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Let $R = \mathbb Z/25\mathbb Z$ and $S = \{1,7,18,24\}$

  1. How many elements of $R$ are non-zero-divisors? How many are invertible?
  2. Now consider the subring $RS$. How many elements of $RS$ are non-zero, how many are zero-divisors, how many are invertible?

For (1) I think $R$ has 24 non-zero elements. Also $S$ is set of powers of $7\pmod {25}$. Thanks for your help!

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Exact duplicate of – Zev Chonoles Sep 27 '13 at 2:33
Sorry I lost track of the last post. Please help me if you can. thanks – Steven Sep 27 '13 at 2:40
Why did you drastically change the content of this question after receiving two answers? This is highly inappropriate, and as such I am reverting this question to a previous revision. – arjafi Sep 27 '13 at 19:15

A zero divisor of a ring $R$ is an element $r \neq 0$ of $R$ which satisfies $rs = 0$ for some nonzero $s$ in $R$. So you haven't answered question (1). Yes, there is only one $0$ in $R$ (this is part of the definition of a ring), but some of the $r \neq 0$ in $R$ might still be zero divisors.

Here's a hint: Remember that $R = \mathbb{Z}/25\mathbb{Z}$ is a set of equivalence classes $[m]$ of integers modulo 25, i.e., under the equivalence $m_1 \sim m_2 \iff 25 \mid (m_2-m_1)$. So the integers $n \in \mathbb{Z}$ that are in $[0]$ (i.e., $n$ such that $0 \sim n$) are the integers that are divisible by 25.

If $m \in \mathbb{Z}$ and $[m]$ is a zero divisor in $R$, then $[m][n] = [mn] = [0]$ for some nonzero $n \in \mathbb{Z}$. Think about what this says: $[m]$ is a zero divisor in $R$ if and only if there is a nonzero integer $n$ such that $25$ divides $mn$. For which $m \in \mathbb{Z}$ is this possible? How many equivalence classes do these $m$ fall into?

Here's an example of a zero divisor: Consider $[4]$ in the ring $T = \mathbb{Z}/12\mathbb{Z}$. Since $4 \cdot 3 = 12$, we know that $[4][3] = [12] = [0]$. (Of course $[m]$ is a different set of integers for $T$ than it was for $R$, since these rings have different notions of equivalence.)

Hint 2: Once you know the number of zero divisors, you get the number of invertible elements (a.k.a. "units") for free. Why? (How are they related in $\mathbb{Z}/n\mathbb{Z}$?)

Hint 3: For (2), can you find $a \in \mathbb{Z}$ such that $7a \equiv 1 \pmod{25}$? What about finding such an $a$ for other $s \in S$?

Hint 4: Slightly more abstractly, are the elements of $S$ zero divisors, units, or something else? The answer should give you information about $RS$.

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so you mean for part (1) we have 1.25 = 5.5 = 25.1 = 25. so we have 3 non-zero divisors? they're all invertible – Steven Sep 27 '13 at 3:29
You're getting there, but those are equations, not elements of a ring. Which elements of $\mathbb{Z}/25\mathbb{Z}$ are you saying are zero divisors? Also, it's unclear to me why you stopped at 25. By that reasoning, we would surely have infinitely many zero divisors in our finite ring, since $25\mid 5^{n+2}$ for all $n \geq 0$. What's wrong with this picture? Another hint might be in order: Only one of the elements $[1]$, $[5]$, and $[25]$ in $\mathbb{Z}/25\mathbb{Z}$ is a zero divisor. (Actually, check your notes for the definition you're supposed to use, since (ct'd below) – Dan Sep 27 '13 at 3:55
... there are two zero divisors if they don't restrict to $r \neq 0$ in $R$. Also, are you sure they're all invertible? Give me integers $m$ and $n$ such that $5m \equiv 1 \pmod{25}$ and $25n \equiv 1 \pmod{25}$ (so that $[m] = [5]^{-1}$ and $[n] = [25]^{-1}$). – Dan Sep 27 '13 at 3:58
I'm still quite confused. R has only 24 non-zero elements. So I think there are 2 zero divisors [1] and [25]. both of them are invertible – Steven Sep 27 '13 at 4:05
$[25]=[0]$. So $[25]$ is not a zero divisor. This also means that $[1][25]=[0]$ really just says that $[1][0]=[0]$, which doesn't tell you that $[1]$ is a zero divisor (reread the definition if you don't see why). – Dan Sep 27 '13 at 5:11

If $a$ is not a zero divisor then $a^k = 1$ has a solution. Conversely if $a$ is a unit it is not a zero divisor. So $a$ is a zero divisor iff it's not a unit. So take $|R| - |$ units of R $|. \ $ $a$ is a unit iff $\gcd(a, |R|) = 1$.

Since $S$ contains $1$, $RS = R$. So the problem becomes counting the zero-divisor, non-zero, and unit elements of $R$.

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I know the theory that you mentioned above. but my problem is how to count – Steven Sep 27 '13 at 3:26
All units are roots of unity?! – oldrinb Sep 27 '13 at 3:32
could you, for example, do part (1) so that I can get an idea of counting here? thanks – Steven Sep 27 '13 at 3:35

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