Is it possible to produce a field with three operations?

A Field $$F$$ is defined such that:

• There is a first operation, say $$+$$ (addition), that is closed, associative and commutative over $$F$$, and has an identity element $$0 \in F$$ and an inverse element $$-x \in F$$ for each $$x \in F$$ (that is, $$(F, +)$$ is a commutative group);
• There is a second operation, say $$\cdot$$ (multiplication), that is closed, associative and commutative over $$F \setminus \{0\}$$, and has an identity element $$1 \in F \setminus \{0\}$$ and an inverse $$1/x \in F \setminus \{0\}$$ for each $$x \in F \setminus \{0\}$$ (that is, $$(F\setminus \{0\}, \cdot)$$ is a commutative group);
• The second operation distributes over the first operation.

My question is this:

Is it possible to define a third operation, say $$*$$, for $$F$$ such that $$(F\setminus \{0,1\}, *)$$, is a commutative group and $$*$$ distributes over multiplication?

In other words, is it possible to have $$(F, +, \cdot, *)$$ where both $$(F,+, \cdot)$$ and $$(F\setminus \{0\},\cdot, *)$$ are fields?

My gut feeling is no.
In all fields I am familiar with $$(\mathbb{Q,R,C,Z}_p)$$, the multiplication is provably equivalent to that field's analytic continuation of iterated addition, but the analytic continuation of iterated multiplication (that is, exponentiation) is provably not a commutative group operation on $$(F \setminus \{0,1\})$$ (and catastrophically so, it fails on all counts! It's not even well-defined!)
However, the coincidence that multiplication is equivalent to iterated addition feels like just that, a coincidence, as the definition of field makes no claims about the relation between the two operations other than the distributive property.
Is my gut feeling correct, or am I right to be skeptical?

• Very interesting concept +1! – Don Thousand Nov 12 at 0:23
• You're right that fields seem to have a lot of coincidental properties, but that really is the general situation as far as I know. All finite fields have order $p^n$ for some prime $p$ and are isomorphic to all other finite fields of the same order. Every field of characteristic zero is infinite and contains a copy of the rationals. Any field of characteristic $p \neq 0$ contains a copy of $\mathbb{Z}_p$. In every case, there is at least a subfield where multiplication is in some sense iterated addition. – Charles Hudgins Nov 12 at 0:44
• Fun corollary of the 4 answers: Up to isomorphism, there is a unique field $F$ which admits a 4th operation $\times$ such that $(F, +, \cdot), (F-\{0\}, \cdot, \ast), (F-\{0,1\}, \ast, \times)$ are all fields. It is the field with $4$ elements. – Jason DeVito Nov 15 at 14:18

Among finite fields with characteristic $$\neq 2$$, David's example is unique. Specifically:

Suppose $$F$$ is a finite field with char $$F\neq 2$$ which has an extra operation $$\ast$$ as described by the OP. Then $$F = \mathbb{Z}_3$$ with $$a\ast b = \min\{a,b\}$$.

Proof: Suppose $$F$$ a finite field of characteristic $$\neq 2$$. Then $$-1\neq 1\in F$$. However, $$(-1)\cdot (-1) =1$$. Thus, the field $$(F- \ \{0\}, \cdot, \ast)$$ is characteristic $$2$$, so $$|F -\{0\}| = 2^n$$ for some $$n \geq 1$$.

On the other hand, the multiplicative group of a finite field is cyclic (see, e.g. this MSE question). Since the fields whose additive structure is cyclic are exactly the prime fields, we deduce that $$n = 1$$.

Thus, in this case, $$|F| = 3$$. This already implies that $$F = \mathbb{Z}_3$$. It remains to compute $$\ast$$.

We know that $$\ast$$ must have the property that $$2\ast 2 = 2$$ since $$\ast$$ maps $$F -\{0,1\}$$ to itself, and we also know that $$1\ast 1 = 1$$ because it's the additive identity of the field ($$F-\{0\}, \cdot, \ast).$$

Next, we see that \begin{align*} 2 &= 2\ast 2\\ &= 2\ast(1\cdot 2)\\ &= (2\ast 1) \cdot (2\ast 2)\\ &= (2\ast 1) \cdot 2,\end{align*} so $$2 = (2\ast 1) \cdot 2$$, so $$2\ast 1 = 1$$. Thus, $$2\ast 1 = \min \{1,2\}$$.

Likewise, \begin{align*}2\ast 0 &= 2\ast(0\cdot 2)\\ &= (2\ast 0) \cdot (2\ast 2)\\&= (2\ast 0)\cdot 2.\end{align*} If $$2\ast 0\neq 0$$, we can divide by it to learn $$2 = 1$$, which is absurd, so $$2\ast 0 = 0$$. That is, $$2\ast 0 = \min\{0,2\}$$.

Moreover, \begin{align*} 1\ast 0 &= (2\cdot 2)\ast 0\\ &= (2\ast 0) \cdot (2\ast 0)\\ &= 0.\end{align*}

Finally, \begin{align*} 0\ast 0 &= 0\ast (0\cdot 1)\\ &= (0\ast 0)\cdot (0\ast 1)\\ &= 0,\end{align*} so $$0\ast 0 = \min\{0,0\}$$.

Thus, $$a\ast b = \min\{a,b\}$$ in all cases. $$\square$$

Suppose $$F_{2^n}$$ is a finite field of characteristic $$2$$ which supports an operation $$\ast$$ as described by the OP. Then $$2^n - 1$$ is a Mersenne prime.

(I am not claiming lonza's operation $$\ast$$ is unique. I have no idea.)

Proof: Suppose $$F_{2^n}$$ is a finite field with $$2^n$$ elements. If $$n = 1$$, (i.e., $$|F| = 2$$), then $$F-\{0,1\}$$ is the empty set, so $$(F-\{0,1\},\ast)$$ doesn't form a group since it doesn't have an identity. Thus, we may assume $$n\geq 2$$.

Now, $$F-\{0\}$$ is again the underlying space of a field, so $$|F-\{0\}| = p^m$$ for some prime $$p$$. Thus, $$p^m + 1 = 2^n$$, so $$1 = 2^n - p^m$$.

By Mihalescu's Theorem, there is no solution to $$2^n - p^m = 1$$ with $$m > 1$$, so $$m = 1$$. Thus, $$2^n - 1 = p$$ is a Mersenne prime. $$\square$$

No infinite field $$F$$ supports such an operation $$\ast$$.

Proof: As Geoffrey notes in his answer, $$F$$ must have characteristic $$0$$, so it contains a copy of $$\mathbb{Q}$$. Further, since $$-1\neq 1$$ and $$(-1)^2 = 1$$, $$F-\{0\}$$ must be characteristic $$2$$. This means that $$a\cdot a = 1$$ for any $$a\in F - \{0\}$$. But this is false when $$a=2\in \mathbb{Q}-\{0\}\subseteq F-\{0\}.$$

• This is a most thorough answer, thank you. – No Name Nov 15 at 4:15
• @NoName: I'm glad you like it, but it's not clear to me that I deserve the green arrow (though I don't know who should receive it). My answer wouldn't work without the other three answers! – Jason DeVito Nov 15 at 14:15
• I know, and I wish I could give the tick to all of them. As it stands, you reference all the others and you tie it all together, and I think that means you deserve the tick. – No Name Nov 16 at 12:57

I don't know whether this is possible in general, but we can find at least once case of a "generalized" field. Consider the field $$(F=\mathbb{Z}_3,+,\cdot,*)$$ where $$+$$ and $$\cdot$$ are defined as normal, and $$a*b$$ is defined as $$\min\{a,b\}$$, where $$0<1<2$$. This is clearly a commutative operation. Also, $$F\setminus\{0,1\}$$ is the trivial group over $$*$$, since we have $$2*2=2$$. Finally, it's a little tedious, but we can show that $$*$$ is indeed distributive over $$\cdot$$ as follows:

For $$0*x$$:

$$0*(a\cdot b)=0=(0*a)\cdot (0*b),$$

For $$1*x$$:

$$1*(0\cdot a)=1*0 = (1*0) \cdot (1*a),$$

$$1*(1\cdot a) = 1*a = (1*1) \cdot (1*a),$$

$$1*(2\cdot 2) = 1*1=1=(1*2)\cdot(1*2),$$

where we only need to consider $$2\cdot2$$ in the last since the other values of $$a$$ were covered in the previous cases.

For $$2*x$$:

$$2*(0\cdot a)= 2*0 = (2*0) \cdot (2*a),$$

$$2*(1 \cdot a) = 2*a =(2*1) \cdot (2*a),$$

$$2*(2\cdot2) = 2*1 = (2*2)\cdot(2*2).$$

Note that in this generalized field, $$0$$ is still absorptive; i.e. $$0*a=a*0=0$$ for any $$a$$.

Yes, it is possible, at least in the finite case.

Suppose that $$K_1$$ and $$K_2$$ are two fields with the additive group of $$K_1$$ isomorphic to the multiplicative group of $$K_2$$.

If $$K_1$$ and $$K_2$$ are finite, then, since the multiplicative group of a finite field is cyclic, the field $$K_1$$ must be a prime field. Let the orders of $$K_1$$ and $$K_2$$ be $$p$$ and $$q^n$$ respectively, so that $$p=q^n-1$$. If $$n=1$$, then $$p$$ and $$q$$ must be two and three respectively. If $$n>1$$, then $$q$$ must be two and $$p$$ must then be a Mersenne prime.

Hence, $$K_1$$ and $$K_2$$ could be one of the following:

• $$GF(2)$$ and $$GF(3)$$
• $$GF(2^n-1)$$ and $$GF(2^n)$$, where $$2^n-1$$ is a Mersenne prime (of course, $$n$$ must also be prime)

In the infinite case, all the elements of $$K_2$$ other than zero and one must have the same multiplicative order. If this order were finite (lets say its $$n$$), then it would contradict the fact that a nonzero polynomial over a field cannot have more roots than its degree (in this case, the polynomial of interest is $$x^n-1$$). Hence, $$K_2$$ must have a torsion-free multiplicative group, which implies that $$K_1$$ and $$K_2$$ must have characteristic zero and two respectively.

But I do not know whether any infinite examples exist.

If $$\ p=2^n-1\$$ is a Mersenne prime, let $$\ q\$$ be a primitive element of $$\ \mathbb{F}_{2^n}^{^\times}\$$ and define: $$(q^a)*(q^b)=q^{ab}\ .$$ for $$\ a,b=0,1,\dots,p-1\$$. Then \begin{align} q^a*(q^bq^c)&=q^a*q^{b+c}\\ &=q^{a(b+c)}\\ &=q^{ab}q^{ac}\\ &=(q^a*q^b)(q^a*q^c)\ . \end{align} The field $$\ (\mathbb{F}_{2^n}^{^\times},.,*)\$$ is of course just (isomorphic to) the field $$\ \mathbb{F}_p\$$.

Uniqueness: In his answer, Jason DeVito raises the question of whether the operation * is unique in this case. Since $$\ q\$$ could be chosen to be any of the $$\ \varphi(p)=p-1\$$ primitive elements available, it isn't, of course, unique in the strictest sense, but it's not difficult to show that any such operation in such fields must be given by this definition for some primitive element.

Let $$\ e\$$ be the $$*$$-identity. Then $$\ e\ne0\$$ and $$\ e\ne 1\$$ (the identity element for $$\cdot\$$), therefore $$\ e\$$ must be primitive, because the order of $$\ \mathbb{F}_{2^n}^{^\times}\$$ is prime. Now, if $$\ r,s\in\mathbb{F}_{2^n}^{^\times}\$$, then $$\ r=e^a\$$ and $$\ s=e^b\$$ for some $$\ a,b=0,1,\dots,p-1\$$, and $$\ r*s=e^a*e^b=(e^a*e)^b\$$ by the distributivity of $$*$$ across $$\cdot\$$, and $$\ e^a*e=e^a\$$ because $$\ e\$$ is the identity for $$*$$. Therefore, we have $$\ r*s=(e^a)^b=e^{ab}\$$

• Thanks for including the final detail of the classification! – Jason DeVito Nov 13 at 19:52