Exploring Ring Isomorphisms Between Mathbb{Z}oplus otimes and Mathbb{Z}cdot

Understanding ring isomorphisms involves identifying a function that preserves the algebraic structure between two rings. In this article, we explore the ring isomorphism between ( R mathbb{Z} oplus mathbb{Z} otimes ) and ( S mathbb{Z} cdot ).

Understanding the Rings R and S

Before defining a ring isomorphism, we gain a deeper understanding of the rings ( R ) and ( S ). Specifically, we investigate their identity elements with respect to the binary operations ( oplus ) and ( otimes ).

Additive and Multiplicative Identities

For the ring ( R mathbb{Z} oplus mathbb{Z} otimes ), let ( e ) represent the additive identity. We need ( e oplus b b ) for any ( b in mathbb{Z} ). Solving this equation yields ( e 1 ). Similarly, for the multiplicative identity, letting ( e ) be the multiplicative identity in ( R ) leads to ( e otimes b - 1 b ), resulting in ( e 0 ).

For the ring ( S mathbb{Z} cdot ), the additive identity remains ( 0 ), and the multiplicative identity is ( 1 ).

Ring Isomorphism Considerations

For a ring isomorphism ( f: R to S ), it must map the identity elements of one ring to the corresponding identity elements of the other ring. Therefore, any such isomorphism ( f ) must satisfy both ( f(1) 0 ) and ( f(0) 1 ).

One such isomorphism is given by ( f(x) 1 - x ). We will prove that this map is indeed a ring isomorphism by demonstrating that it is both a homomorphism and a bijection.

Proof of Ring Homomorphism

Define the map ( f: R to S ) by ( f(x) 1 - x ). We claim that ( f ) is a ring isomorphism.

Additive Homomorphism

For any ( x, y in R ), we have:

( f(x oplus y) f(xy - 1) 1 - (xy - 1) 2 - xy (1 - x)(1 - y) f(x) cdot f(y) )

Multiplicative Homomorphism

For multiplication, we find:

( f(x otimes y) f(xy - xy) 1 - (xy - xy) 1 - (1 - x)(1 - y) f(x) cdot f(y) )

Proof of Bijection

To show that ( f ) is a bijection, we proceed in two steps: proving that ( f ) is one-to-one and that ( f ) is onto.

Injectivity

Assume ( f(x) f(y) ). This implies ( 1 - x 1 - y ), leading to ( x y ). Therefore, ( f ) is one-to-one.

Surjectivity

Let ( y in S ). Then ( x 1 - y in R ) by closure laws in ( mathbb{Z} ). We have ( f(x) f(1 - y) y ), as required. Therefore, ( f ) is also onto.

Conclusion

Since ( f ) is both a homomorphism and a bijection, we conclude that ( f ) is a ring isomorphism.

Write Down the Isomorphism

To write down the isomorphism explicitly, we note the following:

( x cdot y (x oplus y) - 1 )

This is the reverse direction of the map ( f(x) 1 - x ) that we have already defined.

Checking the ring axioms confirms that both ( R ) and ( S ) satisfy the given operations, and the isomorphism holds.