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To explain how addition and multiplication works in Galois fields, almost all the resources use the example of smallest finite field $\mathrm{GF}(2)$ that has elements $\{0,1\}$.

How can we define these properties for a $\mathrm{GF}(2)$ with elements $\{+1,-1\}$.

Can someone please explain the following statement:

Let $u$ be in $\mathrm{GF}(2)$ with the elements $\{+1,-1\}$, where $+1$ is the "null" element under the $\oplus$ addition.

What are the implications of this.

Reference: Hagenauer, J., "The exit chart - introduction to extrinsic information transfer in iterative processing," in Signal Processing Conference, 2004 12th European , vol., no., pp.1541-1548, 6-10 Sept. 2004

Thank you

Björn Friedrich
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NAASI
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    It's just different names for familiar symbols. "+1" is zero, and "-1" is one. If you like, think of the isomorphism $f(x)=1-2x$ mapping ${0,1}$ into ${+1,-1}$. – vadim123 Dec 28 '15 at 17:17
  • Better question: why.... ? – Dustan Levenstein Dec 28 '15 at 17:23
  • This came up in the context of communication. Since bits {0,1} are always modulated, e.g BPSK modulation 0-->+1 & 1-->-1, before sent over channel, maybe that is why the authors defined GF(2) with elements {+1,-1}. – NAASI Dec 28 '15 at 17:27
  • Okay, makes sense. – Dustan Levenstein Dec 28 '15 at 17:39
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    This is accomplished by using an additive character, i.e. the homomorphism $e: GF(2)\to\Bbb{C}^*$ defined as $$e(x)=(-1)^x.$$ It turns addition in $GF(2)$ into multiplication of usual numbers. There are such homomorphisms for the additive groups of all finite fields. Unfortunately they are not usually injective, so we need to use many of those to extract enough information. This lead to discrete Fourier analysis. This in the form of Walsh-Hadamard transform. – Jyrki Lahtonen Dec 28 '15 at 23:29

1 Answers1

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If you use $+1$ as the additive identity, everything else is determined. Here are the operation tables.

$$\begin{array}{c|cc} \oplus&+1&-1\\ \hline +1&+1&-1 \\ -1&-1&+1\end{array}$$

$$\begin{array}{c|cc} \odot&+1&-1\\ \hline +1&+1&+1 \\ -1&+1&-1\end{array}$$

paw88789
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