Difference between revisions of "Ideal"

Revision as of 14:43, 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 $A$ is a ring, a subset $\mathfrak{a}$ of $A$ is called a left ideal of $A$ if it is a subgroup under addition, and if $xa \in \alpha$ , for all $x\in R$ and $a\in \mathfrak{a}$ . Symbolically, this can be written as $0\in \mathfrak{a}, \qquad \mathfrak{a+a\subseteq a}, \qquad A \mathfrak{a \subseteq a} .$ A right ideal is defined similarly, but with the modification $\mathfrak{a}A \subseteq \mathfrak{a}$ . If $\mathfrak{a}$ 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 $A$ are exactly the left ideals of the opposite ring $A^0$ .

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 $A$ is called maximal if it is a maximal element of the set of (left, right, two-sided) ideals distinct from $A$ .

Examples of Ideals

In the ring $\mathbb{Z}$ , the ideals are the rings of the form $n \mathbb{Z}$ , for some integer $n$ .

In a field $F$ , the only ideals are the set $\{0\}$ and $F$ itself.

In general, if $A$ is a ring and $x$ is an element of $A$ , the set $Ax$ is a left ideal of $A$ . Ideals of this form are known as principle ideals.

Generated Ideals

Let $A$ be a ring, and let $(x_i)_{i\in I}$ be a family of elements of $A$ . The left ideal generated by the family $(x_i)_{i\in I}$ is the set of elements of $A$ of the form $\sum_{i \in I} a_i x_i,$ where $(a_i)_{i \in I}$ is a family of elements of $A$ of finite support, as this set is a left ideal of $A$ , thanks to distributivity, and every element of the set must be in every left ideal containing $(x_i)_{i\in I}$ . Similarly, the two-sided ideal generated by $(x_i)_{i\in I}$ is the set of elements of $A$ of the form $\sum_{i\in I} a_i x_i b_i,$ where $(a_i)_{i\in I}$ and $(b_i)_{i \in I}$ are families of finite support.

If $(\mathfrak{a}_i)_{i\in I}$ is a set of (left, right, two-sided) ideals of $A$ , then the (left, two sided) ideal generated by $\bigcup_{i\in I} \mathfrak{a}_i$ is the set of elements of the form $\sum_i x_i$ , where $x_i$ is an element of $\mathfrak{a}_i$ and $(x_i)_{i\in I}$ is a family of finite support. For this reason, the ideal generated by the $\mathfrak{a}_i$ is sometimes denoted $\sum_{i\in I} \mathfrak{a}_i$ .

Multiplication of Ideals

If $\mathfrak{a}$ and $\mathfrak{b}$ are two-sided ideals of a ring $A$ , then the set of elements of the form $\sum_{i\in I} a_ib_i$ , for $a_i \in \mathfrak{a}$ and $b_i \in \mathfrak{b}$ , is also an ideal of $A$ . It is called the product of $\mathfrak{a}$ and $\mathfrak{b}$ , and it is denoted $\mathfrak{ab}$ . It is generated by the elements of the form $ab$ , for $a\in \mathfrak{a}$ and $b\in \mathfrak{b}$ . Since $\mathfrak{a}$ and $\mathfrak{b}$ are two-sided ideals, $\mathfrak{ab}$ is a subset of both $\mathfrak{a}$ and of $\mathfrak{b}$ , so $\mathfrak{ab \subseteq a \cap b} .$

Proposition 1. Let $\mathfrak{a}$ and $\mathfrak{b_1, \dotsc, b}_n$ be two-sided ideals of a ring $A$ such that $\mathfrak{a+b_i} = A$ , for each index $i$ . Then $A = \mathfrak{a + b}_1 \dotsc \mathfrak{b}_n = \mathfrak{ a} + \bigcap_i \mathfrak{b}_i .$

Proof. We induct on $n$ . For $n=1$ , the proposition is degenerately true.

Now, suppose the proposition holds for $n-1$ . Then $\begin{align*} A = (\mathfrak{a + b}_1 \dotsm \mathfrak{b}_{n-1})(\mathfrak{a} + \mathfrak{b}_n) &= \mathfrak{a \cdot a} + \mathfrak{ab}_n + \mathfrak{b}_1 \dotsm \mathfrak{b}_{n-1} \mathfrak{a} + \mathfrak{b}_1 \dotsm \mathfrak{b}_n \\ &= \mathfrak{a} + \mathfrak{b}_1 \dotsm \mathfrak{b}_n \subseteq \mathfrak{ a} + \bigcap_i \mathfrak{b}_i, \end{align*}$ which proves the proposition. $\blacksquare$

Proposition 2. Let $\mathfrak{b}_1, \dotsc \mathfrak{b}_n$ be two-sided ideals of a ring $A$ such that $\mathfrak{b}_i + \mathfrak{b}_j = A$ , for any distinct indices $i$ and $j$ . Then $\bigcap_{1\le i \le n} \mathfrak{b}_i = \sum_{\sigma \in S_n} \mathfrak{b}_{\sigma(1)} \dotsm \mathfrak{b}_{\sigma(n)} ,$ where $S_n$ is the symmetric group on $\{ 1, \dostc, n\}$ (Error compiling LaTeX. Unknown error_msg).

Proof. It is evident that $\sum_{\sigma \in S_n} \mathfrak{b}_{\sigma(1)} \dotsm \mathfrak{b}_{\sigma(n)} \subseteq \bigcap_{1\le i \le n} \mathfrak{b}_i.$ We prove the converse by induction on $n$ .

For $n=1$ , the statement is trivial. For $n=2$ , we note that 1 can be expressed as $b_1 + b_2$ , where $b_i \in \mathfrak{b}_i$ . Thus for any $b\in \mathfrak{b}_1 \cap \mathfrak{b}_2$ , $b = (b_1+b_2)b = b_1b + b_2b \in \mathfrak{b}_1 \mathfrak{b}_2 + \mathfrak{b}_2 \mathfrak{b}_1 .$

Now, suppose that the statement holds for the integer $n-1$ . Then by the previous proposition, $\mathfrak{b}_n + \bigcap_{1\le i \le n-1} \mathfrak{b}_i = A,$ so from the case $n=2$ , $\begin{align*} \bigcap_{1\le i \le n} \mathfrak{b}_i &= \mathfrak{b}_n \bigcap_{1\le i \le n-1} \mathfrak{b}_i \\ &\subseteq \biggl( \mathfrak{b}_n \sum_{\sigma\in S_{n-1}} \mathfrak{b}_{\sigma(1)} \dotsm \mathfrak{b}_{\sigma(n-1)} \biggr) + \biggl( \sum_{\sigma\in S_{n-1}} \mathfrak{b}_{\sigma(1)} \dotsm \mathfrak{b}_{\sigma(n-1)} \biggr) \mathfrak{b}_n \\ &\subseteq \sum_{\sigma \in S_n} \mathfrak{b}_{\sigma(1)} \dotsm \mathfrak{b}_{\sigma(n)} , \end{align*}$ as desired. $\blacksquare$

Problems

<url>viewtopic.php?t=174516 Problem 1</url>

@@ Line 31: / Line 31: @@
 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>
-The ideals of <math>A</math> constitute a [[pseudo-ring]].
 '''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
@@ Line 39: / Line 38: @@
 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>
+'''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==

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