Re: Is every subring of Z is of the form mZ ?
- From: quasi <quasi@xxxxxxxx>
- Date: Sun, 10 Jun 2007 18:49:56 -0500
On Mon, 11 Jun 2007 00:29:33 +0200, student <na@xxxxxx> wrote:
µ wrote:
ged a écrit :
Some people (actually I believe they're 50%, but that does not matter)
use a ring definition that does not require an identity element.
actually an ideal is a subring, if you don't require an identity
but, even if you call mZ an ideal, the question is: is every ideal of
Z of the form mZ?
Yes. The mZ's are the only ideals, subrings (without unit) and subgroups
of Z.
This is my question: why?
Suppose I is a nonzero ideal of Z.
Since x in I implies -x in I, it follows that I has positive elements.
Let m be the least positive element of I.
Claim I=mZ.
To prove it, consider each of the 2 inclusions.
(1) mZ is a subset of I.
(2) I is a subset of mZ.
Inclusion (1) is easy. Try it.
Inclusion (2) is a little more subtle.
Here's one of the standard way to prove it:
Let x be an arbitrary element of I. By the division algorithm, we can
write x=q*m+r where 0<=r<m. Since x and m are in I, it follows that r
is in I. Since m is the least positive element of I, r can't be
positive, hence r must be 0. It follows that x is in mZ.
quasi
.
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