These are my answers to a past exam question, just wondering if someone could please check whether they are right and also whether I have written enough to gain the full marks for each question?
Thanks in advance!
Is 3 + 15Z a zero divisor on R? [3 marks]
Yes because 3.5 = 0.
More precisely: (3+15Z)(5+15Z) = 15+15Z = 0 + 15Z What is the inverse of 7 + 15Z in R? [3 marks]
The inverse is 13 + 15Z because (7 + 15Z)(13 + 15Z) = 1.
Well Done
Is 7 + 15Z a zero divisor in R? [3 marks]
No because there are no such elements that a.7 = 0, where a is in R.
Generally: "If an inverse exists for an element a, then it cannot be a zero divisor". If a.b = 0(with a and b not zero) and a^{-1} exists, then a^{-1}(a.b) = a^{-1}.0 => b = 0. Contradiction. Is R a field? [2 marks]
No because not every non-zero element has an inverse in R.
Well Done. You should also illustrate a counter example. So you can write "No because not every non-zero element has an inverse in R. For instance, 3 does not have an inverse".
Find an ideal I of R consisting of 3 elements. [4 marks]
An ideal of R consisting of 3 elements is {0, 5, 10}.
What is the number of elements of R/I? [2 marks]
R/I = 15/{0,5,10} = 15/3 = 5 elements.
More precisely: [0] = {0,5,10},[1] = {1,6,11}, [2] = {2,7,12}, [3] = {3,8,13},[4] = {4,9,14}.
Write down the multiplication table of R/I. [4 marks]
X 0 1 2 3 4
0 1 1 2 3 4
1 0 2 4 1 3
2 0 2 4 1 3
4 0 4 3 2 1
Its wrong.The table should look like this:
X [0] [1] [2] [3] [4]
[0] 0 0 0 0 0
[1] 0 1 2 3 4
[2] 0 2 4 6 8
[3] 0 3 6 9 12
[4] 0 4 8 12 1 But 6 belongs to [1], 8 belong to [3], 12 belongs to [2] and 9 belongs to [4]
Thus:
X [0] [1] [2] [3] [4]
[0] 0 0 0 0 0
[1] 0 1 2 3 4
[2] 0 2 4 1 3
[3] 0 3 1 4 2
[4] 0 4 3 2 1
A Note: The point of this exercise was to convince you that R/I is a multiplicative group. Can you see it?
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Originally Posted by hunkydory19 
