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.

  1. Let R be a ring. Then 0Rr = 0R for each element r of R.
    Proof: Since 0R is an additive identity, we know that 0R = 0R + 0R, hence 0Rr = (0R + 0R)r, so, by the distributive property, we have ...

  2. 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)Rr = -r, for each element r of R.
    Proof: Suppose we show that (-1)r + r = 0R. 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 = 0R. But (-1)r + r = (-1)r + 1r = (-1 + 1)r, by the distributive property, so, ...

  3. 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 = 0R. But additive inverses are unique, so ...

  4. 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 = 1R. But we know that 1Rs = s1R = s for every element of R. Thus ...

  5. Let R be a ring such that the additive identity and the multiplicative identity are the same element; i.e., such that 0R = 1R. Then R has only one element; i.e., R = {0R}.
    Proof: Consider 0R1R. We can simplify this two ways. First, from a previous problem, we know that 0R1R = ...