# Difference between revisions of "AAED2"

Let $\displaystyle R\,$ be a Euclidean Domain with a function $\displaystyle \varphi$ . Prove that
(a) $\displaystyle \varphi(1) = \min\{\varphi(a) \ |\ a \in R \backslash \{0\}\}$

For all $\displaystyle a \in R \,\backslash \{0\}$ , the property of the function gives

 $\displaystyle \varphi (1)$ $\displaystyle \leq \varphi (1\cdot a)$ $\displaystyle \varphi (1)$ $\displaystyle \leq \varphi (a)$

Therefore, $\displaystyle \varphi (1)$ has the minimum value.

(b) $\displaystyle R^\times = \{r \in R \backslash \{0\} \ |\ \varphi(r) = \varphi(1)\}$

Let $\displaystyle S = \{r \in R \,\backslash \{0\} \ | \ \varphi (r) = \varphi (1)\}$ .
($\displaystyle R^\times \subseteq S$ ): Let $\displaystyle x \in R^\times$ . Then, $\displaystyle x^{-1} \in R^\times$ . So, by property of the function

 $\displaystyle \varphi (x)$ $\displaystyle \leq \varphi (x\cdot x^{-1})$ $\displaystyle \varphi (x)$ $\displaystyle \leq \varphi (1)$

Since $\displaystyle \varphi (1)$ is the minimum value, $\displaystyle \varphi (x) = \varphi (1)$ . Therefore, $\displaystyle x \in S$ , and so $\displaystyle R^\times \subseteq S$ .

($\displaystyle S \subseteq R^\times$ ): Let $\displaystyle x \in S$ . Then $\displaystyle \varphi (x) = \varphi (1)$ , and so $\displaystyle \varphi (x)$ has the minimum value. Now, since $\displaystyle R\,$ is a Euclidean Domain, there exists $\displaystyle q, r \in R$ such that $\displaystyle 1 = qx + r\,$ , with $\displaystyle r = 0\,$ or $\displaystyle \varphi (r) < \varphi (x)$ . Since $\displaystyle \varphi (x)$ has the minimum value, $\displaystyle \varphi (r) < \varphi (x)$ is impossible, and so $\displaystyle r = 0\,$ . Thus, $\displaystyle 1 = qx\,$ , and $\displaystyle x\,$ is a unit with inverse $\displaystyle q\,$ . Therefore, $\displaystyle x \in R^\times$ , and so $\displaystyle S \subseteq R^\times$ .

Therefore, $\displaystyle R^\times = \{r \in R \,\backslash \{0\} \ | \ \varphi (r) = \varphi (1)\}$ .

(c) Use (b) to determine $\displaystyle \mathbb{Z}^\times, F^\times, F[X]^\times,$ and $\displaystyle \mathbb{Z}[i]^\times$

For $\displaystyle \mathbb{Z}^\times$ , with $\displaystyle \varphi$ the absolute value, $\displaystyle \varphi (1) = |1| = 1$ , and so $\displaystyle \mathbb{Z}^\times = \{\pm 1\}$ .

For $\displaystyle F^\times$ , with $\displaystyle \varphi (x) = 1$ , $\displaystyle \varphi (1) = 1 = \varphi (x)$ for all $\displaystyle x \in F$ . Thus $\displaystyle F^\times = F \,\backslash \{0\}$ .

For $\displaystyle F[x]^\times$ , with $\displaystyle \varphi (p(x)) = \deg p(x)$ , $\displaystyle \varphi (1) = \deg 1 = 0$ , and so $\displaystyle F[x]^\times =$ non-zero polynomials of degree zero $\displaystyle = F \,\backslash \{0\}$ .

For $\displaystyle \mathbb{Z}[i]^\times$ , with $\displaystyle \varphi (a + bi) = a^2 + b^2$ , $\displaystyle \varphi (1) = 1^2 + 0^2 = 1$ , and so $\displaystyle \mathbb{Z}[i]^\times = \{\pm 1, \pm i\}$ .