0
$\begingroup$

Since $[0][1] = [0] \in \mathbb Z_2$, this ring is closed under multiplication.

Let $[1]$ be identity: $[0][1] = [0]$ and $[1][1] = [1]$.

$[0][1] = [0 \cdot 1] = [1 \cdot 0] = [1][0]$ so commutativity holds.

If we assume $\mathbb Z_2$ is a ring, associativity is given.

I am having difficulty showing there's inverse element in $\mathbb Z_2$. What could possibly be the inverse of $[0]$? Please, elaborate on this. Thanks.

$\endgroup$
  • $\begingroup$ $[0][0]=[0]$ and $[0][1]=[0]$. Since $\mathbb{Z}_2=\{[0],[1]\}$, we conclude that $\forall x\in\mathbb{Z}_2,\ [0]x=[0]$. Since $[0]\neq[1]$, we conclude that $\forall x\in\mathbb{Z}_2,\ [0]x\neq[1]$. Doesn't this show that $[0]$ has no inverse? $\endgroup$ – gniourf_gniourf Feb 16 '16 at 18:18
  • 1
    $\begingroup$ The additive zero does not have a multiplicative inverse in a ring. $\endgroup$ – John Smith Feb 16 '16 at 18:19
  • 1
    $\begingroup$ In a field, the non-zero elements form a group under multiplication. (The zero element in a ring only has an inverse in the trivial ring with only one element.) $\endgroup$ – Rob Arthan Feb 16 '16 at 18:22
  • 2
    $\begingroup$ You need to reread the definition of a field. $\endgroup$ – Tobias Kildetoft Feb 16 '16 at 18:23
  • 1
    $\begingroup$ $(\mathbb{Z_2},+)$ isn't the same as $(\mathbb{Z_2^{*}},(*))$ $\endgroup$ – user111750 Feb 16 '16 at 18:25
1
$\begingroup$

That will be very difficult. {[0],[1]} is not a group under multiplication.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.