cryptography - 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?

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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?

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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.