Deceptive Isomorphism

**ThePerfectHacker** · July 18th 2006, 11:36 AM

I have a question which might be useful if true.

If two fields have their additive groups and multiplicative groups isomorphic does it mean that the two fields are isomorphic?

(Remember, all that is know is that there exists bijectives $\phi: <F,+>\to <E,+>$ and $\psi: <F^*,\cdot >\to <E^*, +>$ . But the fact that there exists a $\mu: <F,+>\to <E,+> \mbox{ and } \mu:<F^*,\cdot >\to <E^*, \cdot>$ was never demonstrated.

**rgep** · July 30th 2006, 04:17 AM

No. Consider the fields ${\mathbb Q}(\sqrt{-7})$ and ${\mathbb Q}(\sqrt{-163})$ . In each case the additive group is isomorphic to ${\mathbb Q} \oplus {\mathbb Q}$ and the multiplicative group is isomorphic to $C_2 \oplus {\mathbb Z}^\omega$ , the latter because the ring of integers has two roots of unity and is a UFD, so that the multiplicative group is generated by -1 and the prime elements.

**ThePerfectHacker** · July 30th 2006, 06:32 AM

What does $\oplus$ mean in this case?

**rgep** · July 30th 2006, 08:17 AM

Direct sum: the same thing as Cartesian product in this case.

**ThePerfectHacker** · July 30th 2006, 10:16 AM

Originally Posted by rgep

Direct sum: the same thing as Cartesian product in this case.

But the direct products of two field is not a field.

**rgep** · July 30th 2006, 11:43 AM

Originally Posted by ThePerfectHacker

But the direct products of two field is not a field.

Agreed. But what I'm saying is that the two fields I mention each have additive group isomorphic to ${\mathbb Q} \oplus {\mathbb Q}$ : that is, to two copies of the additive group of the rationals.

**BubbleBrain_103** · July 30th 2006, 12:02 PM

I'm just following this discussion.... rgep, could you explain the multiplicative group you mentioned? What are C_2 and Z^omega, and what are the group isomorphisms between the product of the two and the multiplicative groups of your two fields?

**rgep** · July 30th 2006, 01:33 PM

Since each of the fields I mentioned has the property that the corresponding ring over the integers (its maximal order or ring of integers) has unique factorisation, every element of the field is a unit (and here the only possibilities are +- 1) times a product of prime elements. Since in fact the same is true of the rationals you may prefer to think of that as your example. So every element of the multiplicative group is uniquely expressible as $(-1)^{e_0} \cdot \pi_1^{e_1} \cdot \pi_2^{e_2} \cdots$ , where the $\pi_i$ are the prime elements, $e_0$ is either 0 or 1 (taken mod 2) and the $e_i$ are integers, with the property that only finitely many of them are non-zero.

**topsquark** · July 30th 2006, 03:40 PM

Originally Posted by rgep

Since each of the fields I mentioned has the property that the corresponding ring over the integers (its maximal order or ring of integers) has unique factorisation, every element of the field is a unit (and here the only possibilities are +- 1) times a product of prime elements. Since in fact the same is true of the rationals you may prefer to think of that as your example. So every element of the multiplicative group is uniquely expressible as $(-1)^{e_0} \cdot \pi_1^{e_1} \cdot \pi_2^{e_2} \cdots$ , where the $\pi_i$ are the prime elements, $e_0$ is either 0 or 1 (taken mod 2) and the $e_i$ are integers, with the property that only finitely many of them are non-zero.

You know I am just in awe of algebra because you can take something fantastically complicated and do a short proof of it. (The amazing thing is I was able to follow the above argument!

)

-Dan

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