Difference between revisions of "Ideal"
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If <math>\mathfrak{a}</math> and <math>\mathfrak{b}</math> are two-sided ideals of a ring <math>A</math>, then the set of elements of the form <math>\sum_{i\in I} a_ib_i</math>, for <math>a_i \in \mathfrak{a}</math> and <math>b_i \in \mathfrak{b}</math>, is also an ideal of <math>A</math>. It is called the product of <math>\mathfrak{a}</math> and <math>\mathfrak{b}</math>, and it is denoted <math>\mathfrak{ab}</math>. It is generated by the elements of the form <math>ab</math>, for <math>a\in \mathfrak{a}</math> and <math>b\in \mathfrak{b}</math>. Since <math>\mathfrak{a}</math> and <math>\mathfrak{b}</math> are two-sided ideals, <math>\mathfrak{ab}</math> is a subset of both <math>\mathfrak{a}</math> and of <math>\mathfrak{b}</math>, so | If <math>\mathfrak{a}</math> and <math>\mathfrak{b}</math> are two-sided ideals of a ring <math>A</math>, then the set of elements of the form <math>\sum_{i\in I} a_ib_i</math>, for <math>a_i \in \mathfrak{a}</math> and <math>b_i \in \mathfrak{b}</math>, is also an ideal of <math>A</math>. It is called the product of <math>\mathfrak{a}</math> and <math>\mathfrak{b}</math>, and it is denoted <math>\mathfrak{ab}</math>. It is generated by the elements of the form <math>ab</math>, for <math>a\in \mathfrak{a}</math> and <math>b\in \mathfrak{b}</math>. Since <math>\mathfrak{a}</math> and <math>\mathfrak{b}</math> are two-sided ideals, <math>\mathfrak{ab}</math> is a subset of both <math>\mathfrak{a}</math> and of <math>\mathfrak{b}</math>, so | ||
<cmath> \mathfrak{ab \subseteq a \cap b} . </cmath> | <cmath> \mathfrak{ab \subseteq a \cap b} . </cmath> | ||
− | |||
'''Proposition 1.''' Let <math>\mathfrak{a}</math> and <math>\mathfrak{b_1, \dotsc, b}_n</math> be two-sided ideals of a ring <math>A</math> such that <math>\mathfrak{a+b_i} = A</math>, for each index <math>i</math>. Then | '''Proposition 1.''' Let <math>\mathfrak{a}</math> and <math>\mathfrak{b_1, \dotsc, b}_n</math> be two-sided ideals of a ring <math>A</math> such that <math>\mathfrak{a+b_i} = A</math>, for each index <math>i</math>. Then | ||
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Now, suppose the proposition holds for <math>n-1</math>. Then | Now, suppose the proposition holds for <math>n-1</math>. Then | ||
− | <cmath> A = (\mathfrak{a + b}_1 \dotsm \mathfrak{b}_{n-1})(\mathfrak{a} + \mathfrak{b}_n) = \mathfrak{a \cdot a} + \mathfrak{ab}_n + \mathfrak{b}_1 \dotsm \mathfrak{b}_{n-1} \mathfrak{a} + \mathfrak{b}_1 \dotsm \mathfrak{b}_n = \mathfrak{a} + \mathfrak{b}_1 \dotsm \mathfrak{b}_n \subseteq \mathfrak{ a} + \bigcap_i \mathfrak{b}_i, </cmath> | + | <cmath> \begin{align*} A = (\mathfrak{a + b}_1 \dotsm \mathfrak{b}_{n-1})(\mathfrak{a} + \mathfrak{b}_n) &= \mathfrak{a \cdot a} + \mathfrak{ab}_n + \mathfrak{b}_1 \dotsm \mathfrak{b}_{n-1} \mathfrak{a} + \mathfrak{b}_1 \dotsm \mathfrak{b}_n \\ &= \mathfrak{a} + \mathfrak{b}_1 \dotsm \mathfrak{b}_n \subseteq \mathfrak{ a} + \bigcap_i \mathfrak{b}_i, \end{align*} </cmath> |
which proves the proposition. <math>\blacksquare</math> | which proves the proposition. <math>\blacksquare</math> | ||
+ | |||
+ | '''Proposition 2.''' Let <math>\mathfrak{b}_1, \dotsc \mathfrak{b}_n</math> be two-sided ideals of a ring <math>A</math> such that <math>\mathfrak{b}_i + \mathfrak{b}_j = A</math>, for any distinct indices <math>i</math> and <math>j</math>. Then | ||
+ | <cmath> \bigcap_{1\le i \le n} \mathfrak{b}_i = \sum_{\sigma \in S_n} \mathfrak{b}_{\sigma(1)} \dotsm \mathfrak{b}_{\sigma(n)} ,</cmath> | ||
+ | where <math>S_n</math> is the [[symmetric group]] on <math>\{ 1, \dostc, n\}</math>. | ||
+ | |||
+ | ''Proof.'' It is evident that | ||
+ | <cmath> \sum_{\sigma \in S_n} \mathfrak{b}_{\sigma(1)} \dotsm \mathfrak{b}_{\sigma(n)} \subseteq \bigcap_{1\le i \le n} \mathfrak{b}_i. </cmath> | ||
+ | We prove the converse by induction on <math>n</math>. | ||
+ | |||
+ | For <math>n=1</math>, the statement is trivial. For <math>n=2</math>, we note that 1 can be expressed as <math>b_1 + b_2</math>, where <math>b_i \in \mathfrak{b}_i</math>. Thus for any <math>b\in \mathfrak{b}_1 \cap \mathfrak{b}_2</math>, | ||
+ | <cmath> b = (b_1+b_2)b = b_1b + b_2b \in \mathfrak{b}_1 \mathfrak{b}_2 + \mathfrak{b}_2 \mathfrak{b}_1 . </cmath> | ||
+ | |||
+ | Now, suppose that the statement holds for the integer <math>n-1</math>. Then by the previous proposition, | ||
+ | <cmath> \mathfrak{b}_n + \bigcap_{1\le i \le n-1} \mathfrak{b}_i = A, </cmath> | ||
+ | so from the case <math>n=2</math>, | ||
+ | <cmath> \begin{align*} | ||
+ | \bigcap_{1\le i \le n} \mathfrak{b}_i &= \mathfrak{b}_n \bigcap_{1\le i \le n-1} \mathfrak{b}_i \\ | ||
+ | &\subseteq \biggl( \mathfrak{b}_n \sum_{\sigma\in S_{n-1}} \mathfrak{b}_{\sigma(1)} \dotsm \mathfrak{b}_{\sigma(n-1)} \biggr) + \biggl( \sum_{\sigma\in S_{n-1}} \mathfrak{b}_{\sigma(1)} \dotsm \mathfrak{b}_{\sigma(n-1)} \biggr) \mathfrak{b}_n \\ | ||
+ | &\subseteq \sum_{\sigma \in S_n} \mathfrak{b}_{\sigma(1)} \dotsm \mathfrak{b}_{\sigma(n)} , | ||
+ | \end{align*} </cmath> | ||
+ | as desired. <math>\blacksquare</math> | ||
==Problems== | ==Problems== |
Revision as of 15:43, 20 June 2008
In ring theory, an ideal is a special kind of subset of a ring. Two-sided ideals in rings are the kernels of ring homomorphisms; in this way, two-sided ideals of rings are similar to normal subgroups of groups.
Specifially, if is a ring, a subset
of
is called a left ideal of
if it is a subgroup under addition, and if
, for all
and
. Symbolically, this can be written as
A right ideal is defined similarly, but with the modification
. If
is both a left ideal and a right ideal, it is called a two-sided ideal. In a commutative ring, all three ideals are the same; they are simply called ideals. Note that the right ideals of a ring
are exactly the left ideals of the opposite ring
.
An ideal has the structure of a pseudo-ring, that is, a structure that satisfies the properties of rings, except possibly for the existance of a multiplicative identity.
By abuse of language, a (left, right, two-sided) ideal of a ring is called maximal if it is a maximal element of the set of (left, right, two-sided) ideals distinct from
.
Examples of Ideals
In the ring , the ideals are the rings of the form
, for some integer
.
In a field , the only ideals are the set
and
itself.
In general, if is a ring and
is an element of
, the set
is a left ideal of
. Ideals of this form are known as principle ideals.
Generated Ideals
Let be a ring, and let
be a family of elements of
. The left ideal generated by the family
is the set of elements of
of the form
where
is a family of elements of
of finite support, as this set is a left ideal of
, thanks to distributivity, and every element of the set must be in every left ideal containing
. Similarly, the two-sided ideal generated by
is the set of elements of
of the form
where
and
are families of finite support.
If is a set of (left, right, two-sided) ideals of
, then the (left, two sided) ideal generated by
is the set of elements of the form
, where
is an element of
and
is a family of finite support. For this reason, the ideal generated by the
is sometimes denoted
.
Multiplication of Ideals
If and
are two-sided ideals of a ring
, then the set of elements of the form
, for
and
, is also an ideal of
. It is called the product of
and
, and it is denoted
. It is generated by the elements of the form
, for
and
. Since
and
are two-sided ideals,
is a subset of both
and of
, so
Proposition 1. Let and
be two-sided ideals of a ring
such that
, for each index
. Then
Proof. We induct on . For
, the proposition is degenerately true.
Now, suppose the proposition holds for . Then
which proves the proposition.
Proposition 2. Let be two-sided ideals of a ring
such that
, for any distinct indices
and
. Then
where
is the symmetric group on $\{ 1, \dostc, n\}$ (Error compiling LaTeX. Unknown error_msg).
Proof. It is evident that
We prove the converse by induction on
.
For , the statement is trivial. For
, we note that 1 can be expressed as
, where
. Thus for any
,
Now, suppose that the statement holds for the integer . Then by the previous proposition,
so from the case
,
as desired.
Problems
<url>viewtopic.php?t=174516 Problem 1</url>