Given the ring $\mathbb Z/6\mathbb Z$ find $a$ and $b$, two divisors of zero, such that their sum is a unit.
Given the ring $\mathbb Z/6\mathbb Z = \{0,1,2,3,4,5\}$. So by two divisors of zero what I understand is that $a|6$ and $b|6$, and by their sum is equal to unity $a+b = u$, where $\exists \ v \in \ \mathbb Z/6\mathbb Z $ such that $ uv = vu = 1$.
Pairs of distinct elements $a|6$ and $b|6$, are $1,3$ ; $3,2$ ; $1,2$. Their sum is $1+4 = 5$; $2+3 = 5$; $1+2 = 3$.
$ 5 \cdot 5 = 24 \equiv 1 \mod (6)$. That is, $5$ is the only unit that satisfies the condition.
Did I miss something? Is the approach correct? Do I even understand the exercise? Please feel free to correct me.
Thank you very much. Kind regards.
Best Answer
As pointed out in the comments there seems to be some misconception about the term "zero divisor". To make it perfectly clear: given a ring $R$ an element $a\in R$ is called a zero divisor if there exists $b\in R\setminus\{0\}$ such that $ab=0$. In particular, $0$ is always a zero divisor a rings for which this is the only zero divisor are called integral domains.
Back to $\mathbb Z/6\mathbb Z$. An element $a$ here is a zero divisor iff there is some non-zero $b$ such that $ab=0$. As $\mathbb Z/6\mathbb Z$ is quite a small ring you can simply check all elements to find that $0,2,3,4$ are the only zero divisors (you missed $0$ in your comment; see what I wrote above). Also, $1$ is always a unit and you showed that $5$ is a unit too. So, there are two immediate possiblities: $(a,b)=(2,3)$ or $(a,b)=(3,4)$. As you are only asked to find a pair $(a,b)$ such that $a+b$ is a unit this should suffice.
Two facts to prove which might be a good training for better understanding units and zero divisors:
Fact $\mathbf1$. Let $R$ be a ring. If $u\in R$ is a unit then $u$ is not a zero divisor.
Fact $\mathbf2$. Let $R=\mathbb Z/n\mathbb Z$. Then $R$ decomposes disjointly into its set of units and zero divisors. That means every element of $R$ is either a unit or a zero divisor. (This holds more generally for all finite rings)