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!

  • 3
    $\begingroup$ 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

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 $R$.

  • $\begingroup$ 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

$$R/I\cong\Bbb Z[x]/(2x,2)\cong\Bbb Z_2[x]$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.