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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.

Jimmy
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jasmine
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  • See also https://math.stackexchange.com/questions/2591549/find-the-maximal-subgroups-of-z-nz. – lhf Jan 22 '18 at 12:31

2 Answers2

7

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.

lhf
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4

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.

57Jimmy
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