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 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 $R_1 \times R_2$ of two rings is the set of ordered pairs $(r_1,r_2)$ (ie the Cartesian product) with the operations defined component wise: if $a_1,b_1 \in R_1$ and $a_2,b_2 \in R_2$ then $(a_1,a_2)+(b_1,b_2)=(a_1+b_1,a_2+b_2)$ and $(a_1,a_2) \cdot (b_1,b_2)=(a_1 \cdot b_1,a_2 \cdot b_2)$ The same idea holds for groups, vector spaces, modules, etc. Integrally closed: $\mathbb{Z}$ is integrally closed because if $a/b \in \mathbb{Q}$ is the root of a monic polynomial in $\mathbb{Z}$x$$, then $b=1$ and $a/b \in \mathbb{Z}$. By contrast, $\mathbb{Z}$\sqrt{-3}$$ is not integrally closed because in $\mathbb{Q}$\sqrt{-3}$$, the field of fractions, $\frac{1+\sqrt{-3}}{2}$ is a root of $x^2+x+1$ which has coefficients in $\mathbb{Z}$\sqrt{-3}$$ but $\frac{1+\sqrt{-3}}{2}$ is not an element of the ring. Integral basis: Consider the ring $\mathbb{Z}$i$$. Elements in this ring are of the form $a \cdot 1+b \cdot i$ where $a,b \in \mathbb{Z}$ so $1$ and $i$ form an integral basis, whereas $2$ and $3i$ would not form an integral basis, as there are elements in our ring which are not of the form $a \cdot 2+b \cdot 3i$, such as $1+i$.
 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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