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In [[ring theory]], an '''ideal''' is a special kind of [[subset]] of a [[ring]].  Two-sided ideals in rings are the [[kernel]]s of ring [[homomorphism]]s; in this way, two-sided ideals of rings are similar to [[normal subgroup]]s of [[group]]s.

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

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 <math>\mathbb{Z}</math>, the ideals are the rings of the form <math>n \mathbb{Z}</math>, for some integer <math>n</math>.

In a [[field]] <math>F</math>, the only ideals are the set <math>\{0\}</math> and <math>F</math> itself.

In general, if <math>A</math> is a ring and <math>x</math> is an element of <math>A</math>, the set <math>Ax</math> is a left ideal of <math>A</math>.  Ideals of this form are known as [[principal ideal]]s.

By abuse of language, a (left, right, two-sided) ideal of a ring <math>A</math> is called ''maximal'' if it is a [[maximum |maximal element]] of the set of (left, right, two-sided) ideals distinct from <math>A</math>.  A two-sided ideal <math>\mathfrak{m}</math> is maximal if and only if its [[quotient]] ring <math>A/\mathfrak{m}</math> is a [[field]].

An ideal <math>\mathfrak{p}</math> is called a ''prime'' ideal if <math>ab\in \mathfrak{p}</math> implies <math>a\in \mathfrak{p}</math> or <math>b \in \mathfrak{p}</math>.  A two-sided ideal <math>\mathfrak{p}</math> is prime if and only if its quotient ring <math>A/ \mathfrak{p}</math> is a [[domain (ring theory) |domain]].  In [[commutative algebra]], the notion of prime ideal is central; it generalizes the notion of [[prime number]]s in <math>\mathbb{Z}</math>.

== Generated Ideals ==

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

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

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

== Multiplication of Ideals ==

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>

'''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
<cmath> A = \mathfrak{a + b}_1 \dotsc \mathfrak{b}_n = \mathfrak{ a} + \bigcap_i \mathfrak{b}_i . </cmath>

''Proof.''  We induct on <math>n</math>.  For <math>n=1</math>, the proposition is degenerately true.

Now, suppose the proposition holds for <math>n-1</math>.  Then
<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, \dotsc, 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==
<url>viewtopic.php?t=174516 Problem 1</url>

== See also ==

* [[Quotient ring]]
* [[Subring]]
* [[Normal subgroup]]
* [[Krull's Theorem]]
* [[Pseudo-ring]]

[[Category:Ring theory]]