TUT ALG3 6

Problem 1

Let $R=\mathbb{Z}[\sqrt{-5}]$. Let $\alpha =\sqrt{-5}$. Let ${\mathfrak{p}}_1=(2,1+\alpha )$, ${\mathfrak{p}}_2=(2,1-\alpha )$, ${\mathfrak{p}}_3=(3,1+\alpha )$, and ${\mathfrak{p}}_4=(3,1-\alpha )$.

Exercise 304.1(Exercise).

Show that ${\mathfrak{p}}_i$ are prime ideals for $i=1,2,3,4$. Are they maximal ideals?

We know by Lem 332.3 that $\mathbb{Z}[\sqrt{-5}]\cong \mathbb{Z}[x]/(x^2+5)$. Using Exm 107.16,

$$\begin{array}{r}\frac{\mathbb{Z}[\sqrt{-5}]}{(2,1+\alpha )}\cong \frac{\mathbb{Z}[x]}{(2,1+x,x^2+5)}=\frac{\mathbb{Z}[x]}{(2,1+x)}\cong \frac{{\mathbb{F}}_2[x]}{(1+x)}.\end{array}$$

Clearly, $1+x$ is irreducible in ${\mathbb{F}}_2[x]$. Since ${\mathbb{F}}_2[x]$ is a PID (and hence a UFD by Prp 114.11), the notions of prime, maximal, and irreducible coincide (Prp 114.10, Prp 109.9). Thus, ${\mathbb{F}}_2[x]/(1+x)$ is a field, and ${\mathfrak{p}}_1$ is a maximal ideal. Similarly, ${\mathfrak{p}}_2$, ${\mathfrak{p}}_3$, and ${\mathfrak{p}}_4$ are maximal ideals.

Exercise 304.2(Exercise).

Show that $2,3,7,11,1+\sqrt{-5},1-\sqrt{-5}$ are irreducible in $R$.

2b7447

The field norm (Exr 302.2) $N(\alpha )=\alpha \overline{\alpha }$ defined on $\mathbb{Q}[\sqrt{-5}]$ is multiplicative and takes on positive integer values $N(a+\sqrt{-5}b)=a^2+5b^2$ when restricted to $\mathbb{Z}[\sqrt{-5}]$. If $2=\alpha \beta$, we must have $N(\alpha )N(\beta )=4$. This forces either $\alpha =1$ or $\beta =1$. The remaining elements are irreducible by the same argument.

Exercise 304.3(Exercise).

Show that $(6)={\mathfrak{p}}_1{\mathfrak{p}}_2{\mathfrak{p}}_3{\mathfrak{p}}_4$.

Compute:

$$\begin{array}{rl}{\mathfrak{p}}_1{\mathfrak{p}}_2{\mathfrak{p}}_3{\mathfrak{p}}_4& =(4,2+2\alpha ,2-2\alpha ,6)(9,3+3\alpha ,3-3\alpha ,6)\\ & =(2+2\alpha ,2-2\alpha ,6)(9,3+3\alpha ,3-3\alpha )\\ & =(12(\alpha -2),12(\alpha +2),18(1+\alpha ),18(1-\alpha ),36,54).\end{array}$$

By subtracting the first two generators, we get $48\in {\mathfrak{p}}_1{\mathfrak{p}}_2{\mathfrak{p}}_3{\mathfrak{p}}_4$. It follows that $54-48=6\in {\mathfrak{p}}_1{\mathfrak{p}}_2{\mathfrak{p}}_3{\mathfrak{p}}_4$. Since $6$ divides all the generators obtained above, we have ${\mathfrak{p}}_1{\mathfrak{p}}_2{\mathfrak{p}}_3{\mathfrak{p}}_4=(6)$.

Exercise 304.4(Exercise).

Is $(6)={\mathfrak{p}}_1^2\cap {\mathfrak{p}}_3\cap {\mathfrak{p}}_4$?

${\mathfrak{p}}_1^2=(2)$. Note that ${\mathfrak{p}}_1^2$, ${\mathfrak{p}}_3$, and ${\mathfrak{p}}_4$ are pairwise comaximal. By Thm 111.2,

$$\begin{array}{rl}{\mathfrak{p}}_1^2\cap {\mathfrak{p}}_3\cap {\mathfrak{p}}_4& ={\mathfrak{p}}_1^2{\mathfrak{p}}_3{\mathfrak{p}}_4\\ & =(2)(9,3+3\alpha ,3-3\alpha )\\ & =(18,6+6\alpha ,6-6\alpha )\\ & =(6).\end{array}$$

Exercise 304.5(Exercise).

Let ${\mathfrak{q}}_1=(3+5\alpha ),{\mathfrak{q}}_2=(2+2\alpha ),{\mathfrak{q}}_3=(4+\alpha ,1+2\alpha )$. Determine which of these are prime ideals.

$(3+5\alpha )$ is prime in $\mathbb{Z}[\sqrt{-5}]$ iff $(3+5x,x^2+5)$ is prime in $\mathbb{Z}[x]$. Note that $(2x-1)(5x+3)-10(x^2+5)=x-53$, so $(x-53)\in (3+5x,x^2+5)$. Now,

$$\begin{array}{rl}\frac{\mathbb{Z}[x]}{(3+5x,x^2+5)}& =\frac{\mathbb{Z}[x]}{(x-53,3+5x,x^2+5)}\\ & \cong \frac{\mathbb{Z}}{(268,2814)}\\ & =\frac{\mathbb{Z}}{(134)}.\end{array}$$

Since $134$ is not prime in $\mathbb{Z}$, it follows that $(3+5\alpha )$ is not prime in $\mathbb{Z}[\sqrt{-5}]$.

Warning

Recall that $\mathbb{Z}[\sqrt{-5}]$ is not a UFD, so showing $3+5\alpha$ is irreducible does not help!

$(2+2\alpha )=(2)(1+\alpha )$. By Prp 112.5.3, $(2+2\alpha )$ is not prime.

$(4+\alpha ,1+2\alpha )$ is dealt with similarly:

$$\begin{array}{rl}\frac{\mathbb{Z}[x]}{(4+x,1+2x,x^2+5)}& =\frac{\mathbb{Z}}{(7,21)}=\frac{\mathbb{Z}}{(7)}.\end{array}$$

Since $7$ is prime in $\mathbb{Z}$, $(4+\alpha ,1+2\alpha )$ is prime in $\mathbb{Z}[\sqrt{-5}]$.

Exercise 304.6(Exercise).

Let $I=(3,1+\alpha )$. Show that $I^2$ is a principal ideal.

$I^2=(9,3+3\alpha ,2\alpha -4)$. $\alpha -2\in I^2$, so $I^2=(9,3+3\alpha ,\alpha -2)$. $3+3\alpha =3(\alpha -2)+9$ and $9=(2-\alpha )(2+\alpha )$, so $I=(\alpha -2)$.

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Problem 2

Let $\delta =(1+\sqrt{-19})/2$ and $R=\mathbb{Z}[\delta ]$.

Exercise 304.7(Exercise).

Show that $R\cong \mathbb{Z}[x]/(x^2-x+5)$.

$\delta \in \mathbb{Q}[\sqrt{-19}]$ and $\delta$ satisfies $x^2-x+5=0$. We are done by Lem 332.3.

Exercise 304.8(Exercise).

Determine whether $2$ and $3$ are prime in $R$.

By Clm 302.7, $R$ is a PID, and hence a UFD. It therefore suffices to determine the irreducibility of $2$ and $3$. We have shown in Clm 302.6 that $N(a+b\delta )=a^2+ab+5b^2$. The smallest values this norm takes are $1,4$. It is immediate that $2$ and $3$ are irreducible.

Exercise 304.9(Exercise).

Determine whether the following ideals are prime or maximal in $R[x]$:

1. ${\mathfrak{p}}_1=(7,{\delta }^{3}x+x+1)$
2. ${\mathfrak{p}}_2=(2{\delta }^2+3)$
3. ${\mathfrak{p}}_3=(\delta x+1)$
4. ${\mathfrak{p}}_4=(\delta x+2)$
5. ${\mathfrak{p}}_5=(2x\delta +1)$
6. ${\mathfrak{p}}_6=(11,2x\delta +1)$.

${\mathfrak{p}}_1$

Compute (using Exm 107.16, Thm 111.2, Exm 109.4):

$$\begin{array}{rlr}& & \\ & \frac{R}{(7)R}& a+b\delta +(7)R\\ & \cong \frac{\mathbb{Z}[y]}{(7,y^2-y+5)}& a+by+(7,y^2-y+5)\\ & \cong \frac{{\mathbb{F}}_7[y]}{(y^2-y+\overline{5})}& \overline{a}+\overline{b}y+(y^2-y+\overline{5})\\ & =\frac{{\mathbb{F}}_7[y]}{(y-\overline{2})(y-\overline{6})}& \overline{a}+\overline{b}y+(y-\overline{2})(y-\overline{6})\\ & \cong \frac{{\mathbb{F}}_7[y]}{(y-\overline{2})}\times \frac{{\mathbb{F}}_7[y]}{(y-\overline{6})}& (\overline{a}+\overline{b}y+(y-\overline{2}),\overline{a}+\overline{b}y+(y-\overline{6}))\\ & \cong {\mathbb{F}}_7\times {\mathbb{F}}_7.& (\overline{a+2b},\overline{a+6b}).\end{array}$$

Let $\phi :R/(7)R\to {\mathbb{F}}_7\times {\mathbb{F}}_7$ be the canonical isomorphism. Note that $\phi (a+b\delta +(7)R)=(\overline{a+2b},\overline{a+6b})$.

Next,

$$\begin{array}{rl}\frac{R[x]}{(7,1+({\delta }^3+1)x)}& =\frac{R[x]}{(7,1+(-4\delta -4)x)}\\ & \cong \frac{(R/(7)R)[x]}{((1+(7)R)+(-4\delta -4+(7)R)x)}\\ & \cong \frac{({\mathbb{F}}_7\times {\mathbb{F}}_7)[x]}{(\phi (1)+\phi (-4\delta -4)x)}\\ & =\frac{({\mathbb{F}}_7\times {\mathbb{F}}_7)[x]}{((\overline{1},\overline{1})+(\overline{2},\overline{0})x)}\\ & \cong \frac{{\mathbb{F}}_7[x]\times {\mathbb{F}}_7[x]}{((\overline{1}+\overline{2}x,\overline{1}))}\\ & \cong \frac{{\mathbb{F}}_7[x]}{(\overline{1}+\overline{2}x)}\times \frac{{\mathbb{F}}_7[x]}{(\overline{1})}\\ & \cong \frac{{\mathbb{F}}_7[x]}{(\overline{4}+x)}\\ & \cong {\mathbb{F}}_7.\end{array}$$

Thus, ${\mathfrak{p}}_1$ is a maximal ideal.

${\mathfrak{p}}_2$

$2{\delta }^2+3=2\delta -7$. $N(2\delta -7)=55=5\cdot 11$. There exist elements of norms $5$ and $11$ in $R$, and $2\delta -7$ in fact factorizes as $(-1+\delta )(2+\delta )$. Both factors have norm $5$ and $11$ respectively, and thus are not units by the proof of Clm 302.6. Thus, $2\delta -7$ is reducible in $R$, and since $R$ is a UFD, not prime. By Thm 115.10.1, $2\delta -7$ is not prime in $R[x]$.

${\mathfrak{p}}_3$

Let $\mathcal{R}$ be the field of fractions of $R$. Consider the homomorphism $\phi :R[x]\to \mathcal{R}$ given by $x\mapsto -1/\delta$. Clearly, $\delta x+1\in \ker \phi$. If $f(x)\in \ker \phi$, we have $f(x)=\overline{q}(x)(\delta x+1)+\overline{r}$, where $\overline{q}(x)\in \mathcal{R}[x]$ and $\overline{r}\in \mathcal{R}$. $f(-1/\delta )=0$ forces $\overline{r}=0$, and $(\delta x+1)$ divides $f(x)$ in $R[x]$ by Thm 115.10.4. Thus, $\ker \phi =(\delta x+1)$. Clearly, $\mathrm{im}\phi =R\left[\frac{1}{\delta }\right]$, so we have

$$\begin{array}{r}\frac{R[x]}{(\delta x+1)}\cong R\left[\frac{1}{\delta }\right].\end{array}$$

$R\left[\frac{1}{\delta }\right]$ is a subring of the field $\mathcal{R}$, and hence is an integral domain. However, $R\left[\frac{1}{\delta }\right]$ is not a field. Thus, ${\mathfrak{p}}_3$ is prime but not maximal.

${\mathfrak{p}}_4$

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Problem 3

Exercise 304.10(Exercise).

Let $R=\mathbb{Z}[\sqrt[3]{2}]$. Is it possible to write $(5)$ as a product of prime ideals?

Use Dedekind’s theorem to obtain ${\mathfrak{p}}_1=(5,\alpha -3)$ and ${\mathfrak{p}}_2=(5,{\alpha }^2+3\alpha +4)$. It is easy to show that ${\mathfrak{p}}_1$ and ${\mathfrak{P}}_2$ are maximal and that $(5)={\mathfrak{p}}_1{\mathfrak{p}}_2$.

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Problem 4

Let $R=\mathbb{Z}[\sqrt{-n}]$ where $n⩾3$.

Exercise 304.11(Exercise).

Which of the following are irreducible or prime in $R$?

1. $2$
2. $\sqrt{-n}$
3. $1+\sqrt{-n}$

Exercise 304.12(Exercise).

Show that $R$ is not a UFD.

$2$ is irreducible but not prime.

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Problem 5

Let $R=\mathbb{Z}[\omega ]$ where $\omega =e^{2\pi i/3}$. Let $p\ne 3$ be a prime.

Exercise 304.13(Exercise).

Show that $x^2+x+1$ has a root in $\mathbb{Z}/p\mathbb{Z}$ iff $p\ncong 1\mathrm{mod}3$.