## Math 310: Problem set 12

*Instructions*: This problem set is due
Friday, April 14, 2006. Finish the proofs of each of the given
statements. You may assume the statement of Problem m
when you do Problem n, whenever n > m.

- Let R be a ring. Then 0
_{R}r = 0_{R}
for each element r of R.

**Proof**: Since 0_{R} is an additive identity, we know
that 0_{R} = 0_{R} + 0_{R}, hence
0_{R}r = (0_{R} + 0_{R})r, so, by
the distributive property, we have ...

- Let R be a ring. Then the additive inverse of 1 times
any element r of R is the additive inverse -r of r; i.e.,
(-1)
_{R}r = -r, for each element r of R.

**Proof**: Suppose we show that (-1)r + r = 0_{R}.
This means that (-1)r is an additive inverse of r. But inverses
in a group (note that R is a group under +) are unique, and -r is how we denote that
unique additive inverse, so (-1)r must be -r. So all we have to do
is justify why (-1)r + r = 0_{R}. But
(-1)r + r = (-1)r + 1r = (-1 + 1)r, by the distributive property,
so, ...

- Let R be a ring. Then the additive inverse of
the additive inverse of an element r is r itself;
i.e., -(-r) = r for each element r of R.

**Proof**: By definition of additive inverse
we know that -r + r = 0_{R}. But additive inverses are unique,
so ...

- Let R be a ring. Then the multiplicative identity in R is unique.

**Proof**: Say r is a multiplicative identity in R; i.e.,
rs = sr = s for every element s of R.
We must show that r = 1_{R}.
But we know that
1_{R}s = s1_{R} = s for every element of R.
Thus ...

- Let R be a ring such that the additive identity and the
multiplicative identity are the same element; i.e., such that
0
_{R} = 1_{R}. Then R has only one element; i.e.,
R = {0_{R}}.

**Proof**: Consider 0_{R}1_{R}. We can simplify this two ways.
First, from a previous problem, we know that 0_{R}1_{R} = ...