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ChristianKl comments on A pair of free information security tools I wrote - LessWrong
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A signed PGP message has three parts and thus only three places where additional information could be hidden. 1. The header 2. The message itself 3. The signature
The header is standardized. Any changes to the header itself (especially something as blatant as inserting a private key) would be enormously obvious, and would most likely result in a message that would fail to verify due to formatting issues.
The message itself can be verified by the author of the message. If anything shows up on this field that does not exactly match up with what he or she wrote, it will also be extremely obvious.
The signature itself, firstly, must be reproduced with 100% accuracy in order for the message to verify successfully. Any after-the-fact changes to either the message or the signature, will result in a message that does not verify successfully. (This is, of course, the entire purpose of a digital signature). Furthermore, the signature is generated algorithmically and cannot be manipulated by user input. The only way to change the signature would be to change the message prior to signing. However, as indicated above, this would be extremely obvious to the author.
"Algorithmically" doesn't mean that there exactly one way to create a valid signature. Hash functions quite often have collisions.
I'm downvoting this comment because it's misleading.
First of all, no one has ever found an SHA-2 hash collision yet. Second of all, the chances of two SHA-2 hashes colliding is about 1 in 1 quattuorvigintillion. It's so big I had to look up what the number name was. It's 1 with 77 zeroes after it. We're talking universe-goes-into-heat-death-before-it-happens type odds. Only under the most absurd definition of "quite often" could anyone ever reasonably claim that a cryptographic hash function like SHA-2 "quite often" has collisions.
Not that I disagree with your general point, but... 77 isn't a multiple of 3.
Why does it need to be a multiple of 3?
(SHA-2 = 2^256 = 1*10^77)
You wrote that the odds were 1 in 1 quattuorvigintillion. I was under the impression that all "-illion"s have exponents that are multiples of 3.
Ahhhh. I misread the output on Wolfram Alpha. You're right. I'll leave it in the original post for posterity, but also for the record, it's actually 1 in 100 quattuorvigintillion
(That's what I get for trying to be dramatic)