# Tag Info

## New answers tagged finite-field

1

The general case is that of field extension. Given a field $\mathbb{F}_q$ of $q$ elements (in your case, the field is $\mathbb{Z}_p$, the integers modulo a prime $p$), you want to define and do computations in a field $\mathbb{F}_{q^k}$ of $q^k$ elements for some integer $k > 1$. To do so, one first considers $\mathbb{F}_q[X]$ which is the ring of ...

0

If you use a prime $p \equiv 3 \pmod 4$, then an element from $\mathbb{F}_{p^2}$ can be written as $a+ib$, where one has the following rules of calculus: $(a_1+ib_1)+(a_2+ib_2) = (a_1+a_2)+i(b_1+b_2)$ $(a_1+ib_1)\cdot(a_2+ib_2) = (a_1 a_2 - b_1 b_2)+ i (a_1 b_2+ a_2 b_1)$ $(a+ib)^{-1} = (a-i b) / (a^2+b^2)$ Sine $i^2 = -1$, those rules resemble ...

1

Ok, if there is no standard, then "I do it my way": Let $i$ be a root of a fixed minimum polynomial of $GF(p^2)$ and $x$ be an element in $GF(p^2)$. Then $x:= x_1+x_2*i$ can be represented as a vector $(x_1,x_2) \in GF(p)^2$. Now convert $x_1$ and $x_2$ with the FE2OSP function form IEEE P1363 and concatenate them $FE2OSP(x_1)|| FE2OSP(x_2)$. An ...

-3

William Stalling says RSA is based on exponentiation in a finite (Galois) field over integers modulo a prime Source: http://williamstallings.com/Extras/Security-Notes/lectures/publickey.html

3

I'm not sure if I really believe that this is of any use at attacking a stream cipher, but I'll answer your question anyways. What $gf = 0$ means is that, for any input $x$, we have either $f=0$ or $g=0$. Now, in the case of $g = x_0x_1$, then this $g$ will be 0 unless $x_0 = x_1 = 1$. In that case, the first term of $f$, namely $x_0x_2(x_1+1)$ will be 0 ...

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