Compared with the common lattice-based PQC schemes, the modulus $q$ and dimension $n$ of homomorphic encryption are so large. For example, in Kyber, $n=256, n \times k = \{512,768,1024\}$, $q = 12289$ or $3329$, while in FV or other HE schemes, $n$ could be $2^{14}$, and $q$ might reach $2^{744}$. Both of them are based on lattice problem. Why are the parameters so different?


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All (R)LWE-based schemes have noisy ciphertexts. In general, the if the noise is smaller than a fraction of q (say, less than q/2), then the decryption works.

If you just want to encrypt a message and decrypt it in the future, then you can encrypt using a very big noise, close to q already. But when the noise is large, the LWE problem is harder, so, for the same security level, you can choose smaller parameters.

But for homomorphic schemes, you also want to operate with the ciphertexts and these operations increase the noise, so you have to start with small noise (for example, if fresh ciphertexts have noise close to q/4, then one single addition brings the noise to q/2 and no further homomorphic operation is possible). Usually, the schemes choose the initial noise to be bounded by some small constant, and increase q to have more room for the noise growth generated by the homomorphic operations. But then, the LWE problem becomes easier, so, to reach the desired security level, they have to increase the dimension, $n$.

You will notice that schemes whose noise growth is smaller are able to choose much smaller parameter (e.g., TFHE uses $n = 2^{10}$ and $q = 2^{32}$).

  • $\begingroup$ You could add that for FHE additionally one has to leave "room" for processing the decryption operation homomorphically ("bootstrapping"), making the gap even bigger. $\endgroup$
    – j.p.
    Dec 8, 2021 at 9:07
  • $\begingroup$ Yes, for schemes like BGV, FV, and CKKS to support bootstrapping, they usually need large parameters, because the bootstrapping itself introduces a lot of noise and after it we still want to do some homomorphic computation... But FHEW and TFHE are fully homomorphic and have "small" params (of course, comparing them with the first three schemes is very hard because the type of homomorphic operations they offer is very different...) $\endgroup$ Dec 8, 2021 at 15:58

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