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When multiplying polynomials from $\mathbb{Z}_q[X] / (X^n-1) $, the discrete NTT is used because: $$ f \cdot g = \mathsf{NTT}_n^{-1}\left( \mathsf{NTT}_n\left(f\right) * \mathsf{NTT}_n\left(g\right) \right) $$ However, in virtually all schemes I've seen the negacyclic convolution is used - the ring is $\mathbb{Z}_q[X] / (X^n+1) $ and a trick is used to compute $\mathsf{NTT}_{2n}^{-1}\left( \mathsf{NTT}_{2n}\left(f\right) * \mathsf{NTT}_{2n}\left(g\right) \right) $ using $\mathsf{NTT}_n$ because we have to multiply the polynomials in $\mathbb{Z}_q[X] / (X^{2n}-1) $.
My question is - why bother with $\mathbb{Z}_q[X] / (X^n+1) $ and not simply use $\mathbb{Z}_q[X] / (X^n-1) $, thus applying $\mathsf{NTT}_n$ in a straightforward way with no extra complications?

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2 Answers 2

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The underlying hard problem used to construct the cryptographic primitives requires us to use that ring of polynomials modulo $X^n + 1$.

For example, if the security of your scheme relies on RLWE, then you have to stick with the rings that make the RLWE secure, which means that you cannot use $X^n - 1$, as discussed in this answer.

You have the same situation for the Ring-SIS problem.

In general, you have to be careful when you instantiate any problem with $X^n - 1$ as the modulus, because one can evaluate the polynomials on $1$ to work over integers instead of polynomials. This is discussed, for example, in the section "Polynomial evaluation" of this paper.

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You state (I have fixed your notation, residue classes are taken with a $/$ not $\backslash)$ the former is setminus which is not the same:

However, in virtually all schemes I've seen the negacyclic convolution is used - the ring is $\mathbb{Z}_q[X] / (X^n+1) $ and a trick is used to compute $\mathsf{NTT}_{2n}^{-1}\left( \mathsf{NTT}_{2n}\left(f\right) * \mathsf{NTT}_{2n}\left(g\right) \right) $ using $\mathsf{NTT}_n$ because we have to multiply the polynomials in $\mathbb{Z}_q[X] / (X^{2n}-1) $.

and ask the question:

why bother with $\mathbb{Z}_q[X] / (X^n+1) $ and not simply use $\mathbb{Z}_q[X] \setminus (X^n-1) $?

If we have the polynomial factorization $x^{2n}-1=(x^n-1)(x^n+1)$ then computations modulo $x^{2n}-1$ can be speeded up by convolving (via NTT) with respect to each one of the factors and then combining. So

  1. We have a factorization that leads to a fast transform, so we take this route. The extreme case is complex FFT when we can factor all the way into linear factors $x^n-1=\prod_{i=1}^n (\omega^i-1)$ with $\omega$ a primitive $n^{th}$ root of unity.
  2. The factorization is unique, you cannot use $(x^n+1)$ only since $(x^n+1)^2=(x^{2n}+2 x^n+1)$ and $q$ is in general not characteristic 2. If you meant actually just use $x^{2n}-1$ directly, you'd have no speedup.
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    $\begingroup$ Sorry, my bad with the quotient notation. I think you misunderstood my question (or I have misunderstood your answer). I'm not asking how to multiply polynomials of degree $(2n-1)$ using NTTs of order $n$, and am not asking why it's impossible to multiply polynomials from $\mathbb{Z}_q[X]/(X^n+1)$ with NTTs. I'm asking why use polynomials modolu $X^n+1$ in the first place (which need the trick with $(X^{2n}-1)$), and not use polynomials modolu $X^n-1$ $\endgroup$
    – warforgad
    May 22 at 13:46
  • $\begingroup$ if your length is $N=2n,$ i.e., even, this IS the factorisation you must use to begin with i.e., $x^{2n}-1=(x^n-1)(x^n+1)$ because any $2n^{th}$ root of unity in any field or ring is either order $2n$ (those which are -1 when raised to power $n$) or have order dividing $n$. $\endgroup$
    – kodlu
    May 23 at 13:29
  • $\begingroup$ you start with $N$ as a given. $\endgroup$
    – kodlu
    May 23 at 13:29

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