I have seen this question asked elsewhere on here, but with a different direction than the one I want to go. The question is asking me to give an example to show that a factor ring of an integral domain may be a field. The example I was thinking of was $\Bbb Z/(p\Bbb Z)$, and I have convinced myself that this is a field but am not sure the exact math way that I would prove that this is a field using theorems. Could someone give me a pointer on this?


2 Answers 2


The theorems you're looking for are:

  • if $R$ is a commutative ring with $1$ and $I$ is an ideal of $R$, then $R/I$ is a field if and only if $I$ is maximal.

  • if $R$ is a commutative ring with $1$ and $I$ is an ideal of $R$, then $R/I$ is an integral domain if and only if $I$ is prime.

You can find proofs in every algebra book (e.g. Dummit & Foote, Lang, Hungerford, Fraleigh, etc) or even in questions in this website.

So, since $\Bbb Z$ is commutative with $1$ and $p\Bbb Z$ is a maximal ideal for $p$ prime, $\Bbb Z/p\Bbb Z$ is a field.

  • 2
    $\begingroup$ Good and complete. $\endgroup$
    – Lubin
    Jun 2, 2019 at 23:39

For any prime

$p \in \Bbb P \subset \Bbb Z, \tag 1$

it is easy to see that

$0 < k < p \Longrightarrow \gcd(k, p) = 1; \tag 2$


$p \not \mid k, \; 0 < k < p, \tag 3$

and since the only divisors of $p$ are $1$ and $p$ itself, the only common divisor of $p$ and $k$ is $1$; hence (2).

We next exploit the well-known Bezout's identity, that is

$\exists a, b \in \Bbb Z, \; ap + bk = \gcd(p, k) = 1; \tag 4$

when we reduce this equation modulo $p$, that is, when we form the cosets of $(p) \subset \Bbb Z$, we obtain

$(b +(p))(k + (p)) = (a + (p))(p + (p)) + (b +(p))(k + (p))$ $= (ap + (p)) + (bk + (p)) = (ap + bk) + (p) = 1 + (p), \tag 5$


$(a + (p))(p + (p)) = (a + (p))(0 + (p)) \equiv 0 \mod p; \tag 6$

(5) shows that every $k + (p)$ has a multiplicative inverse $b + (p)$ in the commutative ring $\Bbb Z/(p)$; hence $\Bbb Z / (p)$ is a field.


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