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Ryan
Ryan, Engineer
Category: Calculus and Above
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Experience:  B.S. in Civil Engineering
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I have a question about abstract algebra

Customer Question

i have a question about abstract algebra
Submitted: 1 year ago.
Category: Calculus and Above
Expert:  Ryan replied 1 year ago.
Hi,
Welcome! Thank you for using the site.
Please post your question at your convenience. I'll be happy to take a look at it and help if I can.
Thanks,
Ryan
Customer: replied 1 year ago.
So i have a test coming up and the professor said that all problems are fair game and these are the two that i am at a loss for. I am not sure what I did wrong as the professor only gives you the questions back to you.#6 Rings are an important algebraic structure, and modular arithmetic has that structure.
Recall that for the mod m relation, the congruence class of an integer x is denoted [x]m. For example, the elements of [–5]7 are of the form –5 plus integer multiples of 7, which would equate to {. . . –19, –12, –5, 2, 9, 16, . . .} or, more formally, {y: y = -5 + 7q for some integer q}.
A. Use the definition for a ring to prove that Z7 is a ring under the operations + and × defined as follows:
[a]7 + [b]7 = [a + b]7 and [a]7 × [b]7 = [a × b]7
Note: On the right-hand-side of these equations, + and × are the usual operations on the integers, so the modular versions of addition and multiplication inherit many properties from integer addition and multiplication.
1. State each step of your proof.
2. Provide written justification for each step of your proof.
B. Use the definition for an integral domain to prove that Z7 is an integral domain.
1. State each step of your proof.
2. Provide written justification for each step of your proof.#23 Fields are an important algebraic structure, and complex numbers have that structure.A. Let G be the set of the fifth roots of unity.
1. Use de Moivre’s formula to verify that the fifth roots of unity form a group under complex multiplication, showing all work.
2. Prove that G is isomorphic to Z5 under addition by doing the following:
a. State each step of the proof.
b. Justify each of your steps of the proof.B. Let F be a field. Let S and T be subfields of F.
1. Use the definitions of a field and a subfield to prove that S ∩ T is a field, showing all work.
Expert:  Ryan replied 1 year ago.
Hi,
Ok, no problem. I'll have solutions for these posted for you as soon as possible.
Thanks,
Ryan
Customer: replied 1 year ago.
Thank you!
Expert:  Ryan replied 1 year ago.
You're welcome. :)
Expert:  Ryan replied 1 year ago.
Hi,
Thank you for your patience.
Here is a solution for the first part:
Ring
And here is a solution for the second part:
deMoivres
Both files contain several equations which were composed using MathType (aka Equation Editor). As a result, it is best if you download the files to your own computer and open them using Word. The "view" option on the Mediafire site tends to mangle the equations. If necessary, PDF versions of these files can be provided.
Thanks,
Ryan
Customer: replied 1 year ago.
Hi,
sorry it has been so long but we did not have class. I asked the TA if there was anything missing that the teacher would want proved on the exam and said yes. In number 6 he said it was missing the proof of closure under both operations. Can you show me how to fit that in. And in number 23 he said i had to prove the definition of injective and surjective. which I don't understand because to me it was but he said the definition had to be stated and proved. I understand how to apply this but when it comes to proofs I am not as good. My test is this weekend and I am studying now but this is my week point so if you could let me know when you got this that would be great.
Thank you
Expert:  Ryan replied 1 year ago.
Hi,
I'll take a look at the solution file and I'll get back to you as soon as I can.
Thanks,
Ryan
Customer: replied 1 year ago.
thank you
Expert:  Ryan replied 1 year ago.
Hi again,
Here is how you can show closure under addition and multiplication for the first problem:
RingClosure
Some courses seem to require that closure under those operations be demonstrated, while others don't.
Regarding the other task, I don't understand how you are supposed to "prove the definition of injective and surjective". For that matter, I don't understand how one is supposed to "prove" the definition of anything. A definition is simply an agreed upon explanation of an idea or concept. The definition itself isn't something that can be proved. Perhaps there is some confusion over terminology here.
I don't know what I can add to that part. The definition of an injective function is that each value in the range of the function corresponds to exactly one element of the domain. This means that there cannot be two or more values in the domain that map to the same value in the range. In other words, each (a, f(a)) pair is a unique combination of a value from the domain and a value from the range. With only five values in Z5 and G in the proof, this can be seen by inspection, since every value from G appears once and only once on the right side of the equations that are listed for phi(n) on page 4 of the previous document. As noted in the earlier solution, each element of Z5 maps to one and only one value in G, and it maps to a unique value in G (meaning that no two values from Z5 map to the same value in G).
The definition of a surjective function is that each value in the range is mapped to by some value in the domain. In other words, there are no values in the range which DO NOT have a corresponding value in the domain. Another way of thinking about this is that the range is "covered"
by the domain. Again, with Z5 and G, this can easily be seen since each value in G appears on the right side of an equation on page 4.
Thanks,
Ryan
Customer: replied 1 year ago.
I wanted to thank you for your help. I got an A on my exam today:) I did get one portion wrong though. And I was wondering if you could explain that to me. Just because my class is over does not mean I don't want to understand it. The professor said that I didn't complete the existence of an additive inverse(it wasn't in Z₇, but that didn't matter, I understand how to apply it to others) Can you take a look, please. Also, is there anyway to give you a tip? Thanks
Customer: replied 1 year ago.
this was for number 6
Expert:  Ryan replied 1 year ago.
Hi,
Congratulations!
I'll be happy to take a look at the proof and offer whatever explanation I can.
Be forewarned though that it has been my experience that these kinds of proofs often seem to go along with very specific expectations from the instructors. A line of reasoning that is acceptable to one instructor can be labelled as "deficient" by another. We saw a little bit of that here with the part about closure under addition and multiplication. Some course instructors want that included, so that there are 8 parts in the proof, while others don't want it, reducing the required items to 6. It can be frustrating, to say the least.
Regarding leaving a tip, once you have "rated" the answer, the website should offer you the opportunity to leave a "bonus". These are always appreciated, of course, but are completely optional.
Thanks,
Ryan
Expert:  Ryan replied 1 year ago.
Hi again,
After looking at the proof for the Z₇ case, the only nit-picky issue I can think of is that the proof does not explicitly state that (7 - a)₇ is also in Z₇, for every a₇ in Z₇. It's somewhat obvious though that if a is in the range 0 to 6, then 7 - a will be in the range 1 to 7, which is then in the range 0 to 6 when taken modulo 7. Thus, (7 - a)₇ is an element of Z₇.
I suppose an argument could be made that if it is not established that (7 - a)₇ is in Z₇, then it is not necessarily an additive inverse for a in Z₇. The additive inverse must also be a member of the set that it is being applied to.
Once that point is established, the rest of the proof simply follows from the definition of the addition operation. Adding a₇ and (7 - a)₇ gives you (a + 7 - a)₇, which then gives you (7)₇, which is equal to 0₇ under the mod 7 operation.
The same proof can then be shown in the other direction, where a₇ is added to (7 - a)₇, also resulting in 0.
If you have any feedback from the instructor that is more specific, I would be interested in hearing it.
Ryan