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 Lemma 332.2 that $\mathbb{Z}[\sqrt{-5}]\cong \mathbb{Z}[x]/(x^2+5)$. Using Example 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 Proposition 114.11), the notions of prime, maximal, and irreducible coincide (Proposition 114.10, Proposition 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 (Exercise 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 Theorem 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 Proposition 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 $R\cong \mathbb{Z}[x]/(x^2-x+5)$.