What is the main difference between finite fields and rings?

Question

In the course I'm studying, if I've understood it right, the main difference between the two is supposed to be that finite fields have division (inverse multiplication) while rings don't. But as I remember, rings also had inverse multiplication, so I can't see any difference.

What is the main difference between finite fields and rings?

Answer 1

Since you specified finite fields and other answers didn't talk about it, I am adding the following:

Fields are rings which are commutative and in which all nonzero elements have multiplicative inverse.

But finite fields have another important property that distinguish them from rings: every finite field is completely specified by its order, because they always have exactly $p^n$ elements for some prime $p$ and natural $n$ and any two finite fields of same order are isomorphic.

Notice, saying that two fields are isomorphic means that they are the same, but represented in two different ways.

That is not the case for rings. For instance, this answer provides examples of rings that have $4$ elements but are not isomorphic:

$\mathbb{Z}_4,~ \mathbb{F}_2[x]/(x^2+x),~ \mathbb{F}_2[x]/(x^2+x+1),~ \text{and} ~ \mathbb{F}_2[x]/(x^2)$

Answer 2

In general rings do not have inverse multiplication, as you claimed. Think for example about the integers modulo $4$. In this case, $2$ is not invertible as there is no element that multiplied by $2$ gives you $1$ (the unit of the ring). You can check this easily by exhausting the possibilities: $$0\cdot 2 \equiv 0 \bmod 4, \ \ \ 1\cdot 2 \equiv 2 \bmod 4, \ \ \ 2\cdot 2 \equiv 0 \bmod 4, \ \ \ 3\cdot 2 \equiv 2 \bmod 4$$

You can also rely on the following lemma:

The only elements invertible modulo $n$ are those that are coprime to $n$

If $n$ is prime, then all non-zero elements modulo $n$ are coprime to $n$. But if $n$ is not a prime, then you have non-zero elements modulo $n$ that can share nontrivial divisors with $n$.

Answer 3

Take a set $S$ and an operation, let's call it $\oplus$. Also take another operation, say $\odot$. Also fix an element $\mathcal O\in S$ such that $\forall s\in S: \mathcal O\oplus s=s$ and fix another element $\mathcal I\in S$ such that $\forall s\in S: \mathcal I\odot s=s$.

Now I will list a set of properties and behind some properties I'll write names. If all the previous points apply, then the name can be applied to the named tuple.

1. (closure under addition) $\forall s_1,s_2\in S: (s_1\oplus s_2)\in S$
2. (additive associativity) $\forall s_1,s_2,s_3\in S:(s_1\oplus s_2)\oplus s_3=s_1\oplus(s_2\oplus s_3)$. We can now call $(S,\oplus)$ a semigroup.
3. (additive identity) $\exists \mathcal O\in S:\forall s\in S: \mathcal O\oplus s=s\oplus \mathcal O=s$. We can now call $(S,\oplus,\mathcal O)$ a monoid.
4. (additive inverses) $\forall s\in S:\exists s'\in S: s'+s=s+s'=\mathcal O$. We can now call $(S,\oplus,\mathcal O)$ a group.
5. (additive commutativity) $\forall s_1,s_2\in S:s_1\oplus s_2=s_2\oplus s_1$. We can now call $(S,\oplus,\mathcal O)$ an abelian group.
6. (multiplicative closure) $\forall s_1,s_2\in S:s_1\odot s_2\in S$
7. (multiplicative associativity) $\forall s_1,s_2,s_3\in S:s_1\odot (s_2\odot s_3)=(s_1\odot s_2)\odot s_3$. We can now call (with 6 and 7) $(S,\odot)$ a semigroup.
8. (multiplicative identity) $\exists \mathcal I\in S:\forall s\in S:\mathcal I\odot s=s\odot \mathcal I=s$. We can now call (with 6-8) $(S,\odot,\mathcal I)$ a monoid.
9. (right distributivity) $\forall s_1,s_2,s_3\in S: (s_1\oplus s_2)\odot s_3=(s_1\odot s_3)\oplus (s_2\odot s_3)$
10. (left distributivity) $\forall s_1,s_2,s_3\in S: s_1\odot (s_2\oplus s_3)=(s_1\odot s_2)\oplus (s_1\odot s_3)$. We can now call $(S,\oplus,\odot,\mathcal O,\mathcal I)$ a ring. Note: In some definitions having a multiplicative identity is optional.
11. (multiplicative commutativity) $\forall s_1,s_2\in S:s_1\odot s_2=s_2\odot s_1$. We can now call $(S,\oplus,\odot,\mathcal O,\mathcal I)$ a commutative ring.
12. (multiplicative inverses) $\forall s\in (S\setminus\{\mathcal O\}):\exists s'\in S:s\odot s'=s'\odot s=\mathcal I$. We can now call $(S,\oplus,\odot,\mathcal O,\mathcal I)$ a field.

If you want finiteness as well, then you write finite X in front of any of the names above if $\exists n\in\mathbb N:n=\left|S\right|$, that is, if $S$ is has a finite amount of elements.

As for the main difference between rings and fields: Rings don't require multiplication to be commutative. A common example would be $\mathbb R^{2\times 2}$ (the set of all 2x2 matrices with real-valued entries) with the usual matrix addition and multiplication. This features neither commutativity under multiplication nor can you always find inverse matrices. As for the difference between commutative rings and fields, simply consider $\mathbb Z_n$ with $n=pq$ for some primes $p,q$, ie the set of all nonegative integers smaller than $n$ with addition and multiplication with modular reduction $\bmod n$ at the end. In this ring you cannot find a multiplicative inverse of any number with $\gcd(n,x)>1$, so f.ex. for $x=p$, that is you can't find any $x$ such that $x\cdot p\bmod{pq}=1$. For a very concrete example look at the multiplication table for $n=3\cdot 5=15$ and notice how in the row and the column with the entry $3$ you can only ever get $0,3,6,9,12$ as the result of a multiplication by $3$, but never $1$, so $3$ has no multplicative inverse in $\mathbb Z_{15}$ and thus $\mathbb Z_{15}$ with the above elements and operations is not a field, but a commutative ring.