Construct elements of infinite multiplicative order in some quadratic integer rings

Show that for D \in \{ 3,5,6,7 \} the group of units in \mathbb{Z}[\omega] is infinite by exhibiting an explicit unit of infinite multiplicative order.


We begin with a lemma.

Lemma: If a + b\sqrt{D} \in \mathbb{Q}(\sqrt{D}), where D > 1 is squarefree, a > 1, and b \geq 1, then a + b \sqrt{D} has infinite multiplicative order in \mathbb{Q}(\sqrt{D}). Proof: We prove by induction that if (a+b\sqrt{D})^n = p + q \sqrt{D}, then p > 1 and q \geq 1. The base case n = 1 holds by hypothesis. Now suppose (a+b\sqrt{D})^n = p + q \sqrt{D}, and that p > 1 and q \geq 1. Then (a + b\sqrt{D})^{n+1} = (ap + bqD) + (aq + bp)\sqrt{D}, and we have 1 < p < ap + bqD and 1 \leq q < aq + bp. By induction, (a + b\sqrt{D})^n = 1 + 0\sqrt{D} has no solution n. \square

Using the lemma, it suffices to find, for each D, an element a + b \sqrt{D} such that a and b are integers, a > 1, b \geq 1, and N(a+b\sqrt{D}) = a^2 - Db^2 = 1.

  1. For D = 3, note that N(2 + \sqrt{3}) = 1.
  2. For D = 5, note that N(9 + 4\sqrt{5}) = 1.
  3. For D = 6, note that N(5 + 2 \sqrt{6}) = 1.
  4. For D = 7, note that N(8 + 3 \sqrt{7}) = 1.
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Comments

  • Kreizhn  On June 15, 2011 at 3:59 pm

    I think there might be a typo here. In particular, was is d? Should this be D?

    • nbloomf  On June 15, 2011 at 9:03 pm

      That was a typo. Thanks!

  • Joseph  On July 4, 2011 at 10:46 am

    Is the norm of a+bw not different in the case when D=5 as it is congruent to 1 mod 4.

    • nbloomf  On July 4, 2011 at 4:36 pm

      When we write the elements as a+b\sqrt{D}, the norm is N(a+b\sqrt{D}) is a^2 - Db^2. When D is 1 mod 4, the ring of integers is enlarged to include elements of the form a+b\omega where \omega = \frac{1+\sqrt{D}}{2}. Now N(a+b\omega) looks different.

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