# Find all units, zero divisors, nilpotent elements and idempotent elements in $\Bbb Z[x]/(x^2-1)$

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.

• 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 $\Bbb Z[x]/(x-1)$? Feb 1, 2021 at 22:37
• @Sebastiano thanks for the edit Feb 1, 2021 at 22:42
• It's a pleasure so you can see the changes in MathJaX :-) Feb 1, 2021 at 22:45
• @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 $\mathbb{Z}$. Feb 1, 2021 at 22:45
• 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? Feb 1, 2021 at 22:49

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 :)

• So for zero divisors is it like (x-1)(x+1) and those 2 are zero divisors? Feb 1, 2021 at 23:08
• Yes, since $(x^2 -1)$ is zero in the quotient ring, and $(x-1), (x+1)$ are not Feb 1, 2021 at 23:11
• Although you have to be aware that the elements of the quotient ring are equivalence classes in the initial ring, not really polinomials Feb 1, 2021 at 23:17
• So if x-1 and x+1 are not the only ones, how do you get the other ones Feb 1, 2021 at 23:18
• 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 $g(x) \in \mathbb{Z}$ Feb 1, 2021 at 23:25

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.

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