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GF(3) can be constructed as follows by polynomial $p(x)=x+1$:

$0=0$

$1=1$

$\alpha=2$

GF(5) can be constructed as follows by polynomial $p(x)=x+2$:

$0=0$

$1=1$

$\alpha=3$

$\alpha^2=\alpha\cdot\alpha=3\cdot3\bmod5=4$

$\alpha^3=\alpha^2\cdot\alpha=4\cdot3\bmod5=2$

Question is, what can be defined as element $\alpha$ in GF(2) and what primitive polynomial can be used to construct this field?

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The possible confusion here is that the trivial group (with one element only) doesn't really need a generator. You may use $x-1$ as the generating polynomial in the same way as usual, but there really is nothing to generate :-) –  Jyrki Lahtonen Aug 17 '11 at 4:17
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2 Answers

up vote 1 down vote accepted

A generator of the multiplicative group of a finite field is an element $\alpha$ such that the powers of $\alpha$ include all non-zero elements of the field. The multiplicative group of GF(2) has one element, and thus one generator: $\alpha = 1$.

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Hint: How many element are there in the multiplicative group of $GF(2)$? Based on that can you point at a generator (if you need one)?

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2 elements, 0 and 1, but is it possible to define element $\alpha$ in GF(2)? –  scdmb Aug 16 '11 at 21:11
    
@scdmb: Well, because $\alpha$ has to be in the multiplicative group, it cannot be zero. –  Jyrki Lahtonen Aug 16 '11 at 21:22
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