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February 10th, 2018, 01:56 AM   #1
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Hi, I have attached the question and the solutions to part a and b of this question. Would like someone to verify if I have done anything wrong especially with part b. Greatly appreciate it! Thanks.
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 February 10th, 2018, 03:16 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,917 Thanks: 2200 If you can't achieve clearer images, I would suggest typing your work.
 February 10th, 2018, 07:43 AM #3 Senior Member   Joined: Aug 2017 From: United Kingdom Posts: 313 Thanks: 112 Math Focus: Number Theory, Algebraic Geometry I'd say this is fine except I'm not sure what you mean by "S is a commutative ring for all $[a], [b] \in Z_{10}$." Surely you just mean it's a commutative ring? It's worth saying that if a ring $R$ is commutative, then the commutativity of any subring $T$ follows immediately. Indeed, for any $a, b \in T$, note that $a$ and $b$ are in $R$ so they commute. This sort of argument is why we only need to check a few things to show something is a subring - properties like associativity of +, associativity of x, distributivity, follow immediately by the same sort of argument.
 February 10th, 2018, 09:06 AM #4 Banned Camp   Joined: Apr 2017 From: durban Posts: 22 Thanks: 0 Math Focus: Algebra The subring test is a theorem that states that for any ring R, a subset of R is a subring if it is closed under multiplication and subtraction, and contains the multiplicative identity of R.
 February 10th, 2018, 09:08 AM #5 Banned Camp   Joined: Apr 2017 From: durban Posts: 22 Thanks: 0 Math Focus: Algebra I think you can finish it from here. but looking at the first expression the working is right with Z10
February 10th, 2018, 09:24 AM   #6
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Quote:
 Originally Posted by Ola Lawson The subring test is a theorem that states that for any ring R, a subset of R is a subring if it is closed under multiplication and subtraction, and contains the multiplicative identity of R.
The subset $S \subseteq Z_{10}$ in the question does not contain the multiplicative identity of $Z_{10}$, so it wouldn't actually be a subring under the usual definition (the one you're working with). However, the question still says to prove it's a subring, so they're probably using a different definition.

Last edited by cjem; February 10th, 2018 at 09:27 AM.

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