Ttyjfyk

I am trying to show that if $(R, +)$ is an Abelian group and $(R - 0_R, cdot)$ is an Abelian group, then $(R, +, cdot)$ is not necessarily a field. Note that $0_R$ is the identity element of $(R, +)$. I know that a field is a commutative division ring and one of a ring's properties is that $forall a,b in R, ~ acdot (b + c) = a cdot b + acdot c$. Therefore, I am trying to come up with a set and two binary operations that satisfy the first property, but together do not form a field.

So far, I have come up with a group over polynomials with $+$ being normal addition and $cdot$ being composition, but then $(R - 0_R)$ is not commutative. I would appreciate any help/guidance.

Thanks.

As @LordSharktheUnknown implicitly points out, if you just take a finite set with non-prime-power order (six is the first such integer $ge 2$) and put any group structures you want, it will have to work, because finite fields have prime-power order.

But just to be thorough, you can construct an example $R$ with $|R|=n$ for any $n>3$ (we ignore $|R|=1$ as it's not very interesting). It's more or less obvious that you can't do it with $n=2$, so the only mildly surprising thing is that you can't do it with $|R|=3$. Your hand is forced for the additive structure, and then there are only two options for a multiplicative structure, by choosing a labeling of $R-0$, and either forms a field, because the additive structure of $mathbbZ/3mathbbZ$ is preserved by relabeling $1$ and $2$. And it's perhaps a little bit surprising that you can do it with $n=p>3$ a small prime and $p-1$ having no repeated prime factors (e.g., $n=7$), since in such a case, the structures of the two groups are forced; but it turns out you can label things sufficiently oddly that it fails to form a field (and in fact it usually will fail, especially as $n$ grows large).

If $n>3$, let's construct an example. If $n$ is not a prime power, choose any group structures you like (e.g., cyclic), as we saw. If $n = p^e$ with $p$ prime and $ege 2$, then make $(R,+) cong (mathbbZ/nmathbbZ,+)$, which will work since the additive structure of a finite field is not cyclic unless it has prime order. If $n=p$ is prime, then you are forced to have $(R,+) cong (mathbbZ/pmathbbZ,+)$, so let's just identify them, i.e. take $(R,+) := (mathbbZ/pmathbbZ,+)$. Suppose $pge 7$. Define $cdot$ on $(mathbbZ/pmathbbZ)-0$ to be cyclic generated by $2$ so that the powers of $2$ are $2^1 = 2$, $2^2 = 1$, and $2^k = k$ for $3le k le p-1$. Now, doing addition first, we have
$$2cdot(1+1) = 2cdot 2 = 2^2 = 1,$$
but distributing first, we have
$$2cdot(1+1) = 2cdot 1 + 2cdot 1 = 2cdot 2^2 + 2cdot 2^2 = 2^3 + 2^3 = 3+3=6 ne 1$$
in $mathbbZ/pmathbbZ$. If $p=5$, you can define $3^1 = 3$, $3^2 = 2$, $3^3 = 4$, and $3^4 = 1$, and then, doing addition first, we have
$$3cdot(1+1) = 3cdot 2 = 3cdot 3^2 = 3^3 = 4,$$
but, distributing first, we have
$$3cdot(1+1) = 3cdot 1 + 3cdot 1 = 3cdot 3^4 + 3cdot 3^4 = 3^1 + 3^1 = 6 = 1.$$

You can also do it with any infinite set. Pretty much anything you try will work, provided you let loose a bit. For example, take $R = mathbbZ$, with $+$ being regular addition. Let $S = mathbbZsetminus 0$, and let $phi:S to R$ be the bijection which shifts negative numbers up by one and is constant on positive numbers. Now define $acdot b = phi^-1(phi(a)+phi(b))$. We're just relabeling $S$ to be $mathbbZ$ again and then doing regular addition. Now, doing addition first, we have
$$-2cdot(1+1) = -2cdot 2 = phi^-1(-1+2) = 1,$$
but distributing first, we have
$$-2cdot(1+1) = -2cdot 1 + -2cdot 1 = phi^-1(-1+1) + phi^-1(-1+1) = -1+(-1) = -2.$$

In terms of guidance, you should expect that you'll need to do something perverse like this, because most of the examples you'll think of where two binary operations already exist are rings, where distributivity necessarily holds.

Here is a concrete example, inspired by LStU:

The set is $0,1,2,3,4,5$. Addition is just addition mod $6$.

Multiplication is defined by
$$
acdot b = left{ beginarraycl 0& a=0 \ 0 & b=0 \
1 & a = b= 5 \
5 & a=5 wedge b in [1,4]\
5 & b=5 wedge a in [1,4]\
ab pmod5& mboxotherwiseendarray right.
$$
or as a table
$$
beginarraycccccc cdot&0&1&2&3&4&5 \ hline
0 & 0&0&0&0&0&0 \
1 & 0&1&2&3&4&5 \
2 & 0&2&4&1&3&5 \
3 & 0&3&1&4&2&5 \
4 & 0&4&3&2&1&5 \
5 & 5&5&5&5&5&1
endarray
$$
The group properties, as well as commutativity, are easily checked.

Now consider $$ (1+4)cdot 5 = 5cdot 5 = 1 \
1cdot 5 + 4 cdot 5 = 5+5 = 4 neq 1
$$

Here's a simple counting argument showing that being distributive is "rare".

First, if the cardinal $pge 2$ of $R$ is not prime, choose any group structure on $R$ (with neutral 0) which is not elementary abelian. Then this cannot be the additive group of a field, and hence for no group structure $cdot$ on $R-0$ we can get a field, i.e., any choice of multiplication (with 0 absorbing element and making $R-0$ a group) is non-distributive. This already applies when $p=4$ (concrete example: standard addition on $mathbfZ/4mathbfZ$ and any of the three group structures on $1,2,3$ works).

Hence let me stick to prime cardinal $p$.

Let $R$ be a set with $p$ elements, $p$ prime, with a fixed element $0$. It has, up to permutation fixing $0$, a unique group structure, whose automorphism group has order $p-1$. Hence it has $(p-2)!$ group structures with neutral element $0$.

It has, up to permutation, a unique field structure with $0$ as zero, whose automorphism group is trivial. Hence it has $(p-1)!$ field structures.

It remains to count the number of group structures on $R-0$. Let's just count the cyclic ones: the automorphism group being of order $varphi(p-1)$, on $p-1$ elements there are $(p-1)!/varphi(p-1)$ cyclic group structures.

Hence the number of pairs $(+,cdot)$ on $R$ with $+$ a group structure with $0$ neutral and $R$ a cyclic group structure on $R-0$ is $(p-2)!(p-1)!varphi(p-1)$, while among them, only $(p-1)!$ are field structures. So all the other ones are non-distributive.

On has $(p-2)!(p-1)!varphi(p-1)>(p-1)!$ as soon as $(p-2)!varphi(p-1)>1$, that is, for $pge 5$ (for $p=2,3$, since the only group structures on $p-1$ elements are cyclic this reproves that distributivity is automatic). For instance, for $p=5$ one has 288 possible $(+,cdot)$, among which only $24$ are distributive.