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I need a projection from $\mathbb{F}_{2^{n}}$ to $\mathbb{F}_{2^{n-1}}$. I was thinking in a projection of vector spaces, but i want to know if there is a "canonical" projection or something like that, because tha will be very helpful.

Maybe i should reformulate my question: i would like to find some kind of mapping from $\mathbb{F}_{2^{n}}$ to $\mathbb{F}_{2^{n-1}}$ x $\mathbb{F}_{2}$

Dimitri
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    Projection as fields? No such exists, as any homomorphism from a field is injective (or into the $0$-ring), and clearly there is no injective map to a smaller finite field. – Tobias Kildetoft Feb 27 '13 at 21:27
  • i mean a projection of the vector spaces. I need something like that to somehow "split" the field in 2 parts. For example the elements with last coordinate 1 and the elemnts with last coordinate 0 – Dimitri Feb 27 '13 at 21:31
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    Dimitri: At present your question is underdetermined. Please say a little more about what you are looking for. – Pete L. Clark Feb 27 '13 at 22:19
  • Agree with everything that Msrs Kildetoft, Clark and Lubin have said. But I'm curious as to why you think such a mapping would be helpful? I cannot think of a natural reason for that, and I've played with these fields in my time quite a bit. What exactly do you want to achieve with this? – Jyrki Lahtonen Feb 28 '13 at 07:25
  • I am working with two different description of Kerdock codes, and i want to show that they coincide, but...is a little complicated to give more details in a comment – Dimitri Feb 28 '13 at 17:48
  • My favorite description of Kerdock codes uses the ring of Witt vectors of length two. Those rings are characteristic four. The Nechaeev permutation also has a nice description in that language. My first PhD student used that language in his dissertation. A relevant point related to this thread may be that inside the length 2 Witt vectors we have cyclic groups that are twice the sized of the multiplicative group of the finite field. – Jyrki Lahtonen Feb 28 '13 at 22:38
  • @JyrkiLahtonen Thanks very much for the reference :) it looks really interesting – Dimitri Mar 01 '13 at 04:58

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You must remember that $\mathbb F(p^{n-1})$ is not contained in $\mathbb F(p^n)$, except of course in the case $n=2$. So I wonder just what you meant by the word “projection”. Did you just mean a surjective vector-space morphism (as $\mathbb F(p)$-spaces)? That’s all you can hope for, and there’s certainly nothing canonical, except to try to insure that the elements of the intersection-field $\mathbb F(p)$ are left fixed.

Lubin
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Yes, there is a canonical projection, which forgets the last coordinate.

Note that there can be only such a projection that is a homomorphism of vector spaces and not that of fields, because every ring homomorphism $K\to R$ from a field is injective or constant $0$.

Berci
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    How is forgetting a coordinate canonical? It relies on picking a basis. – Tobias Kildetoft Feb 27 '13 at 21:09
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    @dimitri And, of course, forgetting any single coordinate will do just as well to produce a projection. – rschwieb Feb 27 '13 at 21:10
  • Yes, i was going to say the same thing, forget a coordinate was my first thought, but then i see the problem on picking a basis. Anyway i was thinking if there is another way of define an "standard" projection – Dimitri Feb 27 '13 at 21:13
  • @rschwieb: Yes, but they are a bit 'less canonical':) Tobias: In ${\Bbb F_2}^n$ there is already a canonical basis: $(1,0,0,...)$, $(0,1,0,..)$, ... – Berci Feb 27 '13 at 21:15
  • @Berci Why is the last coordinate so special? Why not the first coordinate? They all seem the same to me :) – rschwieb Feb 27 '13 at 21:34
  • You are right. However, they are usually defined as $A^n={(a_1,a_2,..a_{n-1},a_n) }$ and so $A^{n-1}={(a_1,a_2,..,a_{n-1}) }$. Thus, the coordinate $a_n$ is lost. But it starts being rather philosophic.. Also, many times erasing the first coordinate better fits the context.. – Berci Feb 27 '13 at 21:44