Product of rings

In mathematics, the direct product or Cartesian product is a component-wise operation between two sets, groups, rings, or fields. The order of the structure of the direct product always matters. The direct product between two sets, R and S is written as R X S. Take notice that R X S ≠ S X R.

i.e. Let R and S be sets such that r_i ∈ R and s_i ∈ S for i ∈ N. Then (r₁,s₁), (r₂,s₂) ∈ R X S and (s₁,r₁), (s₂,r₂) ∈ S X R. For this example the reader should take notice how the order of the structure of the direct product always matters. R X S and S X R are distinct sets both with the same cardinality. Direct product is a similar operation to multiplying polynomials, except order matters. (r₁,s_i) ∈ R X S ∀ s_i ∈ S. This pattern continues for r₂,r₃, ...,r_n; until all possible combinations between the two sets are satisfied.

i. e. Let (R,+,·) and (S,#,*) be rings. The direct product (R X S,⊕,⊗) is also a ring under component-wise addition and multiplication. If (R,·) and (S,*) are commutative then R X S is commutative. (r_i,s_i) ∈ R X S ∀ r_i ∈ R and ∀ s_i ∈ S.

(r₁,s₁) ⊕ (r₂,s₂) = (r₁ + r₂,s₁ # s₂)

(r₁,s₁) ⊗ (r₂,s₂) = (r₁ · r₂,s₁ * s₂)

The notation examples shown would follow for the component-wise operations on the direct product of two groups or two fields. ΠR_i represents the multiplication of elements of a set, but when used with reference to sets, groups, rings, or fields ΠR_i represents the direct product. In advanced Abstract Algebra texts ΠR_i will typically be the notation referenced for direct product.

Further Explanation: It is possible to combine several rings into one large product ring. This is done as follows: if I is some index set and R_i is a ring for every i in I, then the Cartesian product Π_{i in I} R_i can be turned into a ring by defining the operations coordinatewise, i.e.

(a_i) + (b_i) = (a_i + b_i)

(a_i) · (b_i) = (a_i · b_i)

The resulting ring is called a direct product of the rings R_i. The direct product of finitely many rings R₁,...,R_k is also written as R₁ × R₂ × ... × R_k or R₁ ⊕ R₂ ⊕ ... ⊕ R_k, and can also be called the direct sum (and sometimes the complete direct sum) of the rings R_i.

Examples

An important example is the ring Z/nZ of integers modulo n. If n is written as a product of prime powers (see fundamental theorem of arithmetic):

$n=p_{1}^{n_{1}}\ p_{2}^{n_{2}}\ \cdots \ p_{k}^{n_{k}}$

where the p_i are distinct primes, then Z/nZ is naturally isomorphic to the product ring

$\mathbf {Z} /p_{1}^{n_{1}}\mathbf {Z} \ \times \ \mathbf {Z} /p_{2}^{n_{2}}\mathbf {Z} \ \times \ \cdots \ \times \ \mathbf {Z} /p_{k}^{n_{k}}\mathbf {Z}$

This follows from the Chinese remainder theorem.

Properties

If R = Π_{i in I} R_i is a product of rings, then for every i in I we have a surjective ring homomorphism p_i: R → R_i which projects the product on the i-th coordinate. The product R, together with the projections p_i, has the following universal property:

if S is any ring and f_i: S → R_i is a ring homomorphism for every i in I, then there exists precisely one ring homomorphism f: S → R such that p_i o f = f_i for every i in I.

This shows that the product of rings is an instance of products in the sense of category theory. However, despite also being called the direct sum of rings when I is finite, the product of rings is not a coproduct in the sense of category theory. In particular, if I has more than one element, the inclusion map R_i → R is not ring homomorphism as it does not map the identity in R_i to the identity in R.

If A_i in R_i is an ideal for each i in I, then A = Π_{i in I} A_i is an ideal of R. If I is finite, then the converse is true, i.e. every ideal of R is of this form. However, if I is infinite and the rings R_i are non-zero, then the converse is false; the set of elements with all but finitely many nonzero coordinates forms an ideal which is not a direct product of ideals of the R_i. The ideal A is a prime ideal in R if all but one of the A_i are equal to R_i and the remaining A_i is a prime ideal in R_i. However, the converse is not true when I is infinite. For example, the direct sum of the R_i form an ideal not contained in any such A, but the axiom of choice gives that it is contained in some maximal ideal which is a fortiori prime.

An element x in R is a unit if and only if all of its components are units, i.e. if and only if p_i(x) is a unit in R_i for every i in I. The group of units of R is the product of the groups of units of R_i.

A product of more than one non-zero rings always has zero divisors: if x is an element of the product all of whose coordinates are zero except p_i(x), and y is an element of the product with all coordinates zero except p_j(y) (with i ≠ j), then xy = 0 in the product ring.

Proofs

A) Let (R,+,·) and (S,#,*) be rings. Prove the direct product (R X S,⊕,⊗), with component-wise addition and multiplication, contains a multiplicative identity iff R and S contain multiplicative identities. Note: Additive identities are always assumed for rings since (R,+) and (S,#) are groups by the definition of a ring.

Proof: Let r_i ∈ R, s_i ∈ S, and (r_i,s_i) ∈ R X S ∀ i ∈ N. If 1_R ∈ R such that (r₁ · 1_R) = r₁ = (1_R · r₁) ∀ r₁ ∈ R then 1_R is the unity of R. If 1_S ∈ S such that (s₁ * 1_S) = s₁ = (1_S * s₁) ∀ s₁ ∈ S then 1_S is the unity of S.

(→) (r₁,s₁) ⊗ (1_R,1_S) = (r₁ · 1_R,s₁ * 1_S) = (r₁,s₁). (1_R,1_S) ⊗ (r₁,s₁) = (1_R · r₁,1_S * s₁) = (r₁,s₁). Hence, (1_R,1_S) = 1_{R X S} iff 1_R ∈ R and 1_S ∈ S.

(←) If (u,v) = 1_{R X S} then (u,v) ⊗ (r₁,s₁) = (u · r₁,v * s₁) = (r₁,s₁) and (r₁,s₁) ⊗ (u,v) = (r₁ · u ,s₁ * v) = (r₁,s₁) ∀ r₁ ∈ R and ∀ s₁ ∈ S. Thus, u = 1_R and v = 1_S. Therefore, R X S contains a multiplicative identity iff R and S contain multiplicative identities.

B) Prove (r₁,s₁) is a unit in (R X S,⊕,⊗) iff r₁ is a unit in (R,+,·) and s₁ is a unit in (S,#,*).

Proof: If r₁ is a unit in (R,+,·) then ∃ r^-1
₁ ∈ R such that r₁ · r^-1
₁ = 1_R = r^-1
₁ · r₁. If s₁ is a unit in (S,#,*) then ∃ s^-1
₁ ∈ S such that s₁ · s^-1
₁ = 1_S = s^-1
₁ · s₁.

(→) (r₁,s₁) ⊗ (r^-1
₁,s^-1
₁) = (r₁ · r^-1
₁,s₁ * s^-1
₁) = (r^-1
₁ · r₁, s^-1
₁ * s₁) = (1_R,1_S) = 1_{R X S}.

(←) Suppose (r₁,s₁) is a unit in R X S. Then ∃ (r^-1
₁,s^-1
₁) ∈ R X S such that (r₁,s₁) ⊗ (r^-1
₁,s^-1
₁) = (r₁ · r^-1
₁,s₁ * s^-1
₁) = (r^-1
₁ · r₁, s^-1
₁ * s₁) = (1_R,1_S) = 1_{R X S}.

Thus, (r₁ · r^-1
₁) = (r^-1
₁ · r₁) = 1_R and (s₁ · s^-1
₁) = (s^-1
₁ · s₁) = 1_S. Hence r₁ and s₁ are units in R and S respectively.

See also

• Direct product

References

• Herstein, I.N. (2005) [1968], Noncommutative rings (5th ed.), Cambridge University Press, ISBN 978-0-88385-039-8
• Lang, Serge (2002), Algebra, Graduate Texts in Mathematics, vol. 211 (Revised third ed.), New York: Springer-Verlag, p. 91, ISBN 978-0-387-95385-4, MR 1878556, Zbl 0984.00001