Testwiki:Reference desk/Archives/Mathematics/2024 October 25

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October 25

Why does splitting extension field’s elements into several subfields doesn’t help solving discrete logs despite it helps computing exponentiations and multiplications ?

Let’s say I have 2 finite fields elements $A$ and $B$ in $G{F}_p^6$ having their discrete logarithm belonging to a large semiprime' suborder/subgroup $s$ such as $p>s$.

$A$ and $B$ can be represented as the cubic extension of $G{F}_p^2$ by splitting their finite field elements. This give $A.x\in G{F}_p^2$ ; $A.\in G{F}_p^2$ ; $A.z\in G{F}_p^2$ ; and $B.x\in G{F}_p^2$ ; $B.y\in G{F}_p^2$ ; $B.z\in G{F}_p^2$. This is useful for simplifying computations on $A$ or $B$ like multiplying or squaring by peforming such computations component wise. An example of which can be found here : https://github.com/ethereum/go-ethereum/blob/24c5493becc5f39df501d2b02989801471abdafa/crypto/bn256/cloudflare/gfp6.go#L94

However when the suborder/subgroup $s$ from $G{F}_p^6$ doesn’t exists in $G{F}_p^2$, why does solving the 3 discrete logarithm between each subfield element that are :

1. dlog of $A.x$ and $B.x$
2. dlog of $A.y$ and $B.y$
3. dlog of $A.z$ and $B.z$

doesn’t help establishing the discrete log of the whole $A$ and $B$ ? 82.66.26.199 (talk) 13:30, 25 October 2024 (UTC)

Supposing that you can solve the discrete log in GF(q), the question is to what extent this helps to compute the discrete log in GF(q^k). Let g be a multiplicative generator of $GF(q^k{)}^{\times }$. Then Ng is a multiplicative generator of $GF(q{)}^{\times }$, when N is the norm map down to GF(q). Given A in $GF(q^k{)}^{\times }$, suppose that we have x such that $NA=N{g}^x$. Then $A{g}^{-x}$ belongs to the kernel of the norm map, which is the cyclic group of order (q^k-1)/(q-1) generated by g^{q-1}. Therefore it is required to solve an additional discrete log problem in this new group, the kernel of the norm map. When the degree k is composite, we can break the process down iteratively by using a tower of norm maps. If (a big if) each of the norm one groups in the tower has order a product of small prime factors, then Pohlig-Hellman can be used in each of them. Tito Omburo (talk) 14:53, 25 October 2024 (UTC)

And when the order contains a 200‒bits long prime too large for Pohlig‑Hellman ? 82.66.26.199 (talk) 15:39, 25 October 2024 (UTC)

Well, the basic idea is that if k is composite, then the towers are "relatively small", so they would be smoother than the original problem, and might be a better candidate for PH than the original problem. It seems unlikely that a more powerful method like the function field sieve would be accelerated by having a discrete log oracle in the prime field. The prime field in that case is usually very small already. For methods with p^n where p is large, an oracle for the discrete log in the prime field also doesn't help much (unless you can do Pohlig-Hellman). Tito Omburo (talk) 16:06, 25 October 2024 (UTC)