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SVP (shortest vector problem) is equivalent to SIVP (shortest independent vectors problem) in ring lattice (ideal lattice). How to prove this? Could someone explain it to me? Thanks!

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Consider the ring $R = \mathbb{Z}[x]/(f(x))$ for $f(x)$ an irreducible monic polynomial of degree $n$. Fix a basis $\{1,x,\dots,x^{n-1}\}$, and define the following transformation: $$L(x^i) = x^{i+1}$$ Extend this transformation to all of $R$ via linearity. Cayley-Hamilton theorem [1] states that the characteristic polynomial $p(t) = \det(L - tI)$ is such that: $$p(L) = 0$$ One can explicitly compute the square matrix corresponding to $L$ (it is the companion matrix of $f(x)$), and it is known that:

  • The characteristic polynomial of $L$
  • The minimal polynomial of $L$
  • $f(x)$

Are all identical. It follows that $f(L) = 0$, and this is the smallest polynomial $f(x)$ for which this holds, so among other things $\{1,x,\dots,x^{n-1}\}$ are linearly independent over $R$.

Now let $v(x)\in R$ be any vector (including a candidate "shortest vector"). We have that $\{v(x), xv(x), \dots, x^{n-1}v(x)\}$ are linearly independent over $R$. This is because if there were some linear relation between them: $$\sum_i a_iv(x)x^i = 0\implies v(x)\sum_i a_ix^i = 0\implies \sum_i a_i x^i = 0$$ Contradicting the linear independence. As $\| x^i v(x)\|$ are all the same, we've exhibited a basis of vectors of the same norm. So finding one short vector is enough to find all $n$ short vectors.


[1] Technically we need to make sure that Cayley-Hamiltonian holds over modules over general rings. Fortunately it is known to hold for commutative rings, so we're fine.

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    $\begingroup$ It’s not true that $\| x^i v(x) \|$ are all the same (in general), because there is “wraparound” modulo $f$, which can affect the norm. For certain $f$ it is true, but in general, the choice of $f$ affects the loss factor between the SVP and SIVP problems. $\endgroup$ Apr 17 '20 at 22:20

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