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Author: Ahmed Alharbi
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Name: Ahmed Alharbi

Student Number: 1901974

Leiden University

Q.9

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

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

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

Now since: $L/\mathfrak{p} \cong \mathfrak{p}/\mathfrak{p}^{2}\cong \mathfrak{p}^{2}/\mathfrak{p}^{3}$ $4 = \left| L/\mathfrak{p} \right| =\left| \mathfrak{p}/\mathfrak{p}^{2} \right|=\left| \mathfrak{p}^{2}/\mathfrak{p}^{3} \right|$

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

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

Again we have:

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

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

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

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

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

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

Q.10

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

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

• Now to show the number ring $R = \mathbb{Z}[\alpha]$ 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 $\mathfrak{p}$ a prime ideal of $\mathbb{Z}[\alpha]$ lies over a prime number $p$. If $f$ has multiple factors mod  $p$ then $f$ and $f'$ has a common factor mod  $p$. Consider the linear combination $x^3 -x -1 - \frac{x}{3} (3x^2 -1) = -(\frac{2}{3}x +1)$ which has the root $x = -\frac{3}{2}$. Since it is multiple root then $\overline{0}=\overline{f'(-\frac{3}{2})}= \overline{\frac{23}{4}} \text{ mod } p$ iff $p=23, x=$. However, $f(x) = (x-10)g + r$ where $r$ is constant, hence $r = 989 = f(10) = 0\cdot g + r$ but $43^2 \not | 989=23\cdot 43$, 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 $\mathfrak{p}$ can't lie over more than one prime $p$ of $\mathbb{Z}$ due to $\mathbb{Z}[\alpha]/\mathfrak{p} \cong \mathbb{F}[\overline{\alpha}]$ (Since $\alpha \bigcap \mathbb{Z} = (p) \text{ some prime}$)

Now by trying all possible solutions for the polynomial $x^3 -x -1$ 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))


$\overline{f}= (\alpha^{3} + \alpha + 1) \ mod\ 2$
$\overline{f}= (\alpha^{3} + 2 \alpha + 2) \ mod\ 3$
$\overline{f}= (\alpha + 3) \cdot (\alpha^{2} + 2 \alpha + 3) \ mod\ 5$
$\overline{f}= (\alpha + 2) \cdot (\alpha^{2} + 5 \alpha + 3) \ mod\ 7$
$\overline{f}= (\alpha + 5) \cdot (\alpha^{2} + 6 \alpha + 2) \ mod\ 11$
$\overline{f}= (\alpha^{3} + 12 \alpha + 12) \ mod\ 13$
$\overline{f}= (\alpha + 12) \cdot (\alpha^{2} + 5 \alpha + 7) \ mod\ 17$
$\overline{f}= (\alpha + 13) \cdot (\alpha^{2} + 6 \alpha + 16) \ mod\ 19$
$\overline{f}= (\alpha + 20) \cdot (\alpha + 13)^{2} \ mod\ 23$
$\overline{f}= (\alpha^{3} + 28 \alpha + 28) \ mod\ 29$

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

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

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.

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

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

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