November 20th, 2010, 09:09 PM  #1 
Member Joined: Nov 2010 Posts: 53 Thanks: 0  Some Definitions
Hi! There are some definitions I really feel I can't quite get the grip on. It would be very nice if anyone could help me a bit to better understand these: Direct sum/product: If I f.ex look at the elements in Z/2 x Z/3 (which have 6 elements) and want to find a generator (if one exists) say (1,1) then; (1,1) = (1,1), 2(1,1) = (1,1) + (1,1) = (0,2), 3(1,1) = (1,1) + (0,2) = (1,0),... Is this the same as taking the direct product? Integrally closure/closed: It would be really nice if someone could show me an example to illustrate this. Integral basis: Same here, I know (sort of) the definition, but it would be so nice to have an example to think of. I would really grateful if someone could help me with these definitions. 
November 21st, 2010, 03:42 PM  #2 
Senior Member Joined: Aug 2010 Posts: 195 Thanks: 5  Re: Some Definitions
Direct Product: Set theoretically, the direct product of two rings is the set of ordered pairs (ie the Cartesian product) with the operations defined component wise: if and then and The same idea holds for groups, vector spaces, modules, etc. Integrally closed: is integrally closed because if is the root of a monic polynomial in , then and . By contrast, is not integrally closed because in , the field of fractions, is a root of which has coefficients in but is not an element of the ring. Integral basis: Consider the ring . Elements in this ring are of the form where so and form an integral basis, whereas and would not form an integral basis, as there are elements in our ring which are not of the form , such as . 
November 22nd, 2010, 03:36 AM  #3 
Member Joined: Nov 2010 Posts: 53 Thanks: 0  Re: Some Definitions
Hi! Thank you, it really helped. 
November 24th, 2010, 05:57 AM  #4 
Global Moderator Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4  Re: Some Definitions
For operations in the group Z/2 x Z/3 We can define the group operation on a,b ? Z/2 x Z/3 a ? b = a + b Keeping in mind that each element is an ordered pair. And you perform the addition just like Turgul said. So (1,0) "?" (1,0) = (1,0) + (1,0) = (2,0) = (0,0). Not a generator, but we've found its inverse! (0,1) "?" (0,1) = (0,1) + (0,1) = (0,2). Add it again to get (0,3) = (0,0)... Since the order of this group is 6, any generator would also have order 6... 
November 24th, 2010, 06:55 AM  #5 
Member Joined: Nov 2010 Posts: 53 Thanks: 0  Re: Some Definitions
Ah, ok. Thanks for the info 

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