how to find units in a ring?

[logistic_guy]

here is the question

Let \(\displaystyle R = \mathbb{Z}/12\mathbb{Z}\), the ring of integers modulo \(\displaystyle 12\).

(a) Identify all the zero divisors in the ring \(\displaystyle R\).
(b) Verify that for each pair of zero divisors \(\displaystyle a\) and \(\displaystyle b\) (with \(\displaystyle a,b \neq 0\)), the product \(\displaystyle a \cdot b \equiv 0 \ (\)mod \(\displaystyle 12)\).
(c) If a nonzero element \(\displaystyle a\) in \(\displaystyle \mathbb{Z}/12\mathbb{Z}\) has a multiplicative inverse, does it have any zero divisors? Justify your answer.


my attemb
i can answer (a) and (b) easily
then (c) will be answered automatically
i mean if i find the zero divisors, the remaining elements must be the units and of course they have no zero divisors
my question is how to find the units without finding the zero divisors? is there a way to do that?

 

[fresh_42]

Let [imath] a [/imath] be a unit in [imath] \mathbb{Z}_n. [/imath] Then [imath] a\cdot b\equiv 1\pmod{n} [/imath] or [imath] ab=qn+1 [/imath] for some integer [imath] q. [/imath]

If [imath] p\,|\,a [/imath] and [imath] p\,|\,n [/imath] for a number [imath] 1<p\le a [/imath] then [imath] 0\equiv 1\pmod{p} [/imath] which is impossible. Hence [imath] a,n [/imath] are relatively prime, i.e. coprime, i.e. have no common divisors.

If conversely [imath] a,n [/imath] are coprime, then there are integers [imath] \alpha,\beta [/imath] such that [imath] 1=\alpha\cdot a+ \beta \cdot n [/imath] by Bézout's lemma. That means [imath] 1\equiv \alpha \cdot a \pmod{n}[/imath] and [imath] a [/imath] is a unit in [imath] \mathbb{Z}_n. [/imath]

This means that the units in [imath] \mathbb{Z}_n [/imath] are exactly the numbers that are coprime to [imath] n. [/imath]

Those are for [imath] n=12 [/imath] all numbers [imath] \{1,5,7,11\}. [/imath]

Btw., this is neither statistics nor probability theory.

 

[logistic_guy]

thank

[imath] 1=\alpha\cdot a+ \beta \cdot n [/imath]

according to wiki, that's just the greatest common divisor between \(\displaystyle a\) and \(\displaystyle n\)

for example gcd\(\displaystyle (7,12) = 1\), so the element \(\displaystyle 7\) is a unit
gcd\(\displaystyle (8,12) = 4 \neq 1\), so the element \(\displaystyle 8\) isn't a unit, it's then a zero divisor by default

am i understand this correctly?

Btw., this is neither statistics nor probability theory.

of course

 

[fresh_42]

according to wiki, that's just the greatest common divisor between \(\displaystyle a\) and \(\displaystyle n\)

Yes.

for example gcd\(\displaystyle (7,12) = 1\), so the element \(\displaystyle 7\) is a unit
gcd\(\displaystyle (8,12) = 4 \neq 1\), so the element \(\displaystyle 8\) isn't a unit, it's then a zero divisor by default

am i understand this correctly?

Yes, although I wouldn't have said: "by default".

I haven't worked out an example, but I can imagine rings with elements that are neither a unit nor a zero divisor.

 

[fresh_42]

A polynomial ring is such an example. The units in [imath] \mathbb{R}[x] [/imath] are [imath] \mathbb{R}\setminus \{0\} [/imath] and [imath] 0 [/imath] is the only zero divisor.

Some matrix rings might have zero divisors other than [imath] 0[/imath], units, and elements that are neither.

 

[logistic_guy]

thank fresh_42 very much

A polynomial ring is such an example. The units in [imath] \mathbb{R}[x] [/imath] are [imath] \mathbb{R}\setminus \{0\} [/imath] and [imath] 0 [/imath] is the only zero divisor.

Some matrix rings might have zero divisors other than [imath] 0[/imath], units, and elements that are neither.

i miss the idea of thinking in other type of rings. that make sense it's not necessary other elements is zero divisors by default
you've a nice think

i appreciate your valuable information

 

[fresh_42]

i miss the idea of thinking in other type of rings. that make sense it's not necessary other elements is zero divisors by default
you've a nice think

The integers are another example. [imath] \mathbb{Z} [/imath] is a ring and it has no zero divisors. However, the only units are [imath] \pm 1 [/imath] so all other numbers are neither zero divisors nor units.

 

[logistic_guy]

The integers are another example. [imath] \mathbb{Z} [/imath] is a ring and it has no zero divisors. However, the only units are [imath] \pm 1 [/imath] so all other numbers are neither zero divisors nor units.

appreciate it very much

you'll have a bright future with this great knowldge