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I understand how the RSA algorithm works for encryption and decryption purposes but I don't get how signing is done.

Here's what I (think) I know and is common practice:

  • If I have a message that I want to sign, I don't sign the message itself but I create a hash of it and then sign that hash by using my private key.
  • The signature gets attached to the message and both are transferred to the recipient.
  • The recipient recalculates the hash of the message and then uses my public key to verify the signature he received.

Here are the questions:

  • Why is it common practice to create a hash of the message and sign that instead of signing the message directly?
  • The important part and this is where I really started scratching my head: How can the recipient verify that I own the private key if the public key seems to be enough to recreate the signature?
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    $\begingroup$ The private key is the only one that can generate a signature that can be verified by the corresponding public key. The question then becomes how you can trust the public key is the one that was generated for the private key. The answer for that is key management. The most common key management scheme used is PKI using X509 certificates. $\endgroup$ – Maarten Bodewes Aug 21 '13 at 15:20
  • $\begingroup$ I am personally of the opinion that employing hashing in signing is a conceptual load for the user to understand the scheme that could have been spared. (He would have to understand something about hashing .) Hence I coded a signature scheme with RSA without using hashing. See Ex. 4 in s13.zetaboards.com/Crypto/topic/7234475/1/. $\endgroup$ – Mok-Kong Shen Jul 28 '16 at 11:29
  • $\begingroup$ An Online tool for debug RSA Sign Verify 8gwifi.org/rsasignverifyfunctions.jsp $\endgroup$ – anish Oct 27 '18 at 9:31
  • $\begingroup$ RSA should either use RSA-FDH or OAEP ti be secure. $\endgroup$ – kelalaka Jul 22 at 9:08
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Why is it common practice to create a hash of the message and sign that instead of signing the message directly?

Well, the RSA operation can't handle messages longer than the modulus size. That means that if you have a 2048 bit RSA key, you would be unable to directly sign any messages longer than 256 bytes long (and even that would have problems, because of lack of padding).

In contrast, a cryptographical hash can take an arbitrarily long message, and 'compress' it into a short string, in such a way that we cannot find two messages that hash to the same value. Hence, signing the hash is just as good as signing the original message; without the length restrictions we would have if we didn't use a hash.

The important part and this is where I really started scratching my head: How can the recipient verify that I own the private key if the public key seems to be enough to recreate the signature?

What made you think that the public key is enough to recreate the signature? It is sufficient to verify a signature that you're given, but it is not sufficient to generate new ones (or so we hope; if that's not true, the signature scheme is broken).

If you're using RSA, the signature verification process is (effectively) checking whether:

$S^e = \operatorname{Pad}(\operatorname{Hash}(M))\pmod N$

Definitions: $S$ is the signature; $M$ is the message; $e$ and $N$ are the public exponent and modulus from the public key; $\pmod N$ means that equality is checked modulo $N$; $\operatorname{Pad}$ is the padding function; and $\operatorname{Hash}$ is the hashing function. Note I say "effectively" because sometimes the padding method is nondetermanistic; that makes this check slightly different, but not in a way that matters for this discussion.

Now, if we were trying to forge a signature for a message $M'$ (with only the public key), we could certainly compute $P' = \operatorname{Pad}(\operatorname{Hash}(M'))$; however, then we'd need to find a value $S'$ with:

$S'^e = P' \pmod N$

and, if $N$ is an RSA modulus, we don't know how to do that.

The holder of the private key can do this, because he has a value $d$ with the property that:

$(x^e)^d = x \pmod N$

for all $x$. That means that:

$(P')^d = (S'^e)^d = S' \pmod N$

is the signature.

Now, if we have only the public key, we don't know $d$; getting that value is equivalent to factoring $N$, and we can't do that. The holder of the private key knows $d$, because he knows the factorization of $N$.

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    $\begingroup$ @rubo77: you could, if you knew the factorization of $N$. If you don't, well, that's a hard problem (and, in fact, is known as the RSA problem). $\endgroup$ – poncho Aug 15 '14 at 13:44
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    $\begingroup$ @rubo77: it's not just computing the e'th root of P, and then taking that modulo N. Instead, you're trying to find the value X such that $X^e \bmod N = P$. $\endgroup$ – poncho Aug 16 '14 at 3:19
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    $\begingroup$ Ah, I get ist: You should add in your explanation, that $(mod\ N)$ with the brackets around, means the "Primitive root modulo N". It does not mean: "modulo N", modulo or remainder operator which would be trivial to calculate). That's, what I didn't know so my first question "what should be so difficult?" came up $\endgroup$ – rubo77 Aug 16 '14 at 6:25
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    $\begingroup$ @rubo77: I have no idea what primitive roots have to do with RSA; for one, there are no primitive roots modulo $N$; that is, there is no value $g$ where $g^i \bmod N$ takes on all the values relatively prime to $N$ (assuming $N$ is the product of two distinct odd primes) $\endgroup$ – poncho Jul 27 '16 at 15:06
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    $\begingroup$ @Krumelur A message encrypted with your private key can only be decrypted with your public key (used for authentication) and a message encrypted with your public key can only be decrypted with your private key (used for encryption). $\endgroup$ – sdfqwerqaz1 Nov 24 '16 at 13:33
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The important part and this is where I really started scratching my head: How can the recipient verify that I own the private key if the public key seems to be enough to recreate the signature?

You can use public key to "encrypt" (or "decrypt" which is same in "textbook" RSA) the signature and get hashed message. If the hashed message equals hashed message, then you verified the message being correctly signed.

You cannot use public key and message to recreate a signature that can pass the above verification though.

P.S. For "textbook" RSA, I mean https://www.cs.cornell.edu/courses/cs5430/2015sp/notes/rsa_sign_vs_dec.php

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How does RSA signature verification work?

In unencrypted communication between A and B RSA signatures (without utilizing hash functions) are simply the message $m$ encrypted with sender's private key $d$ (not $e$ as in usual encrypted communication): $c \equiv m^d \pmod n$ which enables the receiver to verify a message by 'decrypting' it with public key $e$ and comparing the result to unencrypted message: $m \equiv c^e \pmod n$.

This procedure is supposed to ensure integrity in communication as the receiver (if the message matches the 'decrypted' signature) can be sure the messages is valid as only the sender is capable of computing signatures 'decryptable' using encryption exponent $e$.

Why is it common practice to create a hash of the message and sign that instead of signing the message directly?

RSA signatures can be forged or spoofed in certain scenarios which is shown in the following demonstration:

  • A and B are communicating on an unencrypted channel and attacker E is able to capture their messages $m$. However, they send RSA-signatures $c$ along with every package and therefore forged messages sent by E will be filtered.
  • Assume A sends messages $m_1$ and $m_2$ with signatures $c_{1,2} \equiv m_{1,2}^d \pmod n$ and E captures them.
  • E is now capable of generating the valid signature for a message $m_x$ for which $m_x = m_1 \cdot m_2$ must be true. This is feasible without private key $d$ because signature $c_x$ of $m_x$ can be calculated by multiplying signatures of $m_1$ and $m_2$: $$m_1^dm_2^d \equiv (m_1 \cdot m_2)^d \equiv m_x^d \equiv c_x \pmod n$$. This is possible due to homomorphism of RSA / module: $$(m_1 \cdot m_2) \bmod n = ((m_1 \bmod n \cdot m_2 \bmod n) \bmod n)$$ Simply put homomorphism means $f(xy) = f(x) \cdot f(y)$.
  • Therefore E can send a message to B which will pass integrity and authenticity test possibly causing errors or leading to confusion. However, E is not able to change contents of $m_x$ as $m_x = c_x^e$ must remain true.

Concluding RSA encrypted messages as signatures can be insufficient depending on the scenario, thus hash functions are commonly used in digital signature generation and additionally @poncho's answer is of relevance too.

Wikipedia articles on homomorphic encryption and module homomorphism dive into detail of this aspect of RSA encryption:

https://en.wikipedia.org/wiki/Homomorphic_encryption

https://en.wikipedia.org/wiki/Module_homomorphism

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