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For example $\mathbb{Z}[x]/(2x)$. The ideal $(2x)$ contains polynomial with even coefficients and without the zero-degree coefficient.

I don't know how to imagine the classes of this kind of quotient ring.

I think that the classes must be represented by all the elements of $\mathbb{Z}[x]$ with zero-degree, taken in (mod2), that is $\mathbb{Z}_2$, but I think there are some errors in my reasoning because I can't solve an exercise that asks:

Let $R=\mathbb{Z}[x]/(2x)$ and let $I$ be the ideal of $R$ generated by $1+1=2$.

a) Prove that $R/I$ is a domain.

Why is the generator written like that? I think it's a tip.

The answer is that $R/I$ is isomorphic to the $\mathbb{Z}_2[x]$ domain. How is it possible if $R$ is isomorphic to $\mathbb{Z}_2$?

I'm doing something wrong. Thank you!

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    $\begingroup$ How do you imagine any ring? :-/ $\endgroup$ Feb 1 '17 at 18:51
  • $\begingroup$ I imagine something, seems like the concept has a shape, but you're right, if I have to describe it, even to myself, it becomes difficult or impossible! I think that "imagine" was a wrong verb in this case, my english is terrible! $\endgroup$
    – pter26
    Feb 2 '17 at 0:19
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You're wrong in thinking that $\mathbb{Z}[x]/(2x)$ is isomorphic to $\mathbb{Z}_2$ (the two element ring).

You can imagine the quotient ring $R=\mathbb{Z}[x]/(2x)$ as formed by “mixed polynomials” of the form $a_0+a_1x+\dots+a_nx^n$ where $a_0\in\mathbb{Z}$ and $a_1,\dots,a_n\in\mathbb{Z}_2$. The addition is done by reducing alike terms, using standard addition for the constant term and addition in $\mathbb{Z}_2$ on the other terms. For multiplication, the constant term will act on the residue classes modulo $2$ in the obvious way: if $a_0$ is even, the product is zero, if it is odd, then it doesn't change the term. So, for instance, \begin{align} (3+[1]x+[1]x^3)(-2+[1]x^2)&=-6+[1]x^2+[0]x+x^3+[0]x^3+[1]x^5\\ &=-6+[1]x^2+[1]x^3+[1]x^5 \end{align} (terms with zero coefficient are omitted). So you see that the ring is infinite.

Now, the constant “mixed polynomial” $2$ has quite an easy action by multiplication: $2(b_0+b_1x+\dots+b_nx^n)=2b_0$.

You can think of $I$ as $J/(2x)$, for a unique ideal $J$ in $\mathbb{Z}[x]$. Precisely, $J$ is the set of polynomials $a_0+a_1x+\dots+a_nx^n$ such that the associated “mixed polynomial” $a_0+[a_1]x+\dots+[a_n]x^n$ (where $[z]$ denotes the residue class of $z\in\mathbb{Z}$ modulo $2$) belongs to $I$, that is, $$ a_0\text{ is even and }[a_1]=[0], [a_2]=[0],\dots,[a_n]=[0] $$

Now it will be clear that $J$ is the ideal generated by $2$ in $\mathbb{Z}[x]$. The quotient ring is then $\mathbb{Z}_2[x]$. This is a domain and not a field, so $I$ is a prime, but not maximal, ideal of $R$.

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  • $\begingroup$ Thank you! I thought that x and all his powers were zero in the first quotient ring, in fact all the powers of x are in the ideal, that is the element zero of the quotient. Why was I wrong? $\endgroup$
    – pter26
    Feb 1 '17 at 22:01
  • $\begingroup$ @user411485 No, only the powers of $x$ with an even coefficient are. $\endgroup$
    – egreg
    Feb 1 '17 at 22:02
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$$R/I\cong\Bbb Z[x]/(2x,2)\cong\Bbb Z_2[x]$$

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