the number of maximal ideal in $Zp^4$

Question

If $p$ is prime ,$\mathbb{Z}_{p^4}$ denotes the ring of integer modulo $p^4$,then the number of maximal ideal in $\mathbb{Z}_{p^4}$

a)$1$

b)$2$

c)$3$

d)$4$

i think there will be $4$ maximal ideal as here $p^4$ that mean here $p\times p\times p\times p\ldots$$4$ times.

Answer 1

Here is a roadmap for the general case:

• The maximal ideals of $\mathbb Z$ are exactly $p\mathbb Z$ where $p$ is prime.

• By the canonical correspondence $\mathbb Z \to \mathbb Z/n\mathbb Z$, the maximal ideals of $\mathbb Z/n\mathbb Z$ correspond exactly to the maximal ideals of $\mathbb Z$ that contain $n\mathbb Z$.

• The ideals of $\mathbb Z$ that contain $n\mathbb Z$ are exactly $m\mathbb Z$ where $m$ is a divisor of $n$.

• The maximal ideals of $\mathbb Z/n\mathbb Z$ are exactly $p\mathbb Z/n\mathbb Z$ where $p$ is prime divisor of $n$.

In your case, $n=p^4$ has only one prime divisor, $p$, and so $\mathbb Z/p^4\mathbb Z$ has only one maximal ideal.

Answer 2

Consider the ideal $(p)$ generated by $p$. It is certainly not the whole ring, and every non-zero element $a$ that is not in $(p)$ is invertible, because it has gcd$(a,p^4)=1$. Since the units are exactly the elements that are not in any maximal ideal, $(p)$ is the only maximal ideal. Note that $\mathbb{Z}_{p^4}/(p)\cong \mathbb{Z}_{p}$, which is indeed a field.