TUT ALG3 4

Problem 1

Immediate from Proposition 114.2.

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Problem 4

Let $I^e$ be the ideal generated by $\varphi (I)$ in $S$. Note that $x^{2}y=y^3\left(x^2/y^2\right)\in I^e$. Since $x^{2}y\in R$, $x^{2}y\in I^e\cap R=I^{ec}$. However, $x^{2}y\notin I$. Thus, $I^{ec}\ne I$.

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Problem 6

Dummit & Foote (2004, p. 276)

Claim 302.1(Claim).

Let $R$ be a euclidean domain that is not a field. Then, there exists $u\in R\setminus (R^{\times }\cup \{0\})$ such that for every $x\in R$, there is some $z\in R^{\times }\cup \{0\}$ such that $u$ divides $x-z$.

Proof.

Suppose$R$ is Euclidean with respect to some norm $N$ and let $u$ be an element of $R\setminus (R^{\times }\cup \{0\})$ (which is nonempty since $R$ is not a field) of minimal norm. For any $x\in R$, write $x=qu+r$ where $r$ is either $0$ or $N(r)<N(u)$. In either case the minimality of $u$ implies $r\in R^{\times }\cup \{0\}$.□

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Problem 8

Exercise 302.2(Exercise).

Determine all the units in $\mathbb{Z}[\sqrt{2}]$.

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Let $R=\mathbb{Z}[\sqrt{2}]$. Define $N:R\to \mathbb{N}$ by $N(\alpha )=\alpha \overline{\alpha }$. $N$ is clearly multiplicative, since $N(\alpha \beta )=\alpha \beta \overline{\alpha \beta }=\alpha \overline{\alpha }\beta \overline{\beta }=N(\alpha )N(\beta )$. Now, If $\alpha \in \mathbb{Z}[\sqrt{2}]$ is a unit, we must have $N(\alpha )N({\alpha }^{-1})=N(\alpha {\alpha }^{-1})=N(1)=1$, so $N(\alpha )$ must be $\pm 1$. Thus, if $a+b\sqrt{2}$ is a unit, it must satisfy the equation

$$a^2-2b^2=\pm 1$$E1

$\alpha$ is a unit iff $-\alpha$ is a unit, so we can restrict ourselves to positive solutions.

Claim 302.3(Claim).

$\gamma :=1+\sqrt{2}$ is the smallest unit of $\mathbb{Z}[\sqrt{2}]$ greater than $1$.

Proof.

Assume$\beta =c+d\sqrt{2}$ such that $1<\beta <\gamma$ and $N(\beta )=\pm 1$. It is easy to see that $c$ and $d$ cannot be $0$. Clearly, both $c$ and $d$ cannot be positive. If $d$ is negative and $c$ is positive, we have

$$\begin{array}{r}\frac{1}{c-|d|\sqrt{2}}=\frac{c+|d|\sqrt{2}}{N(\beta )}.\end{array}$$

Since $1/\beta >0$, it must be that $N(\beta )=1$. This gives us $\beta =1/(c+|d|\sqrt{2})<1$, a contradiction. If $d$ is positive and $c$ is negative,

$$\begin{array}{r}-|c|+d\sqrt{2}=\frac{-N(\beta )}{|c|+d\sqrt{2}}.\end{array}$$

Here, $N(\beta )$ is forced to be $-1$, and again we obtain $\beta =1/(|c|+d\sqrt{2})<1$.□

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Claim 302.4(Claim).

Every positive unit of $\mathbb{Z}[\sqrt{2}]$ is of the form ${\gamma }^n$ for $n\in \mathbb{Z}$.

Proof.

Suppose there exists a unit$\beta$ such that ${\gamma }^n<\beta <{\gamma }^{n+1}$ for some $n\in \mathbb{Z}$. On dividing through by ${\gamma }^n$, we obtain

$$1<\beta /{\gamma }^n<\gamma .$$

$\beta /{\gamma }^n$ is a unit that lies between $1$ and $\gamma$, contradicting Claim 3.□

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Problem 9

See Wilson (2011)

Exercise 302.5(Exercise).

Show that $R=\mathbb{Z}[(1+\sqrt{-19})/2]$ is a PID but not an ED.

Denote $(1+\sqrt{-19})/2$ by $\omega$.

Claim 302.6(Claim).

$R$ is not a Euclidean domain.

Proof.

Define the field norm on$\mathbb{Q}(\sqrt{-19})$ by $N(\alpha )=\alpha \overline{\alpha }$. Show that it is multiplicative. Show that when it is restricted to $R$, it takes positive integer values $N(a+b\omega )=a^2+ab+5b^2$. So a unit in $R$ must have norm $\pm 1$. It is straightforward to see that the only solutions of $N(a+b\omega )=\pm 1$ are $\{\pm 1\}$.

Note that the smallest nonzero values of $a^2+ab+5b^2$ are $1$ and $4$.

If $R$ is a euclidean domain, it must have a universal side divisor $u\in R-\{\pm 1,0\}$. Take $x=2$, so $u$ must divide $2-0$ or $2\pm 1$, that is, $u$ is a non-unit divisor of $2$ or $3$. Use the norm to argue that the the only possible values for $u$ are $\pm 2$ or $\pm 3$. Now, take $x=\omega$, and note that none of $\omega$, $\omega \pm 1$ are divisible by $\pm 2$ or $\pm 3$, so none of these are universal side divisors.□

Claim 302.7(Claim).

$R$ is a PID.

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Let $I$ be an ideal in $R$. For all $a+b\omega \in R$, the square of the complex norm is a nonnegative integer:

$$|a+b\omega {|}^2=a^2+ab+5b^2.$$

Thus, we can choose $\beta \in I$ such that $|\beta |$ is as small as possible, thanks to the well-ordering of $\mathbb{N}$. We aim to show that $I=⟨\beta ⟩$, so suppose not. Then there is an element $\alpha \in I\setminus ⟨\beta ⟩$.

We will show that there exist $p,q\in R$ such that

$$0<\left|p\left(\frac{\alpha }{\beta }\right)-q\right|<1,$$

where $\alpha /\beta \in \mathbb{Q}(\sqrt{-19})$. This derives a contradiction, since it implies

$$0<\left|p\alpha -q\beta \right|<|\beta |,$$

where $p\alpha -q\beta \in R$.

First, observe that for any $r\in R$, ${\alpha }^{\prime }=\alpha +\beta r\in I\setminus ⟨\beta ⟩$. ${\alpha }^{\prime }/\beta =\alpha /\beta +r$, so we can choose $r$ such that $-\sqrt{19}/4\le \mathrm{Im}{\alpha }^{\prime }/\beta \le \sqrt{19}/4$. If $-\sqrt{3}/2<\mathrm{Im}{\alpha }^{\prime }/\beta <\sqrt{3}/2$, ${\alpha }^{\prime }/\beta$ lies at distance less than $1$ from some integer $n$, so take $p=1$ and $q=n$ (${\alpha }^{\prime }/p$ cannot be equal to some integer $n$, since that would imply ${\alpha }^{\prime }\in ⟨\beta ⟩$). Thus, assume $\sqrt{3}/2\le \mathrm{Im}{\alpha }^{\prime }/\beta \le \sqrt{19}/4$.

$$\begin{array}{rl}& \sqrt{3}/2\le \mathrm{Im}{\alpha }^{\prime }/\beta \le \sqrt{19}/4\\ & \sqrt{3}-\sqrt{19}/2\le \mathrm{Im}(2{\alpha }^{\prime }/\beta -\omega )\le 0.\end{array}$$

Note that $\sqrt{19}<\sqrt{27}=3\sqrt{3}$, so

$$-\sqrt{3}/2<\mathrm{Im}(2{\alpha }^{\prime }/\beta -\omega )\le 0,$$

so $2{\alpha }^{\prime }/\beta -\omega$ lies at a distance less than $1$ from some integer.
It is possible that $2{\alpha }^{\prime }/\beta -\omega$ is equal to some integer $q$, in which case ${\alpha }^{\prime }/\beta =(q+\omega )/2=(1\pm \sqrt{-19})/4+s$ for some $s\in R$. Here, choose $p=(1\mp \sqrt{-19})/2$ and $q=2$:

$$\begin{array}{r}\left|p\left(\frac{{\alpha }^{\prime }}{\beta }-s\right)-q\right|=\left|\frac{5}{2}-2\right|=\frac{1}{2}.\end{array}$$

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References

Dummit, D. S., & Foote, R. M. (2004). Abstract Algebra (3rd ed). Wiley.

Wilson, R. A. (2011). An Example of a PID Which Is Not a Euclidean Domain.