Message from @YourFundamentalTheorum
Discord ID: 503832804757602314
that is an isomorphism phi: F[x]/f1(x) -> F[x]/f2(x) where F[x] is the field of polynomials with coefficients in Z_q and f1(x) and f2(x) are irreducible polynomials in F[x]
@YourFundamentalTheorum by F[x]/f1(x) do you mean a quotient group? If so, what is the group you are doing F[x] modulo with? f1 doesn't generate a group unless I am missing something
@ThisIsChris quotient field
and F[x] is a field of polynomials in x with coefficients in Z_q
@YourFundamentalTheorum thanks. How about for an element of F[x]/f1 called G, take a representative g, phi(G) = the coset of g*f2/f1 . Need to prove that the coset is independent of the choice of representative g
the last part is a given.
On the right hand side I mean g times f2 divided by f1 in the normal sense for poly nomials
F[x]/f1(x) is usually represented by all the polynomials """"less than""""" f1(x)
a lesser degree than f1(x) that is
Hm, what happens in the three cases where f2 is higher, equal, or lesser degree than f1?
assume f1 and f2 are of the same degree
otherwise you don't have the same cardinalities
which means they can't be isomorphic
the Galois fields can't be isomorphic that is
Hm I think you can actually prove it if the degrees mismatch too. Since you assume the rep g has degree strictly less than f1, then g*f2/f1 should have degree strictly less than f2
yes you can prove that
it's quite easy actually
the degrees part htat is
we know the exact # of elements in F[x]/f1(x)
it's going to be q^n
Sure since it's irreducible. Otherwise, hm...
If a polynomial has a non-prime degree, to guarantee reducibility you need to allow complex coefficients, right?
i don't think it matters what the degree of the polynomial is
like if it's prime or not
i think what matters are the coefficients
if you don't have prime coefficients, your polynomials stop being a field
the reason why F[x] is a field is because Z_q is a field when q is prime
if q is not prime, Z_q is not a field
and therefore F[x] is not guaranteed to be a field
Why is Z_q not a field when q is, say, 4?
the very basic example is that say if you have Z_4, the polynomial "2" does not have a multiplicative inverse
Ohh
Neat
Z_4 is not a field because 2 does not have a multiplicative inverse
Yeah haha been a while
and since Z_q is a subfield of F_q[x]
yeah I've literally pullen out my abstract alg. book out today
i need some cryptography knowledge from it
i learned all of this again today 😛
That's neat. I enjoyed abstract algebra but mostly only needed the linear algebra part since then (11 years ago). Rotation groups and some other stuff relevant to geometry, but not much more. Cryptography seems interesting, we did cover RSA at some point, I only remember the main idea, that factoring is hard 😁