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.

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Comments

  • 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.

          Thanks for reading!

          • saigonbettas  On December 6, 2010 at 8:36 pm

            Thanks for your hints, and hope to read your proof soon.
            Honestly, thank you very much for your proofs. I have learned a lot from you.

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