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I am working on some basic problems involving rings and am getting confused.

If I have a ring $\mathbb{Z}_2 \times \mathbb{Z}_4$, would this just be all elements $(a,b)$ such that $a \in \mathbb{Z}_2$ and $b \in \mathbb{Z}_4$ or am I mistaken?

I'm then trying to find zero-divisors and units in this ring.

How would I go about doing this?

Thanks

2 Answers 2

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You're right about the definition. Note that $(a,b)$ is a zero-divisor in $\mathbb Z_2 \times \mathbb Z_4$ if, by definition, there exists $(c,d)\neq(0,0)$ so that $$(a,b)\cdot (c,d)=(a\cdot c, b\cdot d)=(0,0)$$ So you get $a\cdot c=0$ in $\mathbb Z_2$, and $b\cdot d =0$ in $\mathbb Z_4$. That means $a$ is a zero-divisor in $\mathbb Z_2$, and $b$ is a zero-divisor in $\mathbb Z_4$. Can you continue from here?

The same works for finding units.

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    thank you, this makes it perfectly clear! – 2017-02-12
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Yes $$\mathbb Z_2\times \mathbb Z_4=\{(a,b)\mid a\in \mathbb Z_2, b\in\mathbb Z_4\}$$

with the operations $$(a,b)+(c,d)=(a+c,b+d)\quad \text{and}\quad (a,b)\cdot (c,d)=(ac,bd).$$

For the divisor of $0$, find $(a,b)$ and $(c,d)$ s.t. $$(a,b)\cdot (c,d)=(0,0)$$ but neither $(a,b)=(0,0)$ nor $(c,d)=0$.

For unit, find elements $(a,b)$ and $(c,d)$ s.t. $$(a,b)\cdot (c,d)=(1,1).$$

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    thank you, it makes sense now! – 2017-02-12
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    would i be correct in saying that there are no units? – 2017-02-12
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    @ptsgeeg: No, since you have at least $(1,1)$. But in fact, you should remark (and show) that $(\mathbb Z_2\times \mathbb Z_4)^\times =\mathbb Z_2^\times \times \mathbb Z_4^\times $, so there are other units... – 2017-02-12
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    oh yes, i forgot that you can have (a,b) = (c,d). what do you mean by the last bit? sorry, i find ring theory and all a bit difficult. – 2017-02-12
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    @ptsgeeg: Don't worry :-) $\mathbb Z_2^\times =\{[1]\}$ and $\mathbb Z_4^\times =\{[1],[3]\}$. Therefore, $$(\mathbb Z_2\times \mathbb Z_4)^\times =\{([1],[1]),([1],[3])\}.$$ – 2017-02-12
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    What does the $\mathbb {Z}^x_2$ mean? – 2017-02-12
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    @ptsgeeg: Sorry, if $A$ is a ring, then $A^\times$ denote the set of the units of $A$ (in particular, it's a group, but it's not important for your exercise) – 2017-02-12
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    oh right. that makes sense. you have been a great help, thank you! – 2017-02-12