## In a Euclidean Domain, nonzero elements of minimal norm are units

Let $R$ be a Euclidean Domain. Let $m$ be minimal among the norms of nonzero elements of $R$ ($m$ exists because every nonempty set of natural numbers has a least element). Prove that every nonzero element of norm $m$ is a unit. Deduce that a nonzero element of norm zero (if it exists) is a unit.

We denote the Euclidean norm on $R$ by $N$. Choose $b \in R$ with $N(b) = m$ and $b \neq 0$. By the Division Algorithm, we have $1 = qb + r$ for some $q,r \in R$, with $0 \leq N(r) < N(b)$. Since $N(b)$ is minimal among the norms of nonzero elements in $R$, we have $r = 0$. Thus $1 = qb$, so that $b$ is a unit.

Suppose now that some nonzero element has norm zero. Then zero is necessarily minimal among the norms of nonzero elements in $R$, and hence any nonzero element of norm zero is a unit.

• Adrian  On November 21, 2010 at 7:00 pm

If you begin by choosing $a = 1$ the norm condition still holds and the proof takes half as long.

• nbloomf  On November 21, 2010 at 10:59 pm

Good point.

• saigonbettas  On December 2, 2010 at 6:12 pm

Do you have ideas for the next exercise which is exercise number 4 in section 8.1 in the text book.

• nbloomf  On December 3, 2010 at 9:51 am

Try this.

• saigonbettas  On December 6, 2010 at 3:46 pm

Thanks very much. Could you give me some hints for number 10 proving that the quotient ring Z[i]/I is finite for any nonzero ideal I of Z[i].

• nbloomf  On December 6, 2010 at 3:54 pm

That one is actually in the pipeline and will be posted soon, but I can give a hint.

Note that since $\mathbb{Z}[i]$ is a Euclidean domain, the ideal $I$ must be principal. Choose a generator $\alpha$ for $I$. Now show that each coset $x+I$ has a representative whose norm is less than the norm of $\alpha$ (using the Euclidean algorithm). Finally, argue that the number of elements of $\mathbb{Z}[i]$ which have norm less than some constant $M$ is bounded.