroup$
    – 6005
    May 13 '20 at 16:56
  • $\begingroup$ 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. $\endgroup$
    – janusz
    May 13 '20 at 17:03
  • $\begingroup$ 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? $\endgroup$
    – janusz
    May 13 '20 at 17:13
  • $\begingroup$ @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. $\endgroup$
    – 6005
    May 13 '20 at 17:33

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