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I have no clue on how to do unit elements (inversible), zero divisors (might not exist), I know only for stuff like $\Bbb Z_n$, and I guessed that nilpotents you get from $(ax+b)^n=0$ and we have $a=0, b=0$ so only $0$ is nilpotent (like $b^n=0$ so b=0 and $a^nx^n=0$ so $a=0$ as well) and for idempotents, you just solve $(ax+b)^2=ax+b$ and you find $a,b$ which have to be something like $0$ and $1$?

Is this true for any case, like what has my function, $x^2-1$ have to do with any of the nilpotent/idempotent stuff?

Maybe I need like $x=\pm 1$, from the function and I put in the equation for idempotents that $x^2=1$? So this is literally the only information I have in all seminars, tutorials etc. I can't find anything related to this stuff, also I'm very bad at math, simple stuff is good.

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  • $\begingroup$ What is relevant is that $x^2-1=(x+1)(x-1)$. So this ring is, for starters, analogous to $\Bbb Z/(2\cdot 3)$. By the way, do you know what ring you have with $\Bbb Z[x]/(x-1)$? $\endgroup$ Feb 1, 2021 at 22:37
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    $\begingroup$ @Sebastiano thanks for the edit $\endgroup$ Feb 1, 2021 at 22:42
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    $\begingroup$ It's a pleasure so you can see the changes in MathJaX :-) $\endgroup$
    – Sebastiano
    Feb 1, 2021 at 22:45
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    $\begingroup$ @AndreiJarca "analogous to" means "similar to". Ted's point is that $\mathbb{Z}[x]/(x^2-1)$ and $\mathbb{Z}/(6)$ are similar because $(x^2-1) = (x+1)(x-1)$ is the product of two irreducible elements in $\mathbb{Z}[x]$ and $6 = 2\cdot 3$ is the product of two irreducible elements in $\mathbb{Z}$. $\endgroup$ Feb 1, 2021 at 22:45
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    $\begingroup$ No, no, not isomorphic at all. But analogous. Why are $[2]$ and $[3]$ zero-divisors in $\Bbb Z_6$? What do you expect to happen in this ring? $\endgroup$ Feb 1, 2021 at 22:49

2 Answers 2

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When you make a quotient ring out of a ring $R$ and one of its ideals $I$, the elements present in that quotient are subsets of $R$ of the form $a + I := \{ a + b \colon b \in I \}$, where $a \in R$. It forms a ring under the operations $(a + I) + (b + I) := ((a+b) + I)$ and $(a + I)(b + I):= (ab +I)$.

In this ring, it is easy to check that its zero (the addition identity) is $0 + I$. So, what you're looking for in that ring are (in the nilpotent case) elements $(a +I) \in R/I$ with $(a + I)^n = (a^n +I) = (0 + I)$.

This in turn is equivalent to stating that $a^n \in I$. Now you can check from the definition of an ideal that in this case $I := (x^2 -1) = \{ g(x)(x^2-1) \colon g(x)\in \mathbb{Z}[x] \}$.

So for an element of $R/I$ to be nilpotent, it must be of the form $q(x) + I$ with $q(x)^n = g(x)(x^2 -1)$ for some $n \in \mathbb{N}$ and some $g(x) \in \mathbb{Z}$.

To give you an example, $((x-1) + I)((x+1) + I) = ((x^2 - 1) + I) = (0 + I)$ since $x^2 -1 \in I$. Note also that $(x-1), (x+1) \notin I$, so $((x-1) + I), ((x+1) + I) \neq (0 + I)$, that is, they are not zero in the ring $\mathbb{Z}[x]/(x^2-1)$

Hope this helps :)

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  • $\begingroup$ So for zero divisors is it like (x-1)(x+1) and those 2 are zero divisors? $\endgroup$ Feb 1, 2021 at 23:08
  • $\begingroup$ Yes, since $(x^2 -1)$ is zero in the quotient ring, and $(x-1), (x+1)$ are not $\endgroup$
    – Daàvid
    Feb 1, 2021 at 23:11
  • $\begingroup$ Although you have to be aware that the elements of the quotient ring are equivalence classes in the initial ring, not really polinomials $\endgroup$
    – Daàvid
    Feb 1, 2021 at 23:17
  • $\begingroup$ So if x-1 and x+1 are not the only ones, how do you get the other ones $\endgroup$ Feb 1, 2021 at 23:18
  • $\begingroup$ Well, by the reasoning I made above, you can see that having two non-zero elements $(a+I),(b+I) \in \mathbb{Z}/(x^2-1)$ with $((a+I)(b+I) = (0 + I))$ is equivalent to stating that those $a,b \in \mathbb{Z}$ satisfy $(ab \in I)$ and $a,b \notin I$. So what you're concretely looking for are polinomials $q_1(x), q_2 (x) \in \mathbb{Z}$ that are not multiples of $(x^2-1)$ such that $q_1(x)q_2(x) = g(x)(x^2-1)$ for some $g(x) \in \mathbb{Z}$ $\endgroup$
    – Daàvid
    Feb 1, 2021 at 23:25
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You just need to do the same, but instead of $(ax+b)^n$ is zero, you have $(ax+b)^n$ is a multiple of $x^2-1$ (that's an example for nilpotent elements). A polynomial $f$ is a multiple of $x^2-1$ iff $f(1)=f(-1)=0$. Using that, you can easily observe that we have no non-zero nilpotents: $(a+b)^n=0$ and $(-a+b)^n=0$ means $a=b=0$.

For other cases, you'll get similar equations, but you must remember that $x^2=1$ in your ring, so all elements can be reduced to linear functions. For example, in this ring

$$ (ax+b)^2 = a^2x^2 + 2abx + b^2 = 2abx + a^2+b^2, $$

so your idempotents have to satisfy $a=2ab$ and $b=a^2+b^2$. I leave details of the rest of the problem for you.

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  • $\begingroup$ Sir, do you have any idea on how to do the units and zero divisors $\endgroup$ Feb 1, 2021 at 23:03

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