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Let $\frac12 \Bbb Z = \{\frac{a}{2} \mid a \in \Bbb Z\}$. Determine why $\frac12 \Bbb Z$ is not a ring with respect to addition and multiplication.

$\frac12 \Bbb Z$ is an abelian group and it also seems to be a monoid since $1 \in \frac12 \Bbb Z$ and for $a,b,c \in \frac12 \Bbb Z$ we have that $$a(b\cdot c) = \frac{a}{2}\left(\frac{b}{2}\cdot \frac{c}{2}\right)= \frac{a}{2} \left(\frac{bc}{4}\right)= \frac{abc}{8} = \left(\frac{a}{2}\cdot \frac{b}{2}\right) \frac{c}{2} =(a\cdot b)c.$$

Multiplication seems also to be distributive since $$a(b + c) = \frac{a}{2}\left(\frac{b}{2}+ \frac{c}{2}\right) = \frac{ab + ac}{4} = \frac{a}{2}\cdot \frac{b}{2} + \frac{a}{2}\cdot \frac{c}{2} = ab +ac.$$

What might I be missing here?

Shaun
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Zwertp
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    $\frac{1}{2} \cdot \frac{3}{2} = \frac{3}{4} \not\in \frac{1}{2} \mathbb{Z}$. – Charles Hudgins Nov 12 '21 at 10:11
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    What is $\frac{1}{2} \cdot \frac{1}{2}$? Is it in your ring? – Severin Schraven Nov 12 '21 at 10:12
  • Hint: if a subring of $\Bbb Q$ had a least positive element $< 1$ then squaring it would yield a smaller positive element, contradiction. But $\Bbb Z/2$ does have least positive element $,1/2 < 1$. See here for how to use this idea to prove the irrationality of square-roots. – Bill Dubuque Nov 12 '21 at 10:23

1 Answers1

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Indeed you mulitplication is associative, distributive and so. What you are missing is that your multiplication $m$ must take elements of your ring to your ring :

$m : A \times A \to A$

In fact you may have a proposition like this in your course :

Let $(R, + , \cdot )$ be a ring and $A \subset R$. If you have

  • $\forall a, b \in A,\quad a+b \in A$
  • $\forall a, b \in A, \quad a-b \in A$
  • $1 \in A$
  • $\forall a, b \in A, \quad a\cdot b \in A$

then $(A, +_{\vert A}, \cdot_{\vert A})$ is a ring.

InfiniteLooper
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    That proposition really depends on how you define subring. I would say that what you wrote is definition of subring, while, for example, being closed under subtraction and multiplication is characterization. – Ennar Nov 12 '21 at 11:06
  • Indeed, edited. – InfiniteLooper Nov 12 '21 at 11:36