Let $\mathcal R$ be the ring of integers of a power of two cyclotomic field. That is, $\mathcal R = \mathbb Z[x] /\langle x^{2^k}+1\rangle $ for some integer $k$. We denote $\mathcal R / q \mathcal R$ by $\mathcal R_q$. This is a well known setup for ring-LWE.

I think that for randomly chosen element $a \in \mathcal R_q$, as I know, the algebraic norm $N(a)$ is approximate to $q^n$. (But I do not also have any reference for this.) Thus, I additionally guess that the determinant of the anti-circulant matrix generated by $\phi(a)$, where $\phi(a)$ is a coefficient embedding from $\mathcal R$ to $\mathbb Z^n$, is also close to $q^n$ with non-negligible probability.

However, I cannot find any reference for the closeness. Could you give me any reference or the answer that the statement is true or not?


1 Answer 1


Fair warning that

  1. my algebraic number theory is a bit rusty, and
  2. this solely reduces your problem to a (standard) problem, namely showing a lower bound on the shortest vector of a random ideal lattice.

Recall that an ideal lattice in some number field $K$ is a lattice that takes the form $L = I\mathcal{O}_K$, i.e. is an ideal $I$ in the ring of integers of $K$. Ideal lattices, as lattices, satisfy Minkowski's first bound, namely that (in dimension $n$) $\lambda_1(L) \leq \sqrt{n} \sqrt[n]{\det L}$.

Now, given an ideal lattice $I\mathcal{O}_K$, noting that $|N(I)| = |\mathcal{O}_K / I| = \left|\frac{\det I}{\det \mathcal{O}_K}\right|$, we get that $$\lambda_1(I\mathcal{O}_K) \leq \sqrt{n}\sqrt[n]{\det I}\implies \frac{\lambda_1(I\mathcal{O}_K)}{\sqrt{n}\sqrt[n]{\det \mathcal{O}_K}} \leq \sqrt[n]{N(I)}.$$

This reduces your problem to showing that, for average ideals $I$ (or perhaps principle ideals $(a)$), $\lambda_1(I\mathcal{O}_K)$ is large. I know of good references for showing that there exists $I$ such that $\lambda_1(I\mathcal{O}_K)$ is large (namely section 4 of this). A typical reference for discussion that on average integer lattices have $\lambda_1(\mathcal{L}(A))$ large is this or this. I will also briefly give a (directly) proof that $\lambda_1(\mathcal{L}(A))$ is large on average (for uniform $A$). The ones in the references use fairly high-powered tools, namely things like Siegal's integration formula. The direct proof hopefully will show where things break down in the ideal case.

Let $\mathcal{B}_n(c)$ be a (zero-centered) ball of integer vectors of radius at most $c$. Let $\mathcal{L}(A)$ be the lattice generated by a matrix $A$, which we will assume is uniformly random. Then, we have that \begin{align*} \Pr_A[\lambda_1(\mathcal{L}(A)) > c] &= 1 - \Pr_A[\lambda_1(\mathcal{L}(A)) \leq c]\\ & 1 - \Pr_A[\mathcal{L}(A) \cap \mathcal{B}(c) = \{0\}]\\ &\geq 1 - |\mathcal{B}(c)\setminus\{0\}| \max_{z\in\mathcal{B}(c)\setminus\{0\}}\Pr_A[z \in\mathcal{L}(A)]. \end{align*} All that we have done now is rearrange things, and apply the union bound. All that remains is to

  1. estimate $|\mathcal{B}(c)\setminus\{0\}|$, and
  2. compute $\Pr_A[z \in\mathcal{L}(A)]$.

The first is a standard (mathematical) computation, see claim 8.2. The second is also standard (at least for integer lattices) --- one has that $$\max_{z\neq 0}\Pr_A[z \in\mathcal{L}(A)] = \frac{|\mathcal{L}(A)\setminus \{0\}|}{|\mathbb{Z}^n/q\mathbb{Z}^n\setminus\{0\}|} = \frac{q^n-1}{\det \mathcal{L}(A) - 1}.$$ For integer lattices, it is straightforward to compute $\det \mathcal{L}(A)$ (see for example this). But this is (roughly) what we are trying to lower bound in the ideal case, i.e. we reduced lower bounding $N(a)$ to lower bounding $\lambda_1(a\mathcal{O}_K)$ to lower bounding $\det a\mathcal{O}_K \approx N(a)\dots$

  • $\begingroup$ Thank you for your kind answer. I will find the references! $\endgroup$
    – Ring-LWE
    Oct 22, 2022 at 3:28

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