ASKSAGE: Sage Q&A Forum - Individual question feedhttp://ask.sagemath.org/questions/Q&A Forum for SageenCopyright Sage, 2010. Some rights reserved under creative commons license.Sun, 19 Nov 2017 16:23:01 -0600Polynomial Mod Idealhttp://ask.sagemath.org/question/39593/polynomial-mod-ideal/I have a polynomial $R\in\mathbb{Z}[x]$.
I then define $R' = \frac{d}{dx}R$, and look at the splitting field of $R'$, $K$, an algebraic number field.
Now, I want to find a prime ideal $\mathfrak{p}$ of $L$ of absolute degree 1 such that $R\mod\mathfrak{p}$ is irreducible.
To do this, I've set up:
p0, n = 5, 7
L = PolynomialRing(ZZ,'x')
R = L(x^n-p0^(n-1)x+p0)
Rprime = L(n*x^(n-1)-p0^(n-1))
K = NumberField(Rprime, 'z')
for P in K.primes_of_degree_one_iter():
<stuff>
I'm now trying ti find the right thing to do for <stuff>. Namely, how can I reduce the polynomial R with respect to the ideal P?Thu, 16 Nov 2017 00:23:54 -0600http://ask.sagemath.org/question/39593/polynomial-mod-ideal/Comment by orangejake for <p>I have a polynomial $R\in\mathbb{Z}[x]$.
I then define $R' = \frac{d}{dx}R$, and look at the splitting field of $R'$, $K$, an algebraic number field.
Now, I want to find a prime ideal $\mathfrak{p}$ of $L$ of absolute degree 1 such that $R\mod\mathfrak{p}$ is irreducible.</p>
<p>To do this, I've set up:</p>
<pre><code>p0, n = 5, 7
L = PolynomialRing(ZZ,'x')
R = L(x^n-p0^(n-1)x+p0)
Rprime = L(n*x^(n-1)-p0^(n-1))
K = NumberField(Rprime, 'z')
for P in K.primes_of_degree_one_iter():
<stuff>
</code></pre>
<p>I'm now trying ti find the right thing to do for <stuff>. Namely, how can I reduce the polynomial R with respect to the ideal P?</p>
http://ask.sagemath.org/question/39593/polynomial-mod-ideal/?comment=39612#post-id-39612I'm trying to implement Lemma 2.3 from Kedlaya's *A construction of polynomials with squarefree
discriminants* (I don't believe I can post links, its identifier on Arxiv is 1103.5728). This lemma says "Let $p_0$ be a prime not dividing $n(n−1)$. Then there exist infinitely many primes
$p_1$ modulo which the polynomial $R(x) = x^n − p_0^{n−1}x + p_0$ is irreducible and its derivative
$R′(x) = nx^{n−1} − p_0^{n−1}$
splits into distinct linear factors.Thu, 16 Nov 2017 14:42:33 -0600http://ask.sagemath.org/question/39593/polynomial-mod-ideal/?comment=39612#post-id-39612Comment by orangejake for <p>I have a polynomial $R\in\mathbb{Z}[x]$.
I then define $R' = \frac{d}{dx}R$, and look at the splitting field of $R'$, $K$, an algebraic number field.
Now, I want to find a prime ideal $\mathfrak{p}$ of $L$ of absolute degree 1 such that $R\mod\mathfrak{p}$ is irreducible.</p>
<p>To do this, I've set up:</p>
<pre><code>p0, n = 5, 7
L = PolynomialRing(ZZ,'x')
R = L(x^n-p0^(n-1)x+p0)
Rprime = L(n*x^(n-1)-p0^(n-1))
K = NumberField(Rprime, 'z')
for P in K.primes_of_degree_one_iter():
<stuff>
</code></pre>
<p>I'm now trying ti find the right thing to do for <stuff>. Namely, how can I reduce the polynomial R with respect to the ideal P?</p>
http://ask.sagemath.org/question/39593/polynomial-mod-ideal/?comment=39613#post-id-39613The proof is:
"The polynomial $R′$ has splitting field $L = \mathbb{Q}(\zeta_n−1, n^{1/(n−1)})$, in which $p_0$ does not
ramify because $p_0$ does not divide $n(n−1)$. Thus $R$ is an Eisenstein polynomial with respect
to any prime above $p_0$ in $L$; in particular, $R$ is irreducible over $L$. By the Chebotarev density
theorem, there exist infinitely many prime ideals of $L$ of absolute degree 1 modulo which $R$
is irreducible; the norm of any such prime ideal is the prime we want."
I'm not entirely sure what "absolute degree 1" means for a prime ideal, and if `K.primes_of_degree_one_iter()` is giving me something else (namely, fractional ideals), this could be the issue. Do you know how I could find prime ideals of $L$ of absolute degree 1?Thu, 16 Nov 2017 14:45:28 -0600http://ask.sagemath.org/question/39593/polynomial-mod-ideal/?comment=39613#post-id-39613Comment by dan_fulea for <p>I have a polynomial $R\in\mathbb{Z}[x]$.
I then define $R' = \frac{d}{dx}R$, and look at the splitting field of $R'$, $K$, an algebraic number field.
Now, I want to find a prime ideal $\mathfrak{p}$ of $L$ of absolute degree 1 such that $R\mod\mathfrak{p}$ is irreducible.</p>
<p>To do this, I've set up:</p>
<pre><code>p0, n = 5, 7
L = PolynomialRing(ZZ,'x')
R = L(x^n-p0^(n-1)x+p0)
Rprime = L(n*x^(n-1)-p0^(n-1))
K = NumberField(Rprime, 'z')
for P in K.primes_of_degree_one_iter():
<stuff>
</code></pre>
<p>I'm now trying ti find the right thing to do for <stuff>. Namely, how can I reduce the polynomial R with respect to the ideal P?</p>
http://ask.sagemath.org/question/39593/polynomial-mod-ideal/?comment=39604#post-id-39604... continuation:
sage: P
Fractional ideal (71429, -71428/3125*z^5 + 71429/5*z - 13020)
sage: for f, pow in P.factor():
....: print "pow=%s ideal is %s" % ( pow, f )
pow=-1 ideal is Fractional ideal (7, 7/3125*z^5)
pow=1 ideal is Fractional ideal (71429, 7/3125*z^5 - 19711)
Now `f` is the last fractional ideal. Its gens are algebraic integers:
sage: for g in f.gens():
....: print g, 'with norm', ZZ(g.norm()).factor()
71429 with norm 71429^6
7/3125*z^5 - 19711 with norm 2 * 3 * 11311 * 71429 * 12098321599065661
and we may build `K.quotient( f )`, but this is maybe not wanted.
Passing to ZZ / 71429 first, then modulo ... is what you want?
Also, note that `K(R)` has degree one, `-93750/7*z + 5`, we need maybe a new `R` in `K[]`...Thu, 16 Nov 2017 07:55:17 -0600http://ask.sagemath.org/question/39593/polynomial-mod-ideal/?comment=39604#post-id-39604Comment by dan_fulea for <p>I have a polynomial $R\in\mathbb{Z}[x]$.
I then define $R' = \frac{d}{dx}R$, and look at the splitting field of $R'$, $K$, an algebraic number field.
Now, I want to find a prime ideal $\mathfrak{p}$ of $L$ of absolute degree 1 such that $R\mod\mathfrak{p}$ is irreducible.</p>
<p>To do this, I've set up:</p>
<pre><code>p0, n = 5, 7
L = PolynomialRing(ZZ,'x')
R = L(x^n-p0^(n-1)x+p0)
Rprime = L(n*x^(n-1)-p0^(n-1))
K = NumberField(Rprime, 'z')
for P in K.primes_of_degree_one_iter():
<stuff>
</code></pre>
<p>I'm now trying ti find the right thing to do for <stuff>. Namely, how can I reduce the polynomial R with respect to the ideal P?</p>
http://ask.sagemath.org/question/39593/polynomial-mod-ideal/?comment=39603#post-id-39603Let us do the above, and further insert some print:
p0, n = 5, 7
L.<x> = PolynomialRing( ZZ )
R = L( x^n - p0^(n-1)*x + p0 )
Rprime = L( n*x^(n-1) - p0^(n-1) )
K.<z> = NumberField( Rprime )
for P in K.primes_of_degree_one_iter():
print P
The long list ends with:
Fractional ideal (44893, 26936/625*z^5 - 134679/5*z - 8174)
Fractional ideal (49697, -99393/3125*z^5 + 99394/5*z - 24271)
Fractional ideal (71429, -71428/3125*z^5 + 71429/5*z - 13020)
`P` is now the last entry, it is a **fractional** ideal. We can factor it like:
sage: P.factor()
(Fractional ideal (7, 7/3125*z^5))^-1 * (Fractional ideal (71429, 7/3125*z^5 - 19711))
Please explain what should be done in this situation. I need one more comment to paste some code...Thu, 16 Nov 2017 07:34:47 -0600http://ask.sagemath.org/question/39593/polynomial-mod-ideal/?comment=39603#post-id-39603Answer by dan_fulea for <p>I have a polynomial $R\in\mathbb{Z}[x]$.
I then define $R' = \frac{d}{dx}R$, and look at the splitting field of $R'$, $K$, an algebraic number field.
Now, I want to find a prime ideal $\mathfrak{p}$ of $L$ of absolute degree 1 such that $R\mod\mathfrak{p}$ is irreducible.</p>
<p>To do this, I've set up:</p>
<pre><code>p0, n = 5, 7
L = PolynomialRing(ZZ,'x')
R = L(x^n-p0^(n-1)x+p0)
Rprime = L(n*x^(n-1)-p0^(n-1))
K = NumberField(Rprime, 'z')
for P in K.primes_of_degree_one_iter():
<stuff>
</code></pre>
<p>I'm now trying ti find the right thing to do for <stuff>. Namely, how can I reduce the polynomial R with respect to the ideal P?</p>
http://ask.sagemath.org/question/39593/polynomial-mod-ideal/?answer=39631#post-id-39631
The link to the article is
(https://arxiv.org/abs/1103.5728)[https://arxiv.org/abs/1103.5728] .
I think, the following code is relevant - for the comment, not for the posted question:
p0 = 5
n = 7
S.<x> = ZZ[]
R = x^n - p0^(n-1)*x + p0
Rx = n*x^(n-1) - p0^(n-1)
F.<u> = CyclotomicField( n-1 )
L.<a> = F.extension( x^(n-1) - n )
LPrimes = L.primes_of_degree_one_list( 100 )
for LP1 in LPrimes:
p1 = LP1.relative_norm().norm()
Fp1 = GF(p1)
Rp1 = PolynomialRing( Fp1, names='x' )
print "p1 = %s" % p1
print "R mod p1 is %s :: %s" % ( Rp1(R ), 'IRRED' if Rp1(R).is_irreducible() else 'RED' )
print "Rx mod p1 is %s == %s" % ( Rp1(Rx), Rp1(Rx).factor() )
(Have to send, and catch that train, possibly be back again.)
Fri, 17 Nov 2017 03:22:45 -0600http://ask.sagemath.org/question/39593/polynomial-mod-ideal/?answer=39631#post-id-39631Comment by orangejake for <p>The link to the article is
(<a href="https://arxiv.org/abs/1103.5728)[https://arxiv.org/abs/1103.5728]">https://arxiv.org/abs/1103.5728)[http...</a> .</p>
<pre><code> I think, the following code is relevant - for the comment, not for the posted question:
p0 = 5
n = 7
S.<x> = ZZ[]
R = x^n - p0^(n-1)*x + p0
Rx = n*x^(n-1) - p0^(n-1)
F.<u> = CyclotomicField( n-1 )
L.<a> = F.extension( x^(n-1) - n )
LPrimes = L.primes_of_degree_one_list( 100 )
for LP1 in LPrimes:
p1 = LP1.relative_norm().norm()
Fp1 = GF(p1)
Rp1 = PolynomialRing( Fp1, names='x' )
print "p1 = %s" % p1
print "R mod p1 is %s :: %s" % ( Rp1(R ), 'IRRED' if Rp1(R).is_irreducible() else 'RED' )
print "Rx mod p1 is %s == %s" % ( Rp1(Rx), Rp1(Rx).factor() )
</code></pre>
<p>(Have to send, and catch that train, possibly be back again.)</p>
http://ask.sagemath.org/question/39593/polynomial-mod-ideal/?comment=39665#post-id-39665This is exactly what I needed! A few questions:
1. Is there any particular reason that the field $F$ is constructed as an extension of a cyclotomic, instead of just specifying the two generators?
2. Similar to the above, is there any particular reason the norm of p1 is computed as the relative norm, and then the (what I'm assuming is) absolute norm?Sun, 19 Nov 2017 16:23:01 -0600http://ask.sagemath.org/question/39593/polynomial-mod-ideal/?comment=39665#post-id-39665