Let $R_q=\mathbb{Z}_q[x]/\langle f(x)\rangle$ where $f(x)=x^n+1$, as in the ring-LWE problem.

Let $a(x)$ be chosen uniformly at random from $R_q$.

Question: Is there any theorem that lower bounds the distance between any two polynomials of the form $a(x)s_1(s)$ and $a(x)s_2(x)$?

In other words, what is the value of $d$ such that $$||a(x)s_1(x)-a(x)s_2(x)||\geq d$$ except with negligible probability, for any two polynomials $s_1(x),s_2(x)\in R_q$ and where $||\cdot||$ is the usual $L_2$ norm?

  • $\begingroup$ Hello. It is a good question, but the $L_2$ norm is defined over vectors and it is not clear how you are embedding the polynomials in a vector space. Are you just representing the polynomials as vectors with their coefficients? (So, for instance, $2x^3 -1$ becomes the vector $(2, 0, 0, -1)$). $\endgroup$ Commented Nov 24, 2018 at 17:19
  • $\begingroup$ Yes, I am thinking of the canonical embedding $\endgroup$
    – P.B.
    Commented Nov 24, 2018 at 17:26
  • $\begingroup$ Well, the canonical embedding is the one that uses isomorphisms to embed the polynomials. The one I've described is the coefficient embedding... $\endgroup$ Commented Nov 24, 2018 at 17:31
  • $\begingroup$ Sorry. I mean the coefficient embedding then $\endgroup$
    – P.B.
    Commented Nov 24, 2018 at 17:46
  • $\begingroup$ How do you define "negligible probability" in this case? $\endgroup$
    – kodlu
    Commented Nov 24, 2018 at 21:19

1 Answer 1


I'm assuming $n$ is a power of $2$ and that $q$ is an odd prime larger than $n$. I'm discarding the trivial case $s_1 = s_2$.

If you consider everything $\mod q$, then it is most likely over the choice of $a$ that there exists $s_1 \neq s_2$ such that $\|a s_1 - a s_2\| = \sqrt{n}$. Indeed, $a$ is invertible in $R_q$ with probability about $1 - n/q$. Take $s_2 = s_1 - a^{-1}$, then you have $a s_1 - a s_2 = 1 \mod q$ and the embedding norm of $1$ is $\sqrt{n}$.

If you do not consider this $\mod q$, i.e. you work in $R=\mathbb Z[x]/⟨f(x)⟩$, then you are precisely asking for the minimal distance $\lambda_1(\mathfrak I)$ of the ideal lattice $\mathfrak I$ generated by $a$. For such an ideal lattice, we can estimate rather precisely this minimal distance. A simple lower bound is $\lambda_1(\mathfrak I) \geq \Delta_K^{1/2n} \cdot N(a)^{1/n}$, where $N$ denotes the algebraic norm of $a$ (that is, the product of all its embeddings), and $\Delta_K$ is the discriminant of field $K = \mathbb Q(x)/(x^n+1)$. The reason is that the minimal vector $x$ must generate a subideal of $\mathfrak I$, so $N(x) \geq N(a)$, and $\|x\|^n \geq \Delta_K^{1/2} N(x)$. An upper bound is also given by Minkowski's theorem.

  • $\begingroup$ Shouldn't the embedding norm of 1 be 1? $\endgroup$
    – P.B.
    Commented Nov 25, 2018 at 0:26
  • $\begingroup$ Well, each embedding of 1 is one, so we are looking at the norm of (1, 1, ... 1), right ? $\endgroup$
    – LeoDucas
    Commented Nov 25, 2018 at 7:54
  • $\begingroup$ So this is the canonical embedding we are talking about right? In this case, we have that $||as_1-as_2||\geq \sqrt{n}$? Or are you just saying that there are $s_1,s_2$ such that $||as_1-as_2||= \sqrt{n}$ but there could be others $r_1,r_2$ for which the differente between $ar_1$ and $ar_2$ is even lower? $\endgroup$
    – P.B.
    Commented Nov 25, 2018 at 13:36
  • $\begingroup$ There could be $0$ as well ;). BBut nothing in between indeed, at least for that particular ring. $\endgroup$
    – LeoDucas
    Commented Nov 25, 2018 at 16:26
  • $\begingroup$ Thank you for your help! Can you provide me some references about these facts? It would be very useful. $\endgroup$
    – P.B.
    Commented Nov 25, 2018 at 16:52

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