My question is in Ring Learning with Errors, let $a(x)\in \mathbb{Z}_q(x)/(2^n-1)$ be a random polynomial, $s(x),e(x)\in \mathbb{Z}_q(x)/(2^n-1)$ are the secret and error polynomial respectively with co-efficients sampled from a discrete Gaussian distribution $D_\sigma$. Now, the decisional hardness of RLWE says that the samples $a(x)\cdot s(x) + e(x)$ are indistinguishable from a random polynomial $u(x)\in \mathbb{Z}_q(x)/(2^n-1)$.

• Now, my question is whether the polynomial $a(x)\cdot s(x)$ $i.e$ the RLWE sample without error will be polynomially indistinguishable from a random string? As far as I can think, the error polynomial $e(x)$ is only used for the hardness of the problem. Does it also contribute to the polynomial indistinguishability of the samples?

The reason I am asking this question is that, recently I found a new problem LWR in these papers Banerjee et al., Alwen et al Bogdanov et al. which removes the error but introduces an error with rounding each co-efficient of $a(x)\cdot s(x)$, $\lfloor p/q(.)\rceil : \mathbb{Z}_q\rightarrow \mathbb{Z}_p$. They have gone great length to prove the computational indistinguishability of the LWR samples. The computational indistinguishability of RLWR is not proven yet. if $a(x)\cdot s(x)$ is computationally indistinguishable from a random polynomial in $\mathbb{Z}_q$ then can't we say $\lfloor p/q(a(x)\cdot s(x))\rceil$ is also computationally indistinguishable from a random polynomial in $\mathbb{Z}_p/(2^n-1)$ if $p|q$

My question is in Ring Learning with Errors, let $a(x)\in \mathbb{Z}_q(x)/(X^n+1)$ where $n$ is a power of $2$, be a random polynomial, $s(x),e(x)\in \mathbb{Z}_q(x)/(X^n+1)$ are the secret and error polynomial respectively with co-efficients sampled from a discrete Gaussian distribution $D_\sigma$. Now, the decisional hardness of RLWE says that the samples $a(x)\cdot s(x) + e(x)$ are indistinguishable from a random polynomial $u(x)\in \mathbb{Z}_q(x)/(X^n+1)$.

• Now, my question is whether the polynomial $a(x)\cdot s(x)$ $i.e$ the RLWE sample without error will be polynomially indistinguishable from a random string? As far as I can think, the error polynomial $e(x)$ is only used for the hardness of the problem. Does it also contribute to the polynomial indistinguishability of the samples?

The reason I am asking this question is that, recently I found a new problem LWR in these papers Banerjee et al., Alwen et al Bogdanov et al. which removes the error but introduces an error with rounding each co-efficient of $a(x)\cdot s(x)$, $\lfloor p/q(.)\rceil : \mathbb{Z}_q\rightarrow \mathbb{Z}_p$. They have gone great length to prove the computational indistinguishability of the LWR samples. The computational indistinguishability of RLWR is not proven yet. if $a(x)\cdot s(x)$ is computationally indistinguishable from a random polynomial in $\mathbb{Z}_q$ then can't we say $\lfloor p/q(a(x)\cdot s(x))\rceil$ is also computationally indistinguishable from a random polynomial in $\mathbb{Z}_p/(X^n+1)$ if $p|q$

My question is in Ring Learning with Errors, let $a(x)\in \mathbb{Z}_q(x)/(2^n-1)$ be a random polynomial, $s(x),e(x)\in \mathbb{Z}_q(x)/(2^n-1)$ are the secret and error polynomial respectively with co-efficients sampled from a discrete Gaussian distribution $D_\sigma$. Now, the decisional hardness of RLWE says that the samples $a(x)\cdot s(x) + e(x)$ are indistinguishable from a random polynomial $u(x)\in \mathbb{Z}_q(x)/(2^n-1)$.

• Now, my question is whether the polynomial $a(x)\cdot s(x)$ $i.e$ the RLWE sample without error will be polynomially indistinguishable from a random string? As far as I can think, the error polynomial $e(x)$ is only used for the hardness of the problem. Does it also contribute to the polynomial indistinguishability of the samples?

The reason I am asking this question that, recently I found a new problem LWR in these papers Banerjee et al., Alwen et al Bogdano et al. which removes the error but introduces an error with rounding each co-efficient of $a(x)\cdot s(x)$, $\lfloor p/q(.)\rceil : \mathbb{Z}_q\rightarrow \mathbb{Z}_p$. They have gone great length to prove the computational indistinguishability of the LWR samples. The computational indistinguishability of RLWR is not proven yet. if $a(x)\cdot s(x)$ is computationally indistinguishable from a random polynomial in $\mathbb{Z}_q$ then can't we say $\lfloor p/q(a(x)\cdot s(x))\rceil$ is also computationally indistinguishable from a random polynomial in $\mathbb{Z}_p/(2^n-1)$ if $p|q$

My question is in Ring Learning with Errors, let $a(x)\in \mathbb{Z}_q(x)/(2^n-1)$ be a random polynomial, $s(x),e(x)\in \mathbb{Z}_q(x)/(2^n-1)$ are the secret and error polynomial respectively with co-efficients sampled from a discrete Gaussian distribution $D_\sigma$. Now, the decisional hardness of RLWE says that the samples $a(x)\cdot s(x) + e(x)$ are indistinguishable from a random polynomial $u(x)\in \mathbb{Z}_q(x)/(2^n-1)$.

• Now, my question is whether the polynomial $a(x)\cdot s(x)$ $i.e$ the RLWE sample without error will be polynomially indistinguishable from a random string? As far as I can think, the error polynomial $e(x)$ is only used for the hardness of the problem. Does it also contribute to the polynomial indistinguishability of the samples?

The reason I am asking this question is that, recently I found a new problem LWR in these papers Banerjee et al., Alwen et al Bogdanov et al. which removes the error but introduces an error with rounding each co-efficient of $a(x)\cdot s(x)$, $\lfloor p/q(.)\rceil : \mathbb{Z}_q\rightarrow \mathbb{Z}_p$. They have gone great length to prove the computational indistinguishability of the LWR samples. The computational indistinguishability of RLWR is not proven yet. if $a(x)\cdot s(x)$ is computationally indistinguishable from a random polynomial in $\mathbb{Z}_q$ then can't we say $\lfloor p/q(a(x)\cdot s(x))\rceil$ is also computationally indistinguishable from a random polynomial in $\mathbb{Z}_p/(2^n-1)$ if $p|q$