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Let $GF(p^n)$ be a finite field, then the additive group is isomorphic to $\mathbb Z / (p\mathbb Z)^n$, and it is simple to compute in that group. The multiplicative group is always cyclic (a standard result), so we have $GF(p^n)^{\times} \cong \mathbb Z/((p^n - 1)\mathbb Z$. But how to compute in the multiplicative group when the elements are named as in the additive group. For example for $GF(5)$ we see for example here that we could not take just the numbers, multiply them and take the modulos to $4$, but accidently the modulos to $5$ works in this example (a coincidence?). To compute modulo $4$ we had to identify $1$ with the residue class of $\overline 0$; $2$ with the residue class of $\overline 1$; $3$ with the class of $\overline 3$ and $4$ with the class of $\overline 2$.

I find this kinda unintuitive and cumbersome? How do you compute with finite fields in the multiplicative group; as shown above to bring in the "standard multiplication tables" everyone learns at school into play by residue calculus is error-prone as how the elements in the field are numbered need not represent the residue classes.

But maybe I have overlooked something all the years and in the end it is kinda easy to compute in the multiplicative group (as it is in the additive group, as they are just the elementary abelian groups written above)??

StefanH
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  • I don't know what's best, but whenever I use a modest size field, say in a computer program, I begin by building a discrete logarithm table. When $p^n<10^6$ or thereabouts, this is the most efficient. Basically I have a table giving the isomorphism between $GF(p^n)^*$ and $\Bbb{Z}/{p^n-1}\Bbb{Z}$. See this Q&A for examples. If the field is seriously large (as in elliptic curve crypto), then you need something else. I have mostly managed to avoid those :-) – Jyrki Lahtonen Dec 15 '15 at 21:33

1 Answers1

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If $f(X) = X^n + g(X)$ is a monic irreducible polynomial of degree $n$ over $GF(p)$, you can represent $GF(p^n)$ as polynomials $\sum_{j=0}^{n-1} x_j \alpha^j$ where $x_j \in GF(p)$ and $\alpha$ is a root of $f(X)$.
The addition works coordinatewise mod $p$.
For multiplication, expansion of $\left(\sum_{j=0}^{n-1} x_j \alpha^j\right)\left(\sum_{j=0}^{n-1} y_j \alpha^j\right)$ gives an expression of the form $\sum_{j=0}^{2n-2} z_j \alpha^j$, which we can then reduce mod $f(\alpha)$. Precompute the representations of $\alpha^j$ for $n \le j \le 2n-2$ recursively: if $\alpha^j = \sum_{k=0}^{n-1} c(j,k) \alpha^k$, then $$\alpha^{j+1} = \sum_{k=0}^{n-2} c(j,k) \alpha^{k+1} + c(j,n-1) \alpha^n = \sum_{k=0}^{n-2} c(j,k) \alpha^{k+1} - c(j,n-1) g(\alpha)$$ (where the coordinatewise arithmetic is done mod $p$). Then $$ \sum_{j=0}^{2n-2} z_j \alpha^j = \sum_{j=0}^{n-1} z_j \alpha^j + \sum_{j=n}^{2n-2} \sum_{k=0}^{n-1} z_j c(j,k) \alpha^k \mod p$$ which amounts to a vector-matrix multiplication and addition.

EDIT: For example, take $GF(5^3)$. A suitable polynomial is $f(X) = X^3 + X + 1$. We have $$ \eqalign{ \alpha^3 &= 4 \alpha + 4\cr \alpha^4 &= 4 \alpha^2 + 4 \alpha\cr}$$ And then, say, $$ \eqalign{(\alpha^2 + 2 \alpha + 3)(\alpha^2 + 3 \alpha + 4) &= \alpha^4 + 3 \alpha^2 + 2 \alpha + 2\cr &= (4 \alpha^2 + 4 \alpha) + 3 \alpha^2 + 2 \alpha + 2\cr &= 2 \alpha^2 + \alpha + 2\cr}$$

Robert Israel
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  • Thanks for your post! Could you supply an example? Also how could someone reduce mod $f(\alpha)$ if $f(\alpha) = 0$ as this is a root? – StefanH Dec 15 '15 at 17:01
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    What I mean by reduction mod $f(\alpha)$ is, given a polynomial expression $h(\alpha)$, write $h(X) = u(X) f(X) + v(X)$ where $v(X)$ has lower degree than $f(X)$, and so (because $f(\alpha) = 0$) $h(\alpha) = v(\alpha)$. – Robert Israel Dec 15 '15 at 17:35
  • Okay, now I see. Seems not that easy to compute in arbitrary finite fields, for $\mathbb Z/p\mathbb Z$ we could just do ordinary multiplication and then mod $p$, easy to achieve. For fields of order $p^k, k > 1$ it seems not that easy. – StefanH Dec 15 '15 at 19:45
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    If $n$ is large, the real sticking point, I think, is finding an irreducible polynomial of given degree over $\mathbb Z/p \mathbb Z$. – Robert Israel Dec 15 '15 at 23:53
  • Are there any heuristics for finding them? – StefanH Dec 16 '15 at 12:02
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    The number of irreducible monic polynomials of degree $n$ over $\mathbb Z/p\mathbb Z$ is $\dfrac{1}{n} \sum_{d | n} \mu(n/d) p^d$. So take a random monic polynomial of degree $n$ whose constant coefficient is nonzero, and it has roughly probability $1/n$ of being irreducible. Keep going until you get one that is irreducible. – Robert Israel Dec 16 '15 at 16:05