# Thread: Questions Involving Rings

1. ## Questions Involving Rings

1) Let R be a ring with unity 1. Show that (-1)a = -a for all a in R.

2) Show that the set of all real numbers of the form a+bsqrt(2), where a,b in Q, forms a field under ordinary addition and multiplication.

3)
a) Show that every element of Zn (integers mod n) is either a unit or a zero-divisor. (Use the fact that an element of this ring is a unit iff (a,n) = 1.)
b) Which elements of Zn are nilpotent?

4) Show that if R is a ring with unity, then the multiplicative identity element in R is unique.

5) Let R be a ring with unity. R is called Boolean if every element of R is idempotent. Show that if R is Boolean then:
a) 2r = 0 for every r in R (in other words, r = -r) (Hint: Consider (r+r)^2)
b) R is commutative.

2. Originally Posted by Janu42
1) Let R be a ring with unity 1. Show that (-1)a = -a for all a in R.
Since (R,+) is an abelian group, (-1)a + 1a = ((-1) + 1)a=0a=0.
(-1)a + 1a -1a = 0 -1a = -a (Adding -1a to both sides and use the fact that 1 is a unity in R).
Thus, (-1)a = -a.

2) Show that the set of all real numbers of the form a+bsqrt(2), where a,b in Q, forms a field under ordinary addition and multiplication.
I leave it to you to check $\displaystyle R = \{a + b\sqrt{2} | a, b \in Q\}$ is a commutative ring with unity. Then, check $\displaystyle \frac{1}{a + b\sqrt{2}}$ belongs to R.

3)
a) Show that every element of Zn (integers mod n) is either a unit or a zero-divisor. (Use the fact that an element of this ring is a unit iff (a,n) = 1.)
b) Which elements of Zn are nilpotent?
In $\displaystyle Z_{n}$, no nilpotent element can be a unit (excluding a trivial ring {0}). All non-zero nilpotent elements are zero divisors.
Thus, if you remove unit elements (satisfying (a,n)=1) from $\displaystyle Z_n$, then the remaining elements are nilpotents.

4) Show that if R is a ring with unity, then the multiplicative identity element in R is unique.
Suppose both x and y are unity elements in the ring R. Then, x = xy = y. Thus, x = y.

5) Let R be a ring with unity. R is called Boolean if every element of R is idempotent. Show that if R is Boolean then:
a) 2r = 0 for every r in R (in other words, r = -r) (Hint: Consider (r+r)^2)
b) R is commutative.
Since r is idempotent in R, $\displaystyle (r + r )^{2} = r^2 + rr + rr + r^2 = r + r$,
$\displaystyle r + rr + rr +r = r + r$.
We have $\displaystyle rr + rr = 0$. Since r is idempotent, r + r = 0. Thus, 2r = 0.

Let x, y be elements in a boolean ring R.
$\displaystyle (x + y) ^{2} = x^2 + xy + yx + y^2 = x + y,$
Since R is a boolean ring, we have xy = -yx. Let x = y. Then x = -x.
That means every element in R is its own additive inverse.
Thus, xy = -yx = yx. We conclude that R is commutative.