Let $R=\mathbb Z_2\oplus \mathbb Z_3\oplus \mathbb Z_5$. How many zero divisors are there in $R$?

Question

Let $R=\mathbb Z_2\oplus \mathbb Z_3\oplus \mathbb Z_5$.Then Total number of the zero divisors in $R?$

(A) $15$

(B) $10$

(C) $20$

(D) $22$

Solution We know that

$x$ is a unit $\Longleftrightarrow \ gcd(\vert x\vert ,30)=1$
$x$ is a Zero Divisor $\Longleftrightarrow \ gcd(\vert x\vert,30)\neq 1$

Since $R$ has $2.3.5=30$ elements.So number of elements in $R$ whose order is relatively prime to 30 is $\phi(30)=8\implies$ number of elements in $R$ whose order is not relatively prime to 30 are $30-\phi(30)=30-8=22\implies $ $R$ has 22 zero divisors.

But the answer key says the solution is 20

Where have I made a mistake?

You are right, see here. Please say why $20$ should be correct. — Dietrich Burde, Oct 14 '22 at 11:41
Yes, exactly. You made a claim "But the correct answer is 20" without proof. Do you have a scan? — Dietrich Burde, Oct 14 '22 at 11:45
(The point is, whoever told you that 20 is correct has made an error.) — JBL, Oct 14 '22 at 11:45
@DietrichBurde:Actually, I'm solving the text exercises...in the answer key it is given that option (C) is correct...thats why i'm asking — Styles, Oct 14 '22 at 11:46
Yes, this is what I mean. Can you show us the answer key? It is possible to confuse the exercise number, sometimes. — Dietrich Burde, Oct 14 '22 at 11:47
And often one confuses the exercise number. So there is no way of telling what is true for us, except that your answer is correct. — Dietrich Burde, Oct 14 '22 at 11:49
@DietrichBurde:I got it...I cross checked the answer key...key is wrong — Styles, Oct 14 '22 at 11:50
Using the totient function requires showing that $R$ is isomorphic to $\mathbb{Z}_{30}$, which, of course, is true by the Chinese Remainder Theorem. — Geoffrey Trang, Oct 14 '22 at 14:25

score 2 · Answer 1 · answered Oct 14 '22 at 14:23

As we've already settled in the comments, your figure of $22$ is the correct one.

The totient function is a great way to do it, but let's give another way just to crosscheck:

Since $R$ is finite, each element is a unit or a zero divisor (I count $0$ among the zero divisors.) So it suffices to count the units. To be a unit, it has to be a unit on every coordinate. There's only one unit in $F_2$, two in $F_3$ and four in $F_5$, so there ought to be $8$ units. The whole ring has $30$, and $30$ less $8$ is $22$, again.

And even if you don't count 0 as a zero divisor, that would make the answer 21, and still 20 would not be the correct answer. — Daniel Schepler, Oct 14 '22 at 16:58

Let $R=\mathbb Z_2\oplus \mathbb Z_3\oplus \mathbb Z_5$. How many zero divisors are there in $R$?

1 Answers1