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(how/can) i declair this isomorphism

asked 2018-03-16 18:53:34 -0500

TWJ gravatar image

updated 2018-03-17 02:32:52 -0500

Hi,

let $U$ be a square matrix of order $m$ over $\mathbb F_{q}$, more precisely $U$ is the companion matrix of a monic irreducible polynomial over $\mathbb F_{q}$ that define $\mathbb F_{q^m}$ .

let $\alpha$ be a primitive element of $\mathbb F_{q^m}$

I wish to declare this morphism to compute some examples with SAGEMATH

$\psi$: $\mathbb F_{q^m}$ $\rightarrow$ $\mathbb{F}_{q}[U]$

$\alpha$ $\mapsto$ $\psi(\alpha)=U$

thanks in advance;

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Here, what is $\mathbb F_q[U]$? Is it a subspace of the matrices of type $m\times m$ over $\mathbb F_q$? Can we take the whole space instead?

An explicit example of $\alpha$ and $U$ (that match) would make things easier.

dan_fulea gravatar imagedan_fulea ( 2018-03-17 08:51:47 -0500 )edit

https://rua.ua.es/dspace/bitstream/10045/27320/1/Tesis_Diaz_Cardell.pdf (link text) can you please check Theorem 3.3 in page 51

TWJ gravatar imageTWJ ( 2018-03-19 08:08:58 -0500 )edit

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answered 2018-03-21 12:30:41 -0500

dan_fulea gravatar image

A simple case: We let $p=2017$ and look for a "modulus" polynomial $f$ of degree $m=5$ which is irreducible over $F=\mathbb F_p$, associate its companion matrix $U$ as in loc. cit.

https://rua.ua.es/dspace/bitstream/10045/27320/1/Tesis_Diaz_Cardell.pdf ,

and the number field $K=F(a)$ generated by $a$ with $f(a)=0$ in $K$, and define the map $$ \psi:K\to M_{m\times m}(F)\ ,$$ to the full space of matrices, which takes $a$ to $U$. So the codomain is bigger in the implementation, but the true image is the one generated by the powers of $U$. We will test that $\psi$ is a map of rings on some random elements.

The code realizing this situation is as follows:

p = 2017
F = GF(p)
R.<X> = PolynomialRing( F )
m = 5
f = R.irreducible_element(m, algorithm='first_lexicographic')
U = companion_matrix(f, format='right')
K.<a> = GF( p**m, modulus=f )
M = MatrixSpace(F, m, m)
psi = K.Hom(M, Rings())(U)

This is all needed to define $\psi$ in sage. Let us investigate the constructed structure in a dialog with the sage interpreter.

sage: p
2017
sage: F
Finite Field of size 2017
sage: f
X^5 + X + 11
sage: f.is_irreducible()
True
sage: U
[   0    0    0    0 2006]
[   1    0    0    0 2016]
[   0    1    0    0    0]
[   0    0    1    0    0]
[   0    0    0    1    0]
sage: K
Finite Field in a of size 2017^5
sage: K.modulus()
x^5 + x + 11
sage: psi
Ring morphism:
  From: Finite Field in a of size 2017^5
  To:   Full MatrixSpace of 5 by 5 dense matrices over Finite Field of size 2017
  Defn: a |--> [   0    0    0    0 2006]
        [   1    0    0    0 2016]
        [   0    1    0    0    0]
        [   0    0    1    0    0]
        [   0    0    0    1    0]
sage:

Now we will pick some random elements in $K$ and test the compatibility of $F$ w.r.t. the structures of an $F$-algebra on $K$ and $M$.

s  = F.random_element()    # some element in 0, 1, ... , 2016
k1 = K.random_element()    # some random element in K
k2 = K.random_element()    # an other random element in K... maybe

print "s  = %s" % s
print "k1 = %s" % k1
print "k2 = %s" % k2

print "Is psi(s*k1)  == s*psi(k1)         ? %s" % bool( psi(s*k1)  == s*psi(k1)         )
print "Is psi(s*k2)  == s*psi(k2)         ? %s" % bool( psi(s*k2)  == s*psi(k2)         )
print "Is psi(k1+k2) == psi(k1) + psi(k2) ? %s" % bool( psi(k1+k2) == psi(k1) + psi(k2) )
print "Is psi(k1*k2) == psi(k1) * psi(k2) ? %s" % bool( psi(k1*k2) == psi(k1) * psi(k2) )

This gives (this time):

s  = 422
k1 = 1679*a^4 + 611*a^3 + 639*a^2 + 228*a + 1823
k2 = 242*a^4 + 968*a^3 + 1389*a^2 + 1838*a + 84
Is psi(s*k1)  == s*psi(k1)         ? True
Is psi(s*k2)  == s*psi(k2)         ? True
Is psi(k1+k2) == psi(k1) + psi(k2) ? True
Is psi(k1*k2) == psi(k1) * psi(k2) ? True

Above, we have for instance:

sage: psi(k1)
[1823 1701 1347 1039 1526]
[ 228  144 1090  708  811]
[ 639  228  144 1090  708]
[ 611  639  228  144 1090]
[1679  611  639  228  144]

Note that the space $F[U]$ is a subspace of dimension $m$ inside the full matrix space, in our case it is generated by the $m$ matrices:

sage: for power in range(m):
....:     print "U^%s is\n%s\n" % ( power, U^power )
....:    
U^0 is
[1 0 0 0 0]
[0 1 0 0 0]
[0 0 1 0 0]
[0 0 0 1 0]
[0 0 0 0 1]

U^1 is
[   0    0    0    0 2006]
[   1    0    0    0 2016]
[   0    1    0    0    0]
[   0    0    1    0    0]
[   0    0    0    1    0]

U^2 is
[   0    0    0 2006    0]
[   0    0    0 2016 2006]
[   1    0    0    0 2016]
[   0    1    0    0    0]
[   0    0    1    0    0]

U^3 is
[   0    0 2006    0    0]
[   0    0 2016 2006    0]
[   0    0    0 2016 2006]
[   1    0    0    0 2016]
[   0    1    0    0    0]

U^4 is
[   0 2006    0    0    0]
[   0 2016 2006    0    0]
[   0    0 2016 2006    0]
[   0    0    0 2016 2006]
[   1    0    0    0 2016]

sage:

Add-on: A real problem may be to realize a situation as the given one in sage for $q$ not a prime number. Please do not downgrade the answer for the reason of dilution or deviation. (Mathematically, there is one field of dimension $q$ up to isomorphism, so we pick the one. Practically, we need the isomorphism(s)...)

We have to fix the "alpha", which will become again an $a$ and in code an a in my case. And we have to construct it in such a way that we know its minimal polynomial over a choice of a field $\mathbb F_q$. (The situation is not completely trivial when $q$ is not a prime. So my choice will be with $q=2^5=32$.)

We introduce $F=\mathbb F_q=\mathbb F_2(s)$ for an element $s$ of a small degree, $3$ or $5$ say, over $\mathbb F_2$.

Then we pick an irreducible polynomial of degree $m$ over $F$.

m = 3
q = 2**3    # feel free to take an other q
F.<s> = GF(q)
R.<X> = PolynomialRing( F )
f = R.irreducible_element(m, algorithm='first_lexicographic')

This gives so far:

sage: f
X^3 + s*X + s^2

We also want the minimal polynomial of $s$, the "modulus" of $F$, we will maybe need it:

sage: s.minpoly()
x^3 + x + 1

Now we have to introduce the field $K$ with $q^m$ elements. There are two ways to do this in sage:

  • Either as K.<a> = F.extension( modulus ), we know its modulus, but in this case we do not have a method vector_space for K!
  • Or as K.<a> = GF( q^m, modulus=? ), and we have to know the modulus of "the same $a$" over $\mathbb F_2$. In this case we can easily get K.vector_space(), this makes it easy to proceed in our case.

(Real problems when dealing with real situations.) So we will give a solution via K.<a> = GF( q^m, modulus=? ), my choice since the vector_space method makes for me things easier. Which is the modulus, the minimal polynomial over $\mathbb F_q$ of $a$? The following code finds it, using the first method to find the initial data for the second one.

q, m    = 2**3, 3
F.<s>   = GF(q)
R8.<X>  = PolynomialRing( F )
modulus = R.irreducible_element(m, algorithm='first_lexicographic')
K.<a>   = F.extension( modulus )

R2.<y>  = PolynomialRing( GF(2) )
for f, multiplicity in (y^(2^9)-y).factor():
    if f.degree() != 9:
        continue
    if f(a) == 0:
        print f
        break

This gives the polynomial y^9 + y^5 + y^4 + y + 1 very quickly. But in case of a slightly bigger $q,m$ situation, e.g. $q=2^5, m=5$, this is already a test of patience. This is no longer the question, so i will take this case, we still have to find the place of $s$ in the new incarnation of the one field $K=\mathbb F_{2^9}$.

R2.<y>   = PolynomialRing( GF(2) )
K.<a>    = GF( 2**9, modulus = y^9 + y^5 + y^4 + y + 1 )
R512.<S> = PolynomialRing( K )
factor( a^3 + S*a + S^2 )

This gives:

(S + a^7 + a^6 + a^5 + 1) * (S + a^7 + a^6 + a^5 + a + 1)

Note that:

sage: (a^7 + a^6 + a^5 + 1).minpoly()
x^9 + x^7 + x^6 + x^4 + 1
sage: (a^7 + a^6 + a^5 + a + 1).minpoly()
x^3 + x + 1

so the second root has to be taken "as a pendant for $s$".

Now we can realize the morphism $\psi$

R.<y> = PolynomialRing( GF(2) )
K.<a> = GF( 2**9, modulus = y^9 + y^5 + y^4 + y + 1 )
F.<s> = GF( 2**3, modulus = y^3 + y + 1 )
ks    = a^7 + a^6 + a^5 + a + 1

R.<X> = PolynomialRing( F )
f = X^3 + s*X + s^2
U = companion_matrix(f, format='right')
M = MatrixSpace(F, 3, 3)
psi = K.Hom(M, Rings())(U)

We can test again the compatibility with the $F$--algebra structure.

incl = F.Hom(K, Fields())( ks )    # F -> K, s -> ks

t  = F.random_element()    # some element in F
k1 = K.random_element()    # some random element in K
k2 = K.random_element()    # an other random element in K... maybe

kt = incl(t)

print "t  = %s" % t
print "kt = %s" % kt

print "k1 = %s" % k1
print "k2 = %s" % k2

print "Is psi(kt*k1)  == t*psi(k1)        ? %s" % bool( psi(kt*k1) == t*psi(k1)         )
print "Is psi(kt*k2)  == t*psi(k2)        ? %s" % bool( psi(kt*k2) == t*psi(k2)         )
print "Is psi(k1+k2) == psi(k1) + psi(k2) ? %s" % bool( psi(k1+k2) == psi(k1) + psi(k2) )
print "Is psi(k1*k2) == psi(k1) * psi(k2) ? %s" % bool( psi(k1*k2) == psi(k1) * psi(k2) )

getting:

t  = s^2 + s + 1
kt = a^8 + a^5 + a^3 + a^2
k1 = a^6 + a^5 + a^4 + a^3 + a^2
k2 = a^8 + a^7 + a^3
Is psi(kt*k1)  == t*psi(k1)        ? True
Is psi(kt*k2)  == t*psi(k2)        ? True
Is psi(k1+k2) == psi(k1) + psi(k2) ? True
Is psi(k1*k2) == psi(k1) * psi(k2) ? True

Here, again, the codomain is full matrix space, but the image of $\psi$ is generated over $F=\mathbb F_2(s)$ by

sage: U^0, U^1, U^2
(
[1 0 0]  [  0   0 s^2]  [  0 s^2   0]
[0 1 0]  [  1   0   s]  [  0   s s^2]
[0 0 1], [  0   1   0], [  1   0   s]
)
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Asked: 2018-03-16 18:53:34 -0500

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Last updated: Mar 21