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Finish solutions for integral domains chapter
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hdgarrood committed May 5, 2017
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19 changes: 13 additions & 6 deletions appendix/solutions/integral-domains/ex4.rst
Expand Up @@ -3,10 +3,17 @@ Exercise 6.4

Suppose :math:`m \geq 2` and :math:`\mathbb{Z}_m` has a zero-divisor. That is,
there exist integers :math:`a, b` such that :math:`\overline{a} \neq
\overline{0}, \overline{b} \neq \overline{0}`, and :math:`\overline{ab} =
\overline{0}`. That is, neither :math:`a` nor :math:`b` is a multiple of
:math:`m`, but :math:`ab` is.
\overline{0}, \overline{b} \neq \overline{0},` and :math:`\overline{ab} =
\overline{0},` or equivalently, neither :math:`a` nor :math:`b` is a multiple
of :math:`m`, but :math:`ab` is. The only way this can happen is if :math:`m`
is composite i.e. not prime, as in this case there must exist integers :math:`1
< k, l < m` with :math:`kl = m` such that :math:`k` divides :math:`a` and
:math:`l` divides :math:`b`.

Using the hint that all integers greater than or equal to :math:`2` have a
unique prime factorisation, we can write :math:`m = p_1^{k_1} \times p_2^{k_2}
\times \ldots \times p_n^{k_n}` for some integer :math:`n`.
Conversely, suppose :math:`m \geq 2` and :math:`\mathbb{Z}_m` is an integral
domain, i.e. it has no zero-divisors. That is, for any integers :math:`a, b`
with :math:`1 < a, b < m,` we have that :math:`ab` is not a multiple of
:math:`m`. The only way this can happen is if :math:`m` is prime.

Therefore, :math:`\mathbb{Z}_m` is an integral domain if and only if :math:`m`
is prime.
5 changes: 2 additions & 3 deletions integral-domains.rst
Expand Up @@ -106,10 +106,9 @@ we now know that :math:`\mathbb{Z}_2` is an integral domain, but
:math:`\mathbb{Z}_{12}` and :math:`\mathbb{Z}_8` are not.

**Exercise 6.4. (hard)** Try to establish whether :math:`\mathbb{Z}_m` is an
integral domain for a few more choices of :math:`m`. Can you think of a rule
integral domain for a couple more choices of :math:`m`. Can you think of a rule
for determining whether :math:`\mathbb{Z}_m` is an integral domain for any
given :math:`m \geq 2`? Hint: all integers greater than or equal to :math:`2`
have a unique prime factorisation.
given :math:`m \geq 2`?

The cancellation law for integral domains
-----------------------------------------
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