# What does Arnold mean here by $x=\alpha 1 + \beta A$?

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?

• Short for vector spaces over a field of $p$ elements. – Randall Apr 5 '19 at 1:08
• @Randall i also assumed so. but how to make $25 = \alpha 1 + \beta 5$ with $\alpha$ and $\beta$ in $\{0,1,2,3,4\}$? – athos Apr 5 '19 at 1:08
• @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. – athos Aug 12 '19 at 9:03
• 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. – Randall Aug 12 '19 at 13:10
• 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. – Randall Aug 26 '19 at 12:37