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I'm reading Arnold's "Geometry and dynamics of Galois fields" (the russian version is here), he mentions --

Ch1. Construction of the table of a field

Definition. By a Galois field we mean a field with finitely many elements. The simplest example is given by the residue class field $Z_p = Z/(pZ)$ modulo a prime $p$. This field consists of $p$ elements.

The following theorem about such fields is classical.

Theorem. The number of elements in an arbitrary Galois field is equal to $p^a$, where $p$ is a prime and $a$ is a positive integer. A field with this number of elements exists (for any number of the form $p^a$) and is unique (up to isomorphism of fields, of course).

To represent the operations of addition and multiplication in a field, we enumerate its elements as follows. The field has a zero element 0 (which is uniquely determined by the condition that $0 + x = x$ for any element $x$), and the other elements form a group (the multiplicative group $F\backslash 0$ of the field $F$) with respect to the multiplication, that is, they have inverse elements $x^{−1}$ for which $xx^{−1} = 1$ (where the element 1 is uniquely determined by the condition that $y1 = y$ for any element $y$).

It turns out that the multiplicative group of a field is always cyclic (of order $p^\alpha − 1$ for a field with $p^\alpha$ elements), that is, every element of this group can be expressed in terms of one of these elements, say, $A$ (a ‘multiplicative generator’), in the form $$A,A^2, A^3, \dots, A^{p^\alpha−1} = 1.$$

The generator $A$ can be chosen in different ways. Namely, any $B = A^k$ with $k$ prime to $p^\alpha − 1$ can be taken as the generator.

Thus, the number of multiplicative generators is equal to the value of the Euler function, $\varphi(p^\alpha − 1)$.

Example. For a field with $p^\alpha = 49$ elements the number of generators is equal to $\varphi(48) = 16$, namely, the generators are the elements $A^k$ with $$k = 1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37, 41, 43, 47.$$

The multiplicative generators form a multiplicative subgroup (in the residue class ring modulo pa − 1). For instance, if $p^\alpha = 49$, then $$5 ·5 = 25, 7 ·7 = 1, 5 · 11 = 7.$$

With respect to addition, a field of $p^a$ elements is a $p$-vector space (the commutative group $(\mathbb Z_p)^a$), and thus the $p^a$ elements of this field can be inscribed into cells of a ‘finite torus’ of dimension $a$, $$x = x_1e_1 + \cdots + x_ae_a$$, where $(e_1, \cdots, e_a)$ are the additive generators and the coefficients $x_j$ are residues modulo $p$.

Example. A field of 49 elements consists of the combinations of the form $$x = \alpha_1+ \beta A$$, where $1$ is the ‘identity’ element of the field and $A$ is the multiplicative generator discussed above (in the general case one can take the elements $\{A,A^2, \cdots, A^{a−1}, A^a=1\}$ as the additive generators).

For a field of $p^2$ elements we conclude that (1) $$A^2 = \alpha 1 +\beta A $$ with some coefficients $\alpha$ and $\beta$ in $\mathbb Z_p$.

I'm confused here. Taking $49$ as example, $p=7, a=2$, what is a $p$-vector space? Taking $A=5$, how could (1) works?

$$A^2 = 25 = \alpha 1 + \beta 5$$

what are the values of $\alpha$ and $\beta$ here? Seems $\alpha$ and $\beta$ are in $\mathbb Z_5$ so could only take values from $\{0, 1,2,3,4\}$, $\alpha 1 + \beta 5$ won't makes 25?

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    $\begingroup$ Short for vector spaces over a field of $p$ elements. $\endgroup$ – Randall Apr 5 '19 at 1:08
  • $\begingroup$ @Randall i also assumed so. but how to make $25 = \alpha 1 + \beta 5$ with $\alpha$ and $\beta$ in $\{0,1,2,3,4\}$? $\endgroup$ – athos Apr 5 '19 at 1:08
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    $\begingroup$ @Randall thank you so much. now i see where i was lost. I'm reading "Abstract Algebrqa" by Finston and Morandi, a newer version of the script, e.g. its Example 5.27 is the example 5.24 in the notes. A even simpler one is Example 5.26, which is isomorphic to Example 3.30 of $F=\{0,1,a,b\}$, where here $p^a=2$. My mistake was, I thought $F_4$ is just just $Z_4$, $a$ is just $2$ and $b$ is $3$. but actually it is not! Thanks again, I see where i'm lacking in foundation, and will pick those up. $\endgroup$ – athos Aug 12 '19 at 9:03
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    $\begingroup$ Correct. If $F$ is a field of order $p^d$ where $d>1$, $F$ will never be isomorphic to $\mathbb{Z}_{p^d}$. That's because the latter has zero divisors, so is not a field to begin with. $\endgroup$ – Randall Aug 12 '19 at 13:10
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    $\begingroup$ I don't know how you're representing your field, so I can't answer. Though, it doesn't matter. In any field, $e^2=e$ implies either $e=0$ or $e=1$. So, in your case, it is not possible. $\endgroup$ – Randall Aug 26 '19 at 12:37

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