Let $(R, 0, 1, +, ·)$ be a ring. By $R^∗$ we denote the set of units, i.e.

$R^∗$ = {$r ∈ R : (∃s ∈ R : s · r = r · s = 1)$}

(a) Show that that $(−1) · r = −r$, the r.h.s. stands for the additive inverse of $r$.

(b) Show that $(R^∗, 1, ·)$ is a group.

(c) Show that the set of zero-divisors of R is disjoint from $R^∗$.

(d) Describe the zero-divisors and units of the ring of $(2 × 2)$-matrices

I'm having troubles with part a), any help please?

concerning part b): to show that $(R^∗, 1, ·)$ is a group, I know that I have to show the axioms of a group:

R is stable under multiplication. Multiplication is associative, i. e., $a.(b.c)=(a.b).c$. In $R$ exists a neutral element e such that $a.e=e.a=a$. Every $a$$R$ has an inverse element $b$$R$, i. e., $a.b=b.a=e$

Concerning part c): I know the definition of "zero-divisor" but don't know how to apply it in this problem: In a ring $R$, a nonzero element $a\in R$ is said to be a zero divisor if there exists a nonzero $b \in R$ such that $a\cdot b = 0$

Concerning part d): I know the definition of "unit" but don't know how to apply it in this problem: an element $a$$R$ for which there exists $b$ ∈ R such that $a.b=b.a=1$ is called a unit and b is the inverse of $a$.

The collection of all invertible matrices constitutes the general linear group $GL(2, R)$ and $GL(2, R) $is its group of units.

Please I really need help in this problem.


1 Answer 1


Regarding part a):

In a group we have that inverse elements are unique. Thus you can check that $$(-1) \cdot r + r = 0 = r + (-1) \cdot r$$ holds to show that $(-1) \cdot r = -r$. That can be done as follows:

$$(-1) \cdot r + r = (-1) \cdot r + 1 \cdot r = (-1 + 1) \cdot r = 0 \cdot r = 0,$$ where we used that $1 \in R$ is neutral, that distributivity holds and that $0 \cdot r = 0$ for all $r \in R$. Do you know the latter? Otherwise you will need to show that $0 \cdot r = 0$ holds for all $r \in R$. To really understand how we did that you should do the other one, i.e. $r + (-1) \cdot r = 0$, for yourself.

Regarding part b):

Yes, you need to go through the group axioms. You will get associativity and the neutral element from the ring axioms and then you will need to "show" the existence of inverse elements by considering the actual set you are given (the units).

Let me also give more details here. Since $R$ is a ring, you now that the multiplication is associative, i.e. $$a \cdot (b \cdot c) = (a \cdot b) \cdot c$$ holds for all $a,b,c \in R$. But since that holds for all elements in $R$ it also holds for all elements in $R^{\times}$, as $R^{\times} \subset R$. The same type of argument works for the neutral element.

How do you get inverse elements? By definition. A unit is per definition an element that has a multiplicative inverse.

Hint for part c):

Why can zero-divisors not have inverse elements? Assume both properties: So take $a \in R$ with $ a \neq 0$, such that there exists an element $b \in R$ with $b \neq 0$ and $ab = 0$ and an element $c \in R$ such that $ac = 1 = ca$. And now just play around with them a bit and see what happens: $$0 = c \cdot 0 = c \cdot (a \cdot b) = (c \cdot a) \cdot b = 1 \cdot b = b,$$ a contradiction. You can also show that one gets $0 = 1$ for example.

Regarding part d):

A unit is an invertible matrix as units are exactly the elements having a multiplicative inverse. Thus you have found a description of the invertible elements by going back to linear algebra: $$\text{Mat}_n(R)^{\times} = \text{GL}_n(R)$$

Now you need to figure out how to get some zero-divisors. Just check this post if you are having problems. That question was asked quite some times and there are nice alternative solutions.

  • $\begingroup$ For part c) is it like this? $ab=0$ so $cab= c.(ab)=c.0=0$ and $ac=1$ so $cab= b.(ac)=b.1=1$, but $1$ $\not=$ $0$ thus the set of zero-divisors of R is disjoint from $R^∗$. Also, I didn't understand what to do in other parts. Could you please show me how they're done? $\endgroup$
    – JOJO
    Commented Sep 6, 2019 at 20:19
  • $\begingroup$ Why do you get $b \cdot 1 = 1$? But yes one can show that $0 = 1$. So if you assume that you are not considering the zero ring you will not have an element that is both a unit and a zero-divisor. $\endgroup$
    – Con
    Commented Sep 6, 2019 at 20:52
  • $\begingroup$ For part a) when I get to $(-1).r+r=0$, I can directly say that $(-1).r=-r$? $\endgroup$
    – JOJO
    Commented Sep 8, 2019 at 10:43
  • $\begingroup$ And for part b), when I have to show that every $a∈R$ has an inverse element $b∈R$, i. e., $a.b=b.a=e$, I say that $e=1$ and so $s.r=r.s=1$ which is true concerning $R^*$ $\endgroup$
    – JOJO
    Commented Sep 8, 2019 at 10:54
  • $\begingroup$ b) Yes. For every element in the group of units exists per definition an inverse element. $\endgroup$
    – Con
    Commented Sep 8, 2019 at 11:17

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .