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Ideal

Revision as of 20:32, 21 September 2008 by Boy Soprano II (talk | contribs) (added a little more)

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 kinds of 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 existence of a multiplicative identity.

Examples of Ideals; Types 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 principal ideals.

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 $A$. A two-sided ideal $\mathfrak{m}$ is maximal if and only if its quotient ring $A/\mathfrak{m}$ is a field.

An ideal $\mathfrak{p}$ is called a prime ideal if $ab\in \mathfrak{p}$ implies $a\in \mathfrak{p}$ or $b \in \mathfrak{p}$. A two-sided ideal $\mathfrak{p}$ is prime if and only if its quotient ring $A/ \mathfrak{p}$ is a domain. In commutative algebra, the notion of prime ideal is central; it generalizes the notion of prime numbers in $\mathbb{Z}$.

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.

The two-sided ideal generated by a finite family ${a_1, \dotsc, a_n}$ is often denoted $(a_1, \dotsc, a_n)$.

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 $\{ 1, \dotsc, n\}$.

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>

See also

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