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Author: Ahmed Alharbi
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Description: Jupyter notebook ANT/fourth_homework.ipynb

Name: Ahmed Alharbi

Student Number: 1901974

Leiden University

Q.9

Let R=Z[19]R = \mathbb{Z}[\sqrt{-19}] and L=Z[α+12]L = \mathbb{Z}[\frac{\alpha+1}{2}] which is Dedekind by theorem 3.19. For an ideal i\mathfrak{i} we have L/iin/in+1L/\mathfrak{i} \cong \mathfrak{i}^{n}/\mathfrak{i}^{n+1} as discussed during the last lecture. Also, we have ikik1L(L/ik)/(ik/ik+1)L/ik\mathfrak{i}^k \subset \mathfrak{i}^{k-1} \subset L \Rightarrow \left(L/\mathfrak{i}^{k}\right) / \left(\mathfrak{i}^{k}/\mathfrak{i}^{k+1}\right) \cong L/ \mathfrak{i}^{k} by the Third isomorphism thoerem (As modules).

Let p\mathfrak{p} our ideal after lifting it to LL, p=(2,1+19)=2(1,1+192)=2L\mathfrak{p} = (2, 1+ \sqrt{-19}) = 2(1, \frac{1+ \sqrt{-19}}{2}) = 2L. (There might be an abuse of notation)

We know that any element in L=Z[α+12]L = \mathbb{Z}[\frac{\alpha+1}{2}] can be written as a+b1+α2a+ b \frac{1+\alpha}{2}. By direct computations we can see that the norm is 22=42\cdot 2 = 4

Now since: L/pp/p2p2/p3L/\mathfrak{p} \cong \mathfrak{p}/\mathfrak{p}^{2}\cong \mathfrak{p}^{2}/\mathfrak{p}^{3} 4=L/p=p/p2=p2/p34 = \left| L/\mathfrak{p} \right| =\left| \mathfrak{p}/\mathfrak{p}^{2} \right|=\left| \mathfrak{p}^{2}/\mathfrak{p}^{3} \right|

L/p=(L/p2)÷(p/p2)(L/p2)=42=16 \left |L/\mathfrak{p} \right | =\left |\left( L/\mathfrak{p^2}\right) \right | ÷\left | \right( \mathfrak{p}/\mathfrak{p^2} \left) \right | \Rightarrow \left|\left( L/\mathfrak{p^2}\right) \right | = 4^2 = 16

L/p2=(L/p3)÷(p2/p3)(L/p3)=416=64 \left |L/\mathfrak{p^2} \right | =\left |\left( L/\mathfrak{p^3}\right) \right | ÷\left | \right( \mathfrak{p^2}/\mathfrak{p^3} \left) \right | \Rightarrow \left|\left( L/\mathfrak{p^3}\right) \right | = 4\cdot 16 = 64

Again we have:

2=L/R=(L/p)÷(R/p)(R/p)=42=22 = \left |L/R\right | =\left |\left( L/\mathfrak{p}\right) \right | ÷\left | \right( R/\mathfrak{p} \left) \right | \Rightarrow \left|\left( R/\mathfrak{p}\right) \right | = \frac{4}{2} = 2

2=L/R=(L/p)÷(R/p)(R/p2)=162=82 = \left |L/R\right | =\left |\left( L/\mathfrak{p}\right) \right | ÷\left | \right( R/\mathfrak{p} \left) \right | \Rightarrow \left|\left( R/\mathfrak{p^2}\right) \right | = \frac{16}{2} = 8

2=L/R=(L/p)÷(R/p)(R/p3)=642=322 = \left |L/R\right | =\left |\left( L/\mathfrak{p}\right) \right | ÷\left | \right( R/\mathfrak{p} \left) \right | \Rightarrow \left|\left( R/\mathfrak{p^3}\right) \right | = \frac{64}{2} = 32

Finally, from previous set of excercises we found that R/p=2|R/ \mathfrak{p}| = 2 and from calculations above we find that

2=R/p=(R/p2)÷(p/p2)(p/p2)=82=42 = \left |R/ \mathfrak{p} \right | =\left |\left( R/\mathfrak{p^2}\right) \right | ÷\left | \right( \mathfrak{p}/\mathfrak{p^2} \left) \right | \Rightarrow \left|\left( \mathfrak{p}/\mathfrak{p^2}\right) \right | = \frac{8}{2} = 4. Since, the number of elements are different then R/pR/ \mathfrak{p} and p/p2\mathfrak{p}/\mathfrak{p^2} can't be isomorphic.

Finally, for seek of contradiction assume 2Rp2 R_{\mathfrak{p}}contains a power of the the maximal idea p\mathfrak{p}, then as we showed in the last set of excercises that p2=2pby repeated applicationpn+1=2np\mathfrak{p}^2 = 2\mathfrak{p} \overset{\text{by repeated application}}{\Rightarrow} \mathfrak{p}^{n+1} = 2^n \mathfrak{p} and for some power 2mp=pm+1=2Rpintegral domain2n1p=Rp2^m\mathfrak{p} = \mathfrak{p}^{m+1} = 2R_{\mathfrak{p}} \overset{\text{integral domain}}{\Rightarrow} 2^{n-1} \mathfrak{p} = R_{\mathfrak{p}} contradiction with the fact that 2Rpp2 R_{\mathfrak{p}} \subset \mathfrak{p} which is a proper ideal.

Q.10

  • First of all notice that f(x)=x3x1f(x) = x^3 -x -1 is irreducible over Z\mathbb{Z}, consequently(Gauss lemma), irreducible over Q\mathbb{Q}.

Well, for any aR,a1f(a)<0a \in \mathbb{R}, a\leq 1 \Rightarrow f(a)< 0 . Also, aR,a2f(a)>0a \in \mathbb{R}, a\geq 2 \Rightarrow f(a) > 0. The only real solutions θ\theta, by intermediate value theorem, is strictly between 1 and 2. Hence, it has no solutions in Z\mathbb{Z}

  • Now to show the number ring R=Z[α]R = \mathbb{Z}[\alpha] is Dedekind, it is sufficient to to show that each ideal is regular. The main tool that will be used here is the third statement in the Kummer-Dedekind theorem 3.1.

Let p\mathfrak{p} a prime ideal of Z[α]\mathbb{Z}[\alpha] lies over a prime number pp. If ff has multiple factors mod pp then ff and ff' has a common factor mod pp. Consider the linear combination x3x1x3(3x21)=(23x+1)x^3 -x -1 - \frac{x}{3} (3x^2 -1) = -(\frac{2}{3}x +1) which has the root x=32x = -\frac{3}{2}. Since it is multiple root then 0=f(32)=234 mod p\overline{0}=\overline{f'(-\frac{3}{2})}= \overline{\frac{23}{4}} \text{ mod } p iff p=23,x=p=23, x= . However, f(x)=(x10)g+rf(x) = (x-10)g + r where rr is constant, hence r=989=f(10)=0g+rr = 989 = f(10) = 0\cdot g + r but 432∤989=234343^2 \not | 989=23\cdot 43, hence all ideals are regular.

  • If an ideal has a norm at most 30 then it lies over a prime numbers which is at most 30, i.e. [2,3,5,7,11,13,17,19,23,29].
    • Note: A prime ideal p\mathfrak{p} can't lie over more than one prime pp of Z\mathbb{Z} due to Z[α]/pF[α]\mathbb{Z}[\alpha]/\mathfrak{p} \cong \mathbb{F}[\overline{\alpha}] (Since αZ=(p) some prime\alpha \bigcap \mathbb{Z} = (p) \text{ some prime})

Now by trying all possible solutions for the polynomial x3x1x^3 -x -1 over primes in our list above.

In [1]:
from sage.all import * from IPython.display import display, Math, Latex
In [2]:
for i in [p for p in range(30) if is_prime(p)]: K = GF(Integer(i))['alpha']; (x,) = K._first_ngens(1) display(Math("\\overline{f}="+latex(factor(x**3 -x -1))+ "\ mod\ %d"%i))
f=(α3+α+1) mod 2\overline{f}= (\alpha^{3} + \alpha + 1) \ mod\ 2
f=(α3+2α+2) mod 3\overline{f}= (\alpha^{3} + 2 \alpha + 2) \ mod\ 3
f=(α+3)(α2+2α+3) mod 5\overline{f}= (\alpha + 3) \cdot (\alpha^{2} + 2 \alpha + 3) \ mod\ 5
f=(α+2)(α2+5α+3) mod 7\overline{f}= (\alpha + 2) \cdot (\alpha^{2} + 5 \alpha + 3) \ mod\ 7
f=(α+5)(α2+6α+2) mod 11\overline{f}= (\alpha + 5) \cdot (\alpha^{2} + 6 \alpha + 2) \ mod\ 11
f=(α3+12α+12) mod 13\overline{f}= (\alpha^{3} + 12 \alpha + 12) \ mod\ 13
f=(α+12)(α2+5α+7) mod 17\overline{f}= (\alpha + 12) \cdot (\alpha^{2} + 5 \alpha + 7) \ mod\ 17
f=(α+13)(α2+6α+16) mod 19\overline{f}= (\alpha + 13) \cdot (\alpha^{2} + 6 \alpha + 16) \ mod\ 19
f=(α+20)(α+13)2 mod 23\overline{f}= (\alpha + 20) \cdot (\alpha + 13)^{2} \ mod\ 23
f=(α3+28α+28) mod 29\overline{f}= (\alpha^{3} + 28 \alpha + 28) \ mod\ 29

Hence, 2,3,13 and 29 inert. Looking at the quotient by ideals as quotient of modules we find the norm of each is p3p^3. Theorem 3.4

In other words, Z[α]Z×Z×Z\mathbb{Z}[\alpha] \cong \mathbb{Z}\times\mathbb{Z} \times \mathbb{Z} as Zmodule\mathbb{Z}-\text{module} with 1,α,α21, \alpha, \alpha^2 as basis. Hence Z[α]/(p)=p3\left |\mathbb{Z}[\alpha]/(p) \right | = p^3 and since (p)pZ[α]/(p)Z[α]/p(p) \subseteq \mathfrak{p} \Rightarrow \mathbb{Z}[\alpha]/(p) \supseteq \mathbb{Z}[\alpha]/\mathfrak{p}. Where, p\mathfrak{p} is a prime divisor of (p)(p). We are now in a finite ring, by a simple counting argument, using quotients and irreducibility of the polynomial, one can conclude that N(p)=pdeg(p)\mathcal{N}(\mathfrak{p}) = p^{deg(\mathfrak{p})}

So, we exclude 13 and 23 since their norm are bigger than 30. By the same method, we enumerate all all norms that are less than 30 with the ideals.

N((2))=8\mathcal{N}((2)) = 8 N((3))=27\mathcal{N}((3)) = 27 N((5,α+3))=5,N((5,α2+2α+3))=52\mathcal{N}((5,\alpha+3)) = 5,\mathcal{N}\left( (5,\alpha^2+2\alpha+3)\right) = 5^2

N((11,(α+5)))=11\mathcal{N}((11, (\alpha + 5))) = 11 N((17,α+12))=17\mathcal{N}((17, \alpha + 12)) = 17 N((19,α+13))=19\mathcal{N}((19, \alpha + 13)) = 19 N((23,(α10)))=23\mathcal{N}((23,(\alpha - 10))) = 23 N((23,α+13))=23\mathcal{N}((23, \alpha + 13)) = 23

Finally, in Z[α]\mathbb{Z}[\alpha] we have 1=α3α=α(α21)=α(α1)(α+1)1 = \alpha^3 -\alpha = \alpha(\alpha^2-1) = \alpha(\alpha-1)(\alpha+1) . Hence, u=α+1u = \alpha+1 is unit and of infinte order because un=m+kα+hα2u^n = m+ k \alpha + h \alpha^2 where all coefficients are positive and they increase as nn increases due to the fact a3=a+1a^3 = a+1.