On 2015-09-17 12:32, Jean-Pierre Flori wrote:
Don't we have a random keyword as well?
Not a random keyword, but a modulus="random" argument (there are several
more algorithms to choose a modulus)
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On Thursday, September 17, 2015 at 12:24:07 PM UTC+2, Jeroen Demeyer wrote:
>
> On 2015-09-17 11:56, Volker Braun wrote:
> > Two comments:
> >
> > A) There is no such thing as "different finite fields of the same size
> > with the same generator name"
>
> Of course there is:
>
> sage: GF(8,
On 2015-09-17 11:56, Volker Braun wrote:
Two comments:
A) There is no such thing as "different finite fields of the same size
with the same generator name"
Of course there is:
sage: GF(8, 'a') is GF(8, 'a', modulus="adleman-lenstra")
False
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Two comments:
A) There is no such thing as "different finite fields of the same size with
the same generator name"
sage: GF(4, 'a') is GF(4, 'a')
True
sage: GF(4, 'a') is GF(4, 'b')
False
There is no way to cast finite field elements to other finite field
elements apart from the prime subfiel
> The question is really: is the map between representations of the same
> finite field, which differ only in variable name, "canonical"?
I would say that there is no big risk to do that, though I am not at
all knowledgeable on those parts of the code. If it can be dangerous
but can be detected, h
On 2015-09-16 16:43, Jean-Pierre Flori wrote:
Hi,
I guess one of the issue is that there is no canonical map between two
different representations of the same finite field (so no coercion).
The question is really: is the map between representations of the same
finite field, which differ only
>> If it is "doable", then indeed it would solve my problem and (to me)
>> does not seem to induce any tricky behaviour.
>
> In this case:
>
> sage: [x.polynomial() for x in K1]
> [0, a, a^2, a + 1, a^2 + a, a^2 + a + 1, a^2 + 1, 1]
> sage: [K2(x.polynomial()) for x in K1]
> [0, b, b^2, b + 1, b^2
On 16 September 2015 at 15:52, Nathann Cohen wrote:
> Hello,
>
>> In your case it is slightly more specific as it might be the case that K1
>> and K2 are constructed in the same exact way (expect for the string 'a' and
>> 'b') so we could just decide to send "a" onto "b", but I'm not sure it is
Hello,
> In your case it is slightly more specific as it might be the case that K1 and
> K2 are constructed in the same exact way (expect for the string 'a' and 'b')
> so we could just decide to send "a" onto "b", but I'm not sure it is very
> helpful to define an exception in this case.
If it
Hi,
I guess one of the issue is that there is no canonical map between two
different representations of the same finite field (so no coercion).
In your case it is slightly more specific as it might be the case that K1
and K2 are constructed in the same exact way (expect for the string 'a' and
'
>
> GF(4) is not a subfield of GF(8).
>
Oh. Right. Actually, the problem with my code happened when both finite
fields were equal, and I wrote this message without thinking.
sage: K1 = GF(8,'a')
sage: K2 = GF(8,'b')
sage: K2(K1(1))
TypeError: unable to coerce from a finite fiel
GF(4) is not a subfield of GF(8).
John
On 16 September 2015 at 10:16, Nathann Cohen wrote:
> Hello everybody,
>
> I am rather ignorant of Sage matters when it comes to finite fields. I have
> been playing with matrices for a while, working around a bug that I did not
> understand. It is now solv
Hello everybody,
I am rather ignorant of Sage matters when it comes to finite fields. I have
been playing with matrices for a while, working around a bug that I did not
understand. It is now solved, but here is what it boils down to:
sage: K1 = GF(4,'w')
sage: K2 = GF(8,'x')
sage: K2(
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