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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Main Page : Abstract Algebra : Euclidean Domains