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I know that in infinite fields, such as $\mathbb{C}$, the mapping $e^x$ is a homomorphism from the additive group to the multiplicative group, and I was just wondering if in any finite field, there exists a (non trivial) homomorphism between the two.

ASKASK
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    Presumably you mean nontrivial homomorphism? If not, sending $\phi(k) = 1_F$ is a group homomorphism from the additive group $F$ to the multiplicative group $F$. – walkar Feb 18 '15 at 02:24
  • Yes sorry I did mean non trivial – ASKASK Feb 18 '15 at 02:24

2 Answers2

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This is impossible for finite fields.

Consider a finite field of order $q$; then the additive group also has order $q$. However, the multiplicative group has order $q - 1$ which does not share any common factors with $q$. Since the order of the image of an element $x$ under a homomorphism must divide the order of $x$ by Lagrange's theorem, it follows that any such homomorphism must be trivial.

Qudit
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  • Great answer. I was going to try to do something with cyclic versus not cyclic, but I recalled that not all finite fields are cyclic additive groups. – walkar Feb 18 '15 at 02:28
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    I was planning on posting this same argument. Darn! One slight note: You don't actually need the hypothesis that $q$ is a prime power to prove this. $q$ and $q-1$ are coprime for any $q$. (I note this just because proving that the order of a finite field is a prime power requires much heavier machinery than just Lagrange's theorem) – Milo Brandt Feb 18 '15 at 02:30
  • That's true! I made your simplification. – Qudit Feb 18 '15 at 02:32
  • Great answer! Thanks! – ASKASK Feb 18 '15 at 02:36
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    Is there a countable field with such a homomorphism? It doesn't look like power maps do the trick on $\Bbb Q$ and $\Bbb A$, since the power of a rational number is algebraic and the power of an algebraic is transcendental (usually). – Mario Carneiro Feb 18 '15 at 02:42
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    I wondered something similar, so I created this question: http://math.stackexchange.com/questions/1153994/homomorphisms-between-additive-and-multiplicative-groups-of-fields – Qudit Feb 18 '15 at 02:54
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    @Mario To give a simple example of a countable field with a map from additive to multiplicative group, consider the set $S$ to be the smallest subfield of $\mathbb R$ closed under the map $x\mapsto e^x$. This is countable and $x\mapsto e^x$ is a homomorphism. – Milo Brandt Feb 18 '15 at 02:56
  • @Meelo: what guarantees that $S$ is countable? – Alexandre Halm Feb 18 '15 at 04:25
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    @AlexH It is finitely generated, so in correspondence with finite strings over a finite alphabet - every element can be written as a finite tree of applications of the unary operations of taking negatives, taking reciprocals, and exponentiation, the binary operations of addition and multiplication, and the constants $0$ and $1$. – Milo Brandt Feb 18 '15 at 04:33
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Sometimes there is a homomorphism from the additive group of integers modulo N-1 and the multiplicative group of integers modulo N.

For example when N=5, given the additive group of four integers A = {0,1,2,3} and the finite field M = {0,1,2,3,4}, we can map each element a of A to a corresponding element m from the multiplicative group of M with m = 2^a mod 5.

With these groups, whenever z = x + y modulo (N-1),

it is also true that 2^z mod N = 2^x * 2^y modulo N.

(where addition, multiplication, exponentation, modulo, etc. are the common integer operations).

This is true whenever N is a Fermat prime. (This is related to the "invertable multiplication", also called "multiplication IDEA style", used in the IDEA cipher).

I suspect there may be other mappings that work for other prime values of N.

(The additive group of four integers A = {0,1,2,3} is not exactly the same as the finite field of four elements GF(4), but they both have 4 elements, so perhaps this is close to what you are looking for.)

David Cary
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  • @MorganRodgers: In the first example, the finite field A is the additive group of integers modulo 4 -- or in other words, A is the set of four integers {0, 1, 2, 3}. In the second more general example, A is the additive group of of integers modulo N-1, where N is a Fermat prime. (The first example is a special case of the second example). How could I edit this answer to make that more clear? – David Cary Jul 07 '15 at 23:26
  • @MorganRodgers: You are right that the additive group of integers modulo 4 has a completely different addition operator than GF(4), the finite field of four elements {0, 1, a, 1+a}. I don't know what I was thinking. You are probably right that I am "simply" mapping GF(5) to itself using additive notation in A and multiplicative notation in M -- but isn't that almost exactly what the original question asked for? – David Cary Jul 08 '15 at 13:38