Argthtjtr

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.

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
$$
acdot b = left{ begin{array}{cl} 0& a=0 \ 0 & b=0 \
1 & a = b= 5 \
5 & a=5 wedge b in [1,4]\
5 & b=5 wedge a in [1,4]\
ab pmod{5}& mbox{otherwise}end{array} right.
$$
or as a table
$$
begin{array}{c|cccccc} cdot&0&1&2&3&4&5 \ hline
0 & 0&0&0&0&0&0 \
1 & 0&1&2&3&4&5 \
2 & 0&2&4&1&3&5 \
3 & 0&3&1&4&2&5 \
4 & 0&4&3&2&1&5 \
5 & 5&5&5&5&5&1
end{array}
$$
The group properties, as well as commutativity, are easily checked.

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

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). 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 $mathbb{Z}/3mathbb{Z}$ is preserved by relabeling $1$ and $2$. 

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 $(R,+) cong (mathbb{Z}/nmathbb{Z},+)$, 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 $(R,+) cong (mathbb{Z}/pmathbb{Z},+)$, so let's just identify them, i.e. take $(R,+) := (mathbb{Z}/pmathbb{Z},+)$. Suppose $pge 7$. Define $cdot$ on $(mathbb{Z}/pmathbb{Z})-{0}$ 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 $mathbb{Z}/pmathbb{Z}$. 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. Take $R = mathbb{Z}$, with $+$ being regular addition. For example, let $S = mathbb{Z}setminus {0}$, 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 $mathbb{Z}$ 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.