Alice chooses

i) A large prime $p_A$ (say 200 to 300 digits)
ii) A primitive element $\alpha_A$ modulo $p_A$,
iii) A (possibly random) integer $d_A$ with $2 \le d_A \le p_A-2$.

Alice computes

iv) $\beta_A \equiv \alpha_A^{d_A} \pmod {p_A}$.

Alice's public key is $(p_A,\alpha_A,\beta_A)$. Her private key is $d_A$.

What is meant by the notation $\alpha_A$, $p_A$, and what is meant by a 'primitive element', does it mean a small number or something else entirely?

  • $\begingroup$ What do you mean by "a (subscript a)"? There is no a with a subscript anywhere in the quote? Do you mean $\alpha_A$? $\endgroup$ Jun 26, 2013 at 10:09

1 Answer 1


The subscript $A$ indicates that these numbers ($p_A$, $\alpha_A$, etc.) are the ones involved in Alice's key. In a description of a protocol with more participants each having their own key, Bob's public key would be $(p_B, \alpha_B, \beta_B)$, and so on.

A primitive element of a finite field is a generator for the multiplicative group, i.e. the set $\{1, \alpha_A, \alpha_A^2, \alpha_A^3, \ldots, \alpha_A^{p_A-1}\}$ must be the whole set $\{1, 2, 3, \ldots, p-1\}$ of nonzero elements of the field $\mathbb{F}_p = \mathbb{Z}/p\mathbb{Z}$, i.e. any nonzero element of the field can be expressed as a power of $\alpha_A$. Another way of expressing this property is that $\alpha_A$ must have the order $p-1$ (the largest possible order of an element of the field), which means that none of the $\alpha_A^k$ for $1 \le k \le p-2$ are equal to $1$ modulo $p$.


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