# The French Field

• Dec 28th 2005, 02:49 PM
ThePerfectHacker
The French Field
Definition: A french field is a field where the additive group is isomorphic to the multiplicative group.
Question: Can there exist a French Field?
Partial Answer: Since the multiplicative group does not contain zero (additive identity) we have that the multiplicative group is a proper subgroup of the additive group. Thus, if there exists an isomorphism then there exists a bijective map from the additive group to the multiplicative group. Thus, by definition of cardinality the additive group has the same cardinality as the multiplicative group. Since, the multiplicative group is a proper subgroup of the additive group then by the definition of an infinite set we see that the field CANNOT be a finite field.
• Dec 28th 2005, 10:41 PM
rgep
One can show the field has to have characteristic two. Otherwise -1 is an element of order two in the multiplicative group and so there has to be an element of order two in the additive group. But now every element of the additive group has order two, and hence so does every element of the multiplicative group. However, the equation x^2=1 can have at most two roots in a field and so the field is finite, a contradiction.
• Dec 29th 2005, 12:55 PM
ThePerfectHacker
How did you reach the result that every element in the additive group must when added to itself is zero? (Note the use of the word order of two was mathematically incorrect but you probably noticed that, anyway it is not important). Because I reached the result since:
$a+a=0$ for some "a" non-zero element of the additive group. Thus,
$x(a+a)=0$ for all "x" in the additive group.
$ax+ax=0$
$a(x+x)=0$
Thus, $x+x=0$ for all "x" Because by the property of divisors of zero.

Let me just add the important step in this explanation since (-1) is not the multiplicative identity element and it has order of two thus there must be a NON-ZER0 element in the additive group of order two.

Anyways rgep your explanation was excellent. I also was thinking of proving it using the concept of polynomials but was unable to. It that your own proof, if it is I love it.