Number of Distinct ideals of $Z_{60}?$ Tried to count all prime numbers between $0$ to $60$ and adding $(0)$ and (R) to it.
that is total $19$ , but I saw that answer is $18$.
So, Please explain.
 A: $\tau(60)=12$ so there are $12$ ideals
Note: The ideals in $Z_n$ are precisely the sets of the form $<d>$ where $d$ divides $n$, so number of ideals are same as the number of divisors of $n$
A: The ideals of $\mathbb Z/60$ correspond to the ideals of $\mathbb Z$ that contain $60\mathbb Z$ and so correspond to the divisors of $60$. Since $60=2^2\cdot 3 \cdot 5$, it has $(2+1)\cdot(1+1)\cdot(1+1)=12$ divisors.
A: As the other answer list, the number of ideals is actually $12$. One other way to show this is to use the Chinese Remainder Theorem, which gives an isomorphism
$$\mathbb Z\diagup60\mathbb Z \xrightarrow{\sim}
\left(\mathbb Z\diagup4\mathbb Z\right) \times
\left(\mathbb Z\diagup3\mathbb Z\right) \times
\left(\mathbb Z\diagup5 \mathbb Z\right)$$
Hence, the number of ideals in $\mathbb Z\diagup60\mathbb Z$ is the product of the number of
ideals of the three factors. Since $\mathbb Z\diagup3\mathbb Z$ and $\mathbb
Z\diagup 5\mathbb Z$ are fields, they only have $2$ ideals (the zero ideal and the
unit ideal). $\mathbb Z\diagup4\mathbb Z$ additionally has the ideal $(2)$; $(3)$ is easily seen to be identical to the unit ideal. So
the result is $3 \cdot 2 \cdot 2 = 12$.
