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CDH in a group of square matrices

ge flag

This paper says the CDH problem in a group of square matrices can be solved by a generalized Chinese remainder theorem. I wonder how this problem might be solved?

DH protocol in the cyclic group of matrices $\langle M \rangle$, and the matrix $M$ is considered as public information. It is assumed that Alice generates a random index $x$, calculates the matrix $M^x$, and sends it to Bob. In turn, Bob generates a random index $y$, calculates the matrix $M^y$, and sends it to Alice. Then both subscribers raise the matrices obtained from a partner in their secret powers and calculate the sheared matrix (encryption key) $K=M^{xy}$. The matrix $M$ must be a high-order matrix (at least 100); ... However, in [3] it has been proved, that Yerosh-Skuratov protocol can easily be cracked based on the generalized Chinese remainder theorem."

fgrieu avatar
ng flag
In what set are the elements of the matrix considered? If it's the finite field $\mathbb F_p$, I think this [paper](http://theory.stanford.edu/~dfreeman/papers/discretelogs.pdf) applies and shows a reduction of the DLP in $\operatorname{GL}_n(\mathbb F_p)$ to the DLP in $\mathbb F_{p^n}$. But that can't be called "generalized Chinese remainder theorem".
Amir Amir avatar
ge flag
My specific problem is the security of Diffi-hellman in the group $\text{GL}(n,2)$!
Daniel S avatar
ru flag
I think that the Freeman paper could be interpreted as "generalised CRT". The method is to lift into a field $GF(2^m)$ where all of the eigenvalues can be found. In this field, the matrices $M$ and $M^x$ diagonalise to eigenvalues. Then solving $n$ discrete logs with the eigenvalues of $M$ as the generators and the other diagonals as the targets gives $n$ congruences for $x$ modulo the order of the eigenvalue and these can be combined with CRT.
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