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2020-07-29 12:08:02 +0200 commented question Tensor product over polynomial rings

Thanks @slelievre for formatting.

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2020-07-27 16:53:47 +0200 asked a question Tensor product over polynomial rings

I'm writing code for Fully Homomorphic Encryption multiplication routine. I'm trying to calculate tensor product over polynomial rings. Let's say I have elements in R = Z[x]/[x^n+1] where coefficients are modulo q.

I want to calculate tensor product of m1_q = (ct0 + sk * ct1) and m2_q = (ct2 + sk * ct3) and scale it down by (q/2). Each element ct0, ct1, ct2, ct3, sk, m1_q and m2_q are elements of Rq.

Just for m1_q and m2_q, it is working out well: (2/q)*(m1_q tensor_product m2_q). How can I calculate it in terms of ct*? I tried writing it as:

c0 = (2/q)*(ct0 tensor_product ct2)
c1 = (2/q)*( (ct0 tensor_product ct3) + (ct2 tensor_product ct1) )
c2 = (2/q)*(ct1 tensor_product ct3)
m_out = (2/q)*(m1_q tensor_product m2_q)

Then I converted c0, c1, c2 and m_out into Rq.

Now I was testing:

c0 + c1 * sk + c2 * sk^2 == m_out

But these are not matching. Appreciate your help.

2020-04-06 11:32:13 +0200 commented answer j-invariant of Fp2

@Sebastien: thank you so much! This code helps. And, my apologies for not looking into the documentation carefully.

2020-04-06 09:25:34 +0200 commented answer j-invariant of Fp2

Is there a method to get all or a list of j-variants? For example, GF(431^2, 'i') has 37 j-variants. Above function just returns one (e.g. 4). How can we get a subset of 37? Thanks.

2020-04-06 07:43:06 +0200 commented answer j-invariant of Fp2

Thanks a lot, @Sebastien.

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2020-04-04 20:25:16 +0200 asked a question j-invariant of Fp2

Hi,

Can you please give a pointer to how to calculate set of supersingular j-invariants of quadratic extension of prime fields? Specifically for Fp2 = Fp(i) with elements of the form 'u+iv' where u,v belongs to Fp.

Thanks Niranjan