To be a field, there are certain properties that the set has to satisfy. But the set of integers doesn't have a multiplicative inverse, so how can it constitute a field?

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    $\begingroup$ The set of integers isn't a field. But integers mod 3 isn't integers. Do you know how arithmetic works mod 3? $\endgroup$ Commented Jan 21, 2013 at 23:08
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    $\begingroup$ Every non-zero integer modulo $3$ does have a multiplicative inverse: $1^2=1$, so $1^{-1}=1$, and $2^2=1$, so $2^{-1}=2$. $\endgroup$ Commented Jan 21, 2013 at 23:09
  • $\begingroup$ Looking at multiplication with a non-zero integer modulo a prime as a map, it is bijective, and thus has an inverse. $\endgroup$
    – Arthur
    Commented Jan 21, 2013 at 23:12
  • $\begingroup$ To see that modulo $p$ you have inverses of nonzero elements: The point is that if $p\not\mid a$, then $p$ and $a$ are relatively prime, so $pm+an=1$ for some $m,n$. This means that $an\equiv 1\pmod p$, that is, $n$ is a multiplicative inverse of $a$. $\endgroup$ Commented Jan 21, 2013 at 23:13
  • $\begingroup$ $(3)$ is a maximal ideal in $\mathbb{Z}$, thus $\mathbb{Z}/3\mathbb{Z}$ is a field $\endgroup$
    – Amr
    Commented Jan 21, 2013 at 23:17

2 Answers 2


It is true that that $\mathbb{Z}$ does not, in general, contain inverses. However, the set of integers modulo $p$ a prime is a very different structure than $\mathbb{Z}$. For an example, let's use $p=3.$

Let's call the set of integers modulo 3 by $\mathbb{F}_3.$ It has three elements, which we will call $\{\overline{0},\overline{1},\overline{2}\}.$ Don't confuse these with $0, 1,2\in\mathbb{Z},$ as they're quite different! One way to think of them is that each one represents the set of all integers which has the given remainder when divided by 3. For instance, $\overline{2}$ denotes the set of all integers which have remainder $2$ when divided by $3$. Equivalently, $\overline{2}$ denotes the set of integers which are congruent to $2$ modulo $3$.

Now we can perform standard modular arithmetic to determine the addition and multiplication tables for this set. We find that $\overline{1}*\overline{1}=\overline{1},$ and $\overline{2}*\overline{2}=\overline{4}=\overline{1}.$ Thus, both of the nonzero elements have inverses! Note that in computing the multiplication I mentioned a new element $\overline{4}$ that looks out of place. But it's okay, because we know that $4\equiv 1\bmod 3.$ So we can always do normal arithmetic under the bar and then simplify it afterward. This is why modular arithmetic is so easy!

We see that this can happen in a finite field because there isn't room for numbers to keep getting bigger and bigger, as there is in $\mathbb{Z}.$ Instead, multiplication "wraps around" until it gets to $\overline{1}$ and you find your inverse.

But be careful! This trick only works because $p$ was prime. As we know, the product of any two integers with magnitude less than $p$ will never be a multiple of $p$, and thus the product of any two nonzero elements of $\mathbb{F}_p$ will never be zero (do you see why this works?). However, suppose that instead of $p$ we chose some $n=pq$ a product of two primes $p$ and $q$. Then, we would not be able to find an inverse for $p$, because in the integers modulo $n$, we have that $\overline{p}*\overline{q}=\overline{pq}=\overline{0}.$ So if there were some $\overline{m}$ such that $\overline{p}*\overline{m}=1,$ then we would have $\overline{q}=\overline{p}*\overline{m}*\overline{q}=\overline{0},$ a contradiction!

That $p$ does not have any divisors is precisely equivalent to the condition that $\overline{p}=\overline{0}$ does not have divisors in the integers modulo $p$, and we recall that an element has an inverse (i.e., it is a unit) precisely if it does not divide $0$, which in this case is $p$.

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    $\begingroup$ Thank you, your explanation is very detailed and extremely helpful! I did reread the definition of equivalence classes relating to modulo, but I didn't quite understand it... obviously. $\endgroup$ Commented Jan 21, 2013 at 23:33

The set of integers modulo 3 is {0,1,2}. For + we have:

+   0 1 2
0   0 1 2
1   1 2 0
2   2 0 1

Notice that 1 + 2 = 3 = 0 (mod 3).

Also for multiplication:

*  0 1 2
0  0 0 0
1  0 1 2
2  0 2 1

So: -1 = 2 and 1/1 = 1 and 1/2 = 2. It is a Field.


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