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I need the idempotent elements of $Z_{900}$

$2^2\cdot 3^2\cdot 5^2=900$

Of course there's $$0 \pmod 4 \\ 0 \pmod 9 \\ 0 \pmod {25} \\ $$ and $$ 1 \pmod 4 \\ 1 \pmod 9 \\ 1 \pmod {25} \\ $$

I found the answers by making a C++ program to test all numbers, but I don't know how to quick solve on paper.

This is the output of the program $(0, 0, 0) \rightarrow 0 \\ (1, 1, 1) \rightarrow 1 \\ (0, 0, 1) \rightarrow 576 \\ (0, 1, 0) \rightarrow 100 \\ (1, 0, 0) \rightarrow 225 \\ (1, 1, 0) \rightarrow 325 \\ (0, 1, 1) \rightarrow 676 \\ (1, 0, 1) \rightarrow 801 \\$

Parcly Taxel
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Jack R
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2 Answers2

13

As you noticed, $900$ factors into $2^23^25^2$, and the relevance of this is that the Chinese remainder theorem gives us an isomorphism $\Bbb Z_{900}\cong\Bbb Z_{2^2}\times\Bbb Z_{3^2}\times\Bbb Z_{5^2}$. You can easily show that an idempotent of this ring must be of the form $(e_1,e_2,e_3)$ where each of the $e_i$ are idempotent in $\Bbb Z_{p^2}$. This reduces the problem to one in $\Bbb Z_{p^n}$, and then by implementing the Chinese remainder theorem you can lift the idempotents back to elements of $\Bbb Z_{900}$.

Following this strategy takes us past several insightful lemmas that can come in handy for other problems.

Understand $\Bbb Z_{p^n}$

By ideal correspondence, and our knowledge of ideals of $\Bbb Z$, the only proper ideals of $\Bbb Z_{p^n}$ are of the form $(p^k+p^n\Bbb Z)$ where $1\leq k\leq n$, and $(p+p^n\Bbb Z)$ is the unique maximal ideal of the whole ring. That makes it a prime ideal as well! Furthermore, you can see that $x^n=0+p^n\Bbb Z$ for every $x\in (p+p^n\Bbb Z)$, that is, every element there is nilpotent.

Useful facts about idempotents

If $e$ is an idempotent

  1. then $1-e$ is idempotent too.
  2. then $e(1-e)=0$.
  3. and also nilpotent, then it is zero.

Together we get

For now, let me drop the $+p^n\Bbb Z$ in the coset notation to declutter the math. I mean that I'm writing $(p)$ rather than $(p+p^n\Bbb Z)$ in $\Bbb Z_{p^n}$.

Let $e$ be idempotent in $\Bbb Z_{p^n}$. Then since $e(1-e)=0\in (p)$ and $(p)$ is prime, one of $e$ or $1-e$ is in $(p)$.

If $e\in (p)$, then it's both an idempotent and a nilpotent, hence zero.

If instead $1-e\in (p)$, then by the same argument, $1-e=0$. But this is the same as saying $e=1$.

So in conclusion, you can see that the $e_i$ must be either the zero elements or the identity elements of their respective rings. In total that gives you $2^3$ choices of idempotents, each of which you can pull back to $\Bbb Z_{900}$ via your favorite implementation of the Chinese remainder theorem.

rschwieb
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  • Why is $(p+\mathbb{Z}_{p^n})$ an ideal? In the case of $\mathbb{Z}_4={0,1,2,3}$, for example, we would get $2+\mathbb{Z}_4={2,3,0,1}$. Ok, I'm obviously reading this incorrectly - any help? – man_in_green_shirt Jul 23 '15 at 17:55
  • @man_in_green_shirt Actually, it looks like the notation is a longstanding typo... – rschwieb Jul 23 '15 at 18:26
  • So as it is now, going back to the example I gave, would $2+4\mathbb{Z}={2,6,10,14,...}$ be an ideal in $\mathbb{Z}_4={0,1,2,3}$? Does that make sense? – man_in_green_shirt Jul 23 '15 at 18:34
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    @man_in_green_shirt You're mixing coset notations. To write consistently, you would say that $2\Bbb Z_4={0+4\Bbb Z, 2+4\Bbb Z}\lhd {0+4\Bbb Z,1+4\Bbb Z,2+4\Bbb Z,3+4\Bbb Z}=\Bbb Z_4$. $2+4\Bbb Z$ is not an ideal at all, it is an element of $\Bbb Z_4$. Surrounding it with parentheses is standard notation for the ideal generated by this element. – rschwieb Jul 23 '15 at 18:43
  • Ok, I'm still unsure about what the set $(p+p^n\mathbb{Z})$ would look like in this simple case – man_in_green_shirt Jul 23 '15 at 18:53
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    @man_in_green_shirt In the example I just gave, $2\Bbb Z_4=(2+4\Bbb Z)$. Probably the source of confusion is the mixing of notations. If you pick one notation you would like to use, I can help you sort it out. – rschwieb Jul 23 '15 at 18:54
  • Ok, got it, thanks for your patience/help – man_in_green_shirt Jul 23 '15 at 19:00
  • Are there really more than one implementation of CRT, as you suggest at the end? Seems like there might only be one, natural one, though I guess you could always compose with an automorphism. –  Mar 20 '20 at 01:18
  • The Bezout coefficients are not unique. Is that your point? –  Mar 20 '20 at 01:38
  • @ChrisCuster No, I was referring to the (at least two) constructive methods, like the ones you can see in the wiki article. All methods are going to give the same solution. – rschwieb Mar 20 '20 at 01:41
  • Thanks for the clarification. –  Mar 20 '20 at 02:24
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Hint $\ \,p\,$ prime, $\,p^n\mid e(1\!-\!e)\,\Rightarrow\, p^n\mid e\,$ or $\,p^n\mid 1\!-\!e,\ $ by $\ 1\!-\!e,\, e\,$ coprime, by $\ (1\!-\!e)+ e = 1$

So $\!\bmod p^n$ the only idempotents (roots of $\,e^2\equiv e)\,$ are the trivial idempotents $\,e \equiv 0,1.\,$

So, by CRT [cf, below] $\,e\,$ is idempotent mod $\,2^2 3^2 5^2\iff e \equiv 0,1\,$ mod $\,2^2,3^2,5^2,\,$ e.g.

$\ e \equiv (\color{#c00}1,0,0)\Rightarrow {\rm mod}\ 2^2\!:\ e\equiv \color{#c00}1 \equiv 3^2 5^2 k \equiv \color{#c00}k,\,$ so $\, e = 15^2k = 15^2(1\! +\! 4j)\equiv 225\pmod{\!900}$

So $\,(0,1,1) = 1-e \equiv 1-225 = 901-225 \equiv 676.\,$ Same for other $\,(e_1,e_2,e_2),\ e_i \in \{0,1\}$

Remark $\ $ Idempotents $\!\bmod n\,$ correspond to splittings of $n$ into $\color{#c00}{\rm co}\color{#0a0}{\rm prime}$ factors, e.g. above the idempotent $\,e = 225 = 3^2 5^2\,$ corresponds to the factorization $\,n = \color{#c00}{2^2}\!\times \color{#0a0}{3^2 5^2}\,$ where $\,\color{#c00}{e\equiv 1\pmod{\!2^2}},\,$ $\color{#0a0}{e\equiv 0\pmod{\!3^2 5^2}}.\,$ In fact some integer factorization algorithms work by searching for nontrivial idempotents mod $\,n,\,$ which immediately yield a factorization of $\,n\,$ (generally we can quickly factor $\,n\,$ given any polynomial which has more roots mod $\,n\,$ than its degree, so a nontrivial idempotent (or square-root) $\Rightarrow$ quadratic has $> 2$ roots, which splits $n$).

Generally we can find modular roots of polynomials using CRT as explained below, where above is the special case $f(x) = x^2 -x = x(x-1)$.

By CRT, each combination of a root $\,r_i\,$ mod $\,m\,$ and a root $\,s_j\,$ mod $\,n\,$ corresponds to a unique root $\,t_{ij}\,$ mod $\,mn\,$ i.e.

$$\begin{eqnarray} f(x)\equiv 0\pmod{mn}&\overset{\rm CRT}\iff& \begin{array}{}f(x)\equiv 0\pmod m\\f(x)\equiv 0\pmod n\end{array} \\ &\iff& \begin{array}{}x\equiv r_1,\ldots,r_k\pmod m\phantom{I^{I^{I^I}}}\\x\equiv s_1,\ldots,s_\ell\pmod n\end{array}\\ &\iff& \left\{ \begin{array}{}x\equiv r_i\pmod m\\x\equiv s_j\pmod n\end{array} \right\}_{\begin{array}{}1\le i\le k\\ 1\le j\le\ell\end{array}}^{\phantom{I^{I^{I^I}}}}\\ &\overset{\rm CRT}\iff& \left\{ x\equiv t_{i j}\!\!\pmod{mn} \right\}_{\begin{array}{}1\le i\le k\\ 1\le j\le\ell\end{array}}\\ \end{eqnarray}\qquad\qquad$$

Bill Dubuque
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  • The part about how some integer factorization algorithms work seems repeated unnecessarily – J. W. Tanner Nov 07 '22 at 17:00
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    @J.W.T Thanks, dupe text removed. – Bill Dubuque Nov 07 '22 at 17:05
  • For those who are finding it difficult to understand the line $e\equiv 1\equiv 3^2 5^2 k \equiv 1$ the op has used $3\times 5 \equiv 1 \pmod 4 $. I had a hard time understanding why this method was not working for other values of $n$ like $2^2 5^2$. – nkh99 Oct 29 '23 at 10:21
  • In the case of $n=100$ and $e\equiv (0 \pmod 4, 1\pmod {25})$ it will be like $4k\equiv1 \pmod {25}$ i.e., $k\equiv 19 \pmod {25}$ since $4\times 19\equiv 1 \pmod {25}$. Thus $e\equiv 4(25l+19)\equiv 76\pmod {100}!$ This is a great answer btw. – nkh99 Oct 29 '23 at 10:26