Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I've lately found about digital signatures with message recovery.

I was wondering, in what degree do these signatures provide confidentiality of the message (if they do at all) ? I've briefly read a couple of papers but they don't mention confidentiality as a security service provided and I wander why.

My theory is that since the message is recovered from the signature, after the verification is done, and for the verification to be done a key is needed, this means that whoever does not have the key (ignoring it is public), cannot recover the message.

Am I correct, from a mathematical point of view? I'd like a cryptographer to comment if possible, or someone who knows the mathematics behind digital signatures with message recovery.

share|improve this question
add comment

1 Answer

The key used to verify a Digital Signature is the public key of the signer, available to everybody. So Digital Signature with message recovery provides no confidentiality at all.

share|improve this answer
    
Hi TonyK. I specifically mentioned "ignoring it is public" to avoid this exact situation. Take this example: An organisation has an internal PKI architecture, and the public key is shared amongst the employees and noone else. If a signature leaks outside the organisation, to someone who does not have the public key, will he be able to infer anything about the message? If not, this proves that there is indeed a confidentiality property.. And I wonder how mathematics back this up. –  john Apr 22 '11 at 14:59
    
That's like using a computer as a doorstop. It might do the job, but it's not what it was designed for. –  TonyK Apr 26 '11 at 15:35
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.