Quotient ring#For algebras over a ring

Given a ring $R$ and a two-sided ideal $I$ in {{tmath|1= R }}, we may define an equivalence relation $\backslash sim$ on $R$ as follows:

: $a\; \backslash sim\; b$ if and only if $a\; -\; b$ is in {{tmath|1= I }}.

Using the ideal properties, it is not difficult to check that $\backslash sim$ is a congruence relation.

In case {{tmath|1= a \sim b }}, we say that $a$ and $b$ are congruent modulo $I$ (for example, $1$ and $3$ are congruent modulo $2$ as their difference is an element of the ideal {{tmath|1= 2 \mathbb{Z} }}, the even integers). The equivalence class of the element $a$ in $R$ is given by:

$$\backslash left[\; a\; \backslash right]\; =\; a\; +\; I\; :=\; \backslash left\backslash lbrace\; a\; +\; r\; :\; r\; \backslash in\; I\; \backslash right\backslash rbrace$$

This equivalence class is also sometimes written as $a\; \backslash bmod\; I$ and called the "residue class of $a$ modulo $I$".

The set of all such equivalence classes is denoted by {{tmath|1= R\ /\ I }}; it becomes a ring, the factor ring or quotient ring of $R$ modulo {{tmath|1 = I }}, if one defines

• {{tmath|1= (a + I) + (b + I) = (a + b) + I }};
• {{tmath|1= (a + I)(b + I) = (ab) + I }}.

(Here one has to check that these definitions are well-defined. Compare coset and quotient group.) The zero-element of $R\backslash \; /\backslash \; I$ is {{tmath|1= \bar{0} = (0 + I) = I }}, and the multiplicative identity is {{tmath|1= \bar{1} = (1 + I) }}.

The map $p$ from $R$ to $R\backslash \; /\backslash \; I$ defined by $p(a)\; =\; a\; +\; I$ is a surjective ring homomorphism, sometimes called the natural quotient map or the canonical homomorphism.

• The quotient ring $R\backslash \; /\backslash \; \backslash lbrace\; 0\; \backslash rbrace$ is naturally isomorphic to {{tmath|1= R }}, and $R\; /\; R$ is the zero ring {{tmath|1= \lbrace 0 \rbrace }}, since, by our definition, for any {{tmath|1= r \in R }}, we have that {{tmath|1= \left[ r \right] = r + R = \left\lbrace r + b : b \in R \right\rbrace }}, which equals $R$ itself. This fits with the rule of thumb that the larger the ideal {{tmath|1= I }}, the smaller the quotient ring {{tmath|1= R\ /\ I }}. If $I$ is a proper ideal of {{tmath|1= R }}, i.e., {{tmath|1= I \neq R }}, then $R\; /\; I$ is not the zero ring.
• Consider the ring of integers $\backslash mathbb\{Z\}$ and the ideal of even numbers, denoted by {{tmath|1= 2 \mathbb{Z} }}. Then the quotient ring $\backslash mathbb\{Z\}\; /\; 2\; \backslash mathbb\{Z\}$ has only two elements, the coset $0\; +\; 2\; \backslash mathbb\{Z\}$ consisting of the even numbers and the coset $1\; +\; 2\; \backslash mathbb\{Z\}$ consisting of the odd numbers; applying the definition, {{tmath|1= \left[ z \right] = z + 2 \mathbb{Z} = \left\lbrace z + 2y : 2y \in 2\mathbb{Z} \right\rbrace }}, where $2\; \backslash mathbb\{Z\}$ is the ideal of even numbers. It is naturally isomorphic to the finite field with two elements, {{tmath|F_{2} }}. Intuitively: if you think of all the even numbers as {{tmath|1= 0 }}, then every integer is either $0$ (if it is even) or $1$ (if it is odd and therefore differs from an even number by {{tmath|1= 1 }}). Modular arithmetic is essentially arithmetic in the quotient ring $\backslash mathbb\{Z\}\; /\; n\; \backslash mathbb\{Z\}$ (which has $n$ elements).
• Now consider the ring of polynomials in the variable $X$ with real coefficients, {{tmath|1= \mathbb{R} [X] }}, and the ideal $I\; =\; \backslash left(\; X^2\; +\; 1\; \backslash right)$ consisting of all multiples of the polynomial {{tmath|1= X^2 + 1 }}. The quotient ring $\backslash mathbb\{R\}\; [X]\backslash \; /\backslash \; (\; X^2\; +\; 1\; )$ is naturally isomorphic to the field of complex numbers {{tmath|1= \mathbb{C} }}, with the class $[X]$ playing the role of the imaginary unit {{tmath|1= i }}. The reason is that we "forced" {{tmath|1= X^2 + 1 = 0 }}, i.e. {{tmath|1= X^2 = -1 }}, which is the defining property of {{tmath|1= i }}. Since any integer exponent of $i$ must be either $\backslash pm\; i$ or {{tmath|1= \pm 1 }}, that means all possible polynomials essentially simplify to the form {{tmath|1= a + bi }}. (To clarify, the quotient ring {{tmath|1= \mathbb{R} [X]\ /\ ( X^2 + 1 ) }} is actually naturally isomorphic to the field of all linear polynomials {{tmath|1= aX + b; a,b \in \mathbb{R} }}, where the operations are performed modulo {{tmath|1= X^2 + 1 }}. In return, we have {{tmath|1= X^2 = -1 }}, and this is matching $X$ to the imaginary unit in the isomorphic field of complex numbers.)
• Generalizing the previous example, quotient rings are often used to construct field extensions. Suppose $K$ is some field and $f$ is an irreducible polynomial in {{tmath|1= K[X] }}. Then $L\; =\; K[X]\backslash \; /\backslash \; (f)$ is a field whose minimal polynomial over $K$ is {{tmath|1= f }}, which contains $K$ as well as an element {{tmath|1= x = X + (f) }}.
• One important instance of the previous example is the construction of the finite fields. Consider for instance the field $F\_3\; =\; \backslash mathbb\{Z\}\; /\; 3\backslash mathbb\{Z\}$ with three elements. The polynomial $f(X)\; =\; Xi^2\; +1$ is irreducible over $F\_3$ (since it has no root), and we can construct the quotient ring {{tmath|1= F_3 [X]\ /\ (f) }}. This is a field with $3^2\; =\; 9$ elements, denoted by {{tmath|1= F_9 }}. The other finite fields can be constructed in a similar fashion.
• The coordinate rings of algebraic varieties are important examples of quotient rings in algebraic geometry. As a simple case, consider the real variety $V\; =\; \backslash left\backslash lbrace\; (x,y)\; |\; x^2\; =\; y^3\; \backslash right\backslash rbrace$ as a subset of the real plane {{tmath|1= \mathbb{R}^2 }}. The ring of real-valued polynomial functions defined on $V$ can be identified with the quotient ring {{tmath|1= \mathbb{R} [X,Y]\ /\ (X^2 - Y^3) }}, and this is the coordinate ring of {{tmath|1= V }}. The variety $V$ is now investigated by studying its coordinate ring.
• Suppose $M$ is a $\backslash mathbb\{C\}^\{\backslash infty\}$-manifold, and $p$ is a point of {{tmath|1= M }}. Consider the ring $R\; =\; \backslash mathbb\{C\}^\{\backslash infty\}(M)$ of all $\backslash mathbb\{C\}^\{\backslash infty\}$-functions defined on $M$ and let $I$ be the ideal in $R$ consisting of those functions $f$ which are identically zero in some neighborhood $U$ of $p$ (where $U$ may depend on {{tmath|1= f }}). Then the quotient ring $R\backslash \; /\backslash \; I$ is the ring of germs of $\backslash mathbb\{C\}^\{\backslash infty\}$-functions on $M$ at {{tmath|1= p }}.
• Consider the ring $F$ of finite elements of a hyperreal field {{tmath|1= ^* \mathbb{R} }}. It consists of all hyperreal numbers differing from a standard real by an infinitesimal amount, or equivalently: of all hyperreal numbers $x$ for which a standard integer $n$ with exists. The set $I$ of all infinitesimal numbers in {{tmath|1= ^* \mathbb{R} }}, together with {{tmath|1= 0 }}, is an ideal in {{tmath|1= F }}, and the quotient ring $F\backslash \; /\backslash \; I$ is isomorphic to the real numbers {{tmath|1= \mathbb{R} }}. The isomorphism is induced by associating to every element $x$ of $F$ the standard part of {{tmath|1= x }}, i.e. the unique real number that differs from $x$ by an infinitesimal. In fact, one obtains the same result, namely {{tmath|1= \mathbb{R} }}, if one starts with the ring $F$ of finite hyperrationals (i.e. ratio of a pair of hyperintegers), see construction of the real numbers.

The quotients {{tmath|1= \mathbb{R} [X] / (X) }}, {{tmath|1= \mathbb{R} [X] / (X + 1) }}, and $\backslash mathbb\{R\}\; [X]\; /\; (X\; -\; 1)$ are all isomorphic to $\backslash mathbb\{R\}$ and gain little interest at first. But note that $\backslash mathbb\{R\}\; [X]\; /\; (X^2)$ is called the dual number plane in geometric algebra. It consists only of linear binomials as "remainders" after reducing an element of $\backslash mathbb\{R\}\; [X]$ by {{tmath|1= X^2 }}. This variation of a complex plane arises as a subalgebra whenever the algebra contains a real line and a nilpotent.

Furthermore, the ring quotient $\backslash mathbb\{R\}\; [X]\; /\; (X^2\; -\; 1)$ does split into $\backslash mathbb\{R\}\; [X]\; /\; (X\; +\; 1)$ and {{tmath|1= \mathbb{R} [X] / (X - 1) }}, so this ring is often viewed as the direct sum {{tmath|1= \mathbb{R} \oplus \mathbb{R} }}.

Nevertheless, a variation on complex numbers $z\; =\; x\; +\; yj$ is suggested by $j$ as a root of {{tmath|1= X^2 - 1 = 0 }}, compared to $i$ as root of {{tmath|1= X^2 + 1 = 0 }}. This plane of split-complex numbers normalizes the direct sum $\backslash mathbb\{R\}\; \backslash oplus\; \backslash mathbb\{R\}$ by providing a basis $\backslash left\backslash lbrace\; 1,\; j\; \backslash right\backslash rbrace$ for 2-space where the identity of the algebra is at unit distance from the zero. With this basis a unit hyperbola may be compared to the unit circle of the ordinary complex plane.

Suppose $X$ and $Y$ are two non-commuting indeterminates and form the free algebra {{tmath|1= \mathbb{R} \langle X, Y \rangle }}. Then Hamilton's quaternions of 1843 can be cast as:

$$\backslash mathbb\{R\}\; \backslash langle\; X,\; Y\; \backslash rangle\; /\; (\; X^2\; +\; 1,\backslash ,\; Y^2\; +\; 1,\backslash ,\; XY\; +\; YX\; )$$

If $Y^2\; -\; 1$ is substituted for {{tmath|1= Y^2 + 1 }}, then one obtains the ring of split-quaternions. The anti-commutative property $YX\; =\; -XY$ implies that $XY$ has as its square:

$$(XY)\; (XY)\; =\; X\; (YX)\; Y\; =\; -X\; (XY)\; Y\; =\; -(XX)\; (YY)\; =\; -(-1)(+1)\; =\; +1$$

Substituting minus for plus in both the quadratic binomials also results in split-quaternions.

The three types of biquaternions can also be written as quotients by use of the free algebra with three indeterminates $\backslash mathbb\{R\}\; \backslash langle\; X,\; Y,\; Z\; \backslash rangle$ and constructing appropriate ideals.

Clearly, if $R$ is a commutative ring, then so is {{tmath|1= R\ /\ I }}; the converse, however, is not true in general.

The natural quotient map $p$ has $I$ as its kernel; since the kernel of every ring homomorphism is a two-sided ideal, we can state that two-sided ideals are precisely the kernels of ring homomorphisms.

The intimate relationship between ring homomorphisms, kernels and quotient rings can be summarized as follows: the ring homomorphisms defined on $R\backslash \; /\backslash \; I$ are essentially the same as the ring homomorphisms defined on $R$ that vanish (i.e. are zero) on {{tmath|1= I }}. More precisely, given a two-sided ideal $I$ in $R$ and a ring homomorphism $f\; :\; R\; \backslash to\; S$ whose kernel contains {{tmath|1= I }}, there exists precisely one ring homomorphism $g\; :\; R\backslash \; /\backslash \; I\; \backslash to\; S$ with $gp\; =\; f$ (where $p$ is the natural quotient map). The map $g$ here is given by the well-defined rule $g([a])\; =\; f(a)$ for all $a$ in {{tmath|1 R }}. Indeed, this universal property can be used to define quotient rings and their natural quotient maps.

As a consequence of the above, one obtains the fundamental statement: every ring homomorphism $f\; :\; R\; \backslash to\; S$ induces a ring isomorphism between the quotient ring $R\backslash \; /\backslash \; \backslash ker\; (f)$ and the image {{tmath|1= \mathrm{im} (f) }}. (See also: Fundamental theorem on homomorphisms.)

The ideals of $R$ and $R\backslash \; /\backslash \; I$ are closely related: the natural quotient map provides a bijection between the two-sided ideals of $R$ that contain $I$ and the two-sided ideals of $R\backslash \; /\backslash \; I$ (the same is true for left and for right ideals). This relationship between two-sided ideal extends to a relationship between the corresponding quotient rings: if $M$ is a two-sided ideal in $R$ that contains {{tmath|1= I }}, and we write $M\backslash \; /\backslash \; I$ for the corresponding ideal in $R\backslash \; /\backslash \; I$ (i.e. {{tmath|1= M\ /\ I = p(M) }}), the quotient rings $R\backslash \; /\backslash \; M$ and $(R\; /\; I)\backslash \; /\backslash \; (M\; /\; I)$ are naturally isomorphic via the (well-defined) mapping {{tmath|1= a + M \mapsto (a + I) + M / I }}.

The following facts prove useful in commutative algebra and algebraic geometry: for $R\; \backslash neq\; \backslash lbrace\; 0\; \backslash rbrace$ commutative, $R\backslash \; /\backslash \; I$ is a field if and only if $I$ is a maximal ideal, while $R\; /\; I$ is an integral domain if and only if $I$ is a prime ideal. A number of similar statements relate properties of the ideal $I$ to properties of the quotient ring {{tmath|1= R\ /\ I }}.

The Chinese remainder theorem states that, if the ideal $I$ is the intersection (or equivalently, the product) of pairwise coprime ideals {{tmath|1= I_1, \ldots, I_k }}, then the quotient ring $R\backslash \; /\backslash \; I$ is isomorphic to the product of the quotient rings {{tmath|1= R\ /\ I_n,\; n = 1, \ldots, k }}.

An associative algebra $A$ over a commutative ring $R$ is a ring itself. If $I$ is an ideal in $A$ (closed under $R$-multiplication), then $A\; /\; I$ inherits the structure of an algebra over $R$ and is the quotient algebra.

• Associated graded ring
• Residue field
• Goldie's theorem
• Quotient module

{{reflist}}

• F. Kasch (1978) Moduln und Ringe, translated by DAR Wallace (1982) Modules and Rings, Academic Press, page 33.
• Neal H. McCoy (1948) Rings and Ideals, §13 Residue class rings, page 61, Carus Mathematical Monographs #8, Mathematical Association of America.
• {{cite book | author=Joseph Rotman | title=Galois Theory| publisher=Springer | pages=21–23 | year=1998 | isbn=0-387-98541-7 | edition=2nd }}
• B.L. van der Waerden (1970) Algebra, translated by Fred Blum and John R Schulenberger, Frederick Ungar Publishing, New York. See Chapter 3.5, "Ideals. Residue Class Rings", pp. 47–51.

• {{springer|title=Quotient ring|id=p/q076920}}
• [http://www.math.niu.edu/~beachy/aaol/rings.html#ideals Ideals and factor rings] from John Beachy's Abstract Algebra Online

Ring

Category:Ring theory