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loida

Member

Registered: 2009-11-12

Posts: 32

algebra(ideals)

a)Find all invertible elements in Z/10Z. Describe a group of invertible elements up to isomorphism. Find all classes associates in Z/10Z.
b)Find all maximal ideals of Z/10Z
c)Find all prime ideals of Z/10Z
d)Find all irreducible elements in Z/10Z
e)Find all prime elements in Z/10Z

pls help!

Last edited by loida (2010-01-07 10:33:47)

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: algebra(ideals)

When ever checking to see if an ideal is maximal or prime in a commutative ring with 1 (like Z/10Z), the best method tends to be modding out.  If you mod out by an ideal and get a field, then the ideal is maximal.  If you mod out and get an integral domain, then it's prime.

Use the isomorphism theorem when you're doing this.

Edit: And it should take no longer than 5 minutes to find every ideal of Z/10Z.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

loida

Member

Registered: 2009-11-12

Posts: 32

Re: algebra(ideals)

tnx Ricky

i only know Z/10Z={10Z, 1+10Z,2+10Z,....9+10Z}
2 is prime, but not irreducible

a) (0)^(-1)=0 , (1)^(-1)=1, (2)^(-1)=8, (3)^(-1)=7, (4)^(-1)=4,(5)^(-1)=5,(6)^(-1)=6,(7)^(-1)=3,(8)^(-1)=2,(9)^(-1)=9 ?
i'm desperate
could u pls help me to figure this out?

Last edited by loida (2010-01-08 09:37:42)

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: algebra(ideals)

Start with the principal ideals:

(1), (2), (3), ..., (9)

Figure out which of these ideals are the same.  For example, (2) = (4) since 4*3 = 12 = 2 (mod 10), and certainly 4 is in (2).

Once that's done, figure out all the ideals that are generated by two elements: (1, 2) for example.  Of course this ideal is really all of Z/10Z, and you'll find that this is the case for most of them.  Also remember that there is no need to check something like (2, 4) since (2) = (2,4).  There aren't many to check.

There aren't any ideals generated by three elements (that don't form the whole ring), so you're done.

Now you can go through and check what happens when you mod out by each ideal.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

loida

Member

Registered: 2009-11-12

Posts: 32

Re: algebra(ideals)

would it be a problem if u put a solution on these exerices and i tell what i don't understand..i'm sorry, i'm just stupid for this things and don't know the basic

tnx once again...i'm really thankful for everything what u've written

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: algebra(ideals)

Sorry, I can't do that.  Write out all the ideals that you can find and we'll go from there.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

loida

Member

Registered: 2009-11-12

Posts: 32

Re: algebra(ideals)

(2) = (6)?
(1),(2),(3),(5),(7),(8),(9) ?
don't know how check with 2 elements, sorry
i'm afraid i have to quit, i don't have much time left and have much things to learn
but thanks anyway...i really appreciate ur effort
i'm ashamed of my ignorance..

Last edited by loida (2010-01-09 09:25:29)

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: algebra(ideals)

(2) = (4) = (6) = (8) (this is easy to prove: 4*3 = 2, 6*2 = 2, 8*4 = 2)
(1) = (3) = (7) = (9) (3*7 = 1, 7*3 = 1, 9*9 = 1)

So there are four principal ideals: (0), (1), (2), and (5).

Each of these numbers are relatively prime, so any ideal generated by two of them must be the entire ring, (1).  Thus, there are only four ideals:

(0), (1), (2), and (5)

Ignoring (0) and (1) because these are trivial, (2) and (5) are easy to check whether they are maximal and/or prime: clearly they are both maximal (there are only 4 ideals!).  Remembering that:

(G/H)/(K/H) = G/K

If we mod out by (2), this is: (Z/10Z)/(2Z/10Z) = Z/2Z.  As this is an integral domain, this ideal is prime.  Same for (5) which would give us Z/5Z.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

loida

Member

Registered: 2009-11-12

Posts: 32

Re: algebra(ideals)

woow!!
thanks Ricky!

Last edited by loida (2010-01-10 08:34:51)

jdskja

Guest

Re: algebra(ideals)