Chaum's credentials (technical question)

[hughes@ah.com (Eric Hughes)]

Hal Finney writes about a paper of Chaum's, and near the end asks:

>In short, I haven't found any interpretation of Chaum's math that makes
>sense.  Can anyone shed any light on this?  

I think you're correct, Hal.  From everything I can tell, Chaum's
confused product and greatest common divisor.

First, though, there's a basic fact about the arithmetic of integers
which everyone who wants to learn more algebra should know.  Z is the
set of integers.

    For every m,n in Z, there is a,b in Z such that a*m + b*n = gcd(m,n).

One calculates the gcd by means of the Euclidean algorithm, and the
coefficients a and b by an extension of that algorithm.  Lots of basic
algorithm books contain descriptions.

From an abstract point of view, this is a simple consequence that Z is
a principal ideal domain.  The ideal (m,n) is composed of the linear
span of m and n.  Since this ideal is principal, by definition there
is a c such that (m,n) = (c).  Clearly c is in the linear span of m
and n, hence coefficients exist.

Note that reducing mod n gives a*m = gcd(m,n) (mod n).  If m and n are
relatively prime (means gcd = 1), the a is m inverse mod n.  Likewise
b is n inverse mod m.  This is a standard algorithm for calculating
modular inverses.

Here's the relevant passage, again:

>   "Suppose an organization X were to require that you have each of two
>   credentials, say both that with public exponents e1 and e2.  You could
>   send X separatley Px^d1 and Px^d2.  It is also possible for you to use
>   the two credentials to form the single credential Px^(d1*d2), which
>   will be called their AND....  To create the AND, you: set g to the
>   multiplicative inverse of d1 modulo d2; set h to the remainder after
>   dividing g*d1-1 by d2; and computing
>   (Px^d1)^g * (Px^d2)^(-h) = Px^(d1*d2)."

I believe that Hal is correct when he points out that h is not a
remainder (which would be zero, as he points out) but the quotient.  I
originally misread this as quotient because I recognized the context.

First, the multiplicative inverse of d1 (mod d2) exists only if the
two are relatively prime.  Hal did not quote the whole article, so I
don't know if this criterion is stated elsewhere.  Let us assume the
d's have this property, since the d's can be so chosen.

The calculation of d1 (mod d2) is exactly the calculation of the
coefficients in the extended Euclidean algorithm above.  Consider a*d1
+ b*d2 = 1.  Reducing mod d2, we have a*d1 = 1 (mod d2).  That means
that a is d1^(-1) (mod d2).  Likewise (a*d1 - 1) / d2 = -b.  Chaum's
description exactly fits the gcd context.

>The other possibility I thought of is that he meant that the signing
>agency would make g and h, as he defined them, public.  

That's my interpretation as well.  After the calculation, as Hal
observes, you just end up with Px, not Px^(d1*d2).  That's totally
useless, since you already knew Px.

It is certainly possible to create coefficients for combining
credentials such that you end up with a product in the exponent.  For
example, the pairs <d2,0> and <0,d1> both work nicely, with the bad
side effect that you've given away a private key.  Let's try blinding
them.

Suppose you have coefficients a and b such that a*d1+b*d2=0; the pair
<d2,-d1> works here.  Then every such pair of product-combining
coefficients can be represented as <d2,0> + r*<a,b>.  Since the
exponents are mod phi(N), we can suppose that the pair <d2+r*a,r*b>
doesn't _directly_ reveal the private keys.  But it's unclear to me
that this pair of coefficients doesn't reveal d1 and d2.  One doesn't
know phi(N), but one may not need to.

Eric