AES encryption how to transport IV

Question

I understand that unique IV is important in encrypting to prevent attacks like frequency analysis. The question: For AES CBC encryption, whats the importance of the IV? has a pretty clear answer explaining the importance of the IV.

Would there be any security holes in sending the IV in clear text? Or would it need to be encrypted with the same public/private key that was used to send the symmetric key?

If the IV needs to be sent encrypted, then why not generate a new symmetric key each time and consider the IV as part of the key? Is it that generating a symmetric key is too costly? Or is it to minimize the amount of data transported?

---

The top answer to Secret vs. Non-secret Initialization Vector states:

A typical key establishment protocol will result in both involve parties computing a piece of data which they, but only they, both know. With Diffie-Hellman (or any Elliptic Curve variant thereof), the said shared piece of data has a fixed length and they have no control over its value (they just both get the same seemingly random sequence of bits).

How do two entities derive the "same seemingly random sequence of bits" without having a shared piece of information? Is the assumption that the shared information was sent encrypted? And, if the shared information is sent encrypted, why not just send the IV encrypted?

---

Because an application needs to transport the symmetric key securely, it would seem that separating the IV from the key itself is essentially an optimization. Or am I missing something?

Answer 1

There is no security hole by sending the IV in cleartext - this is similar to storing the salt for a hash in plaintext: As long as the attacker has no control over the IV/salt, and as long as it is random, there is no problem.

Answer 2

The main difference between initialization vector and key is that the key has to be kept secret, while the IV doesn't have to be - it can be readable by an attacker without any danger to the security of the encryption scheme in question.

The idea is that you can use the same key for several messages, only using different (random) initialization vectors for each, so relations between the plain texts don't show in the corresponding ciphertexts.

That said, if you are using a key agreement scheme like Diffie-Hellman, which gives you a new shared secret for each session anyways, you can also use it to generate the first initialization vector. This does not really give much security advantages compared to choosing the initialization vector directly and sending it with the message, but saves some bits of bandwith, and some bits of entropy from your random source. And it makes the IV a bit more random in case that one of the partners has a bad randomness source (though DH is not really secure in this case, too).

How do two entities derive the "same seemingly random sequence of bits" without having a shared piece of information? Is the assumption that the shared information was sent encrypted? And, if the shared information is sent encrypted, why not just send the IV encrypted?

Diffie-Hellman is based on a group-theoretic problem: Eve knows a (cyclic) group G with generator g and sees the the two values g^a (transmitted from Alice to Bob) and g^b (transmitted from Bob to Alice), where a and b are random large integers chosen by Alice and Bob, and unknown to Eve and even the other partner). The shared secret is then (g^a)^b = g^(a·b) = (g^b)^a. Obviously Bob (who knows b) can calculate the secret as (g^a)^b, while Alice (who knows a) can calculate (g^b)^a. Eve somehow needs to derive this secret to crack the protocol.

In some groups this (known as the computational Diffie-Hellman problem) seems to be a hard problem, and we are using these groups in Cryptography. (In the original DH, we use a subgroup of prime order of the multiplicative group of some large finite prime field, in Elliptic Curve DH we use an elliptic curve group over a finite field. Other groups work, too (but some of them are weak, e.g. in the additive group of a field it is trivial to solve).)

Then both Alice and Bob use a key derivation function to derive the actual keying material (i.e. encryption keys for both directions, MAC keys, and the starting IVs).