Product of Rings (Pt. II)
Point of Post: This is a continuation of this post.
Unit Group of a Product of Rings
The next question we’d like to answer is whether or not we can describe the group of units in terms of the groups of units . The first obvious fact is that, as sets we have:
Theorem: Let be a set of unital rings. Then, as sets
Proof: Let be in and define to be such that . We see then that and for every and so and so is a unit in the product with .
Conversely, if is in then we know that since we know that the canonical projection is a unital homomorphism and so preserves units and .
In particular we see that the product of rings always has a non-zero non-unitRecalling our recasting of the definition of the product when is finite we see in particular that is setwise equal to . The interesting thing though is that comparing the definition of the product of groups we get the following theorem:
Theorem: Let be a set of maps. Then, as groups
We would now like to discuss certain isomorphism issues which come up naturally when discussing products. Things like whether , whether and implies , etc.
We’ve actually secretly done the first of these. Namely:
Theorem: Let be a set of rings with a set of ideals with . Then,
Proof: We’ve already discussed that the product of the canonical epimorphism is an epimorphism which has kernel . The conclusion then follows from the first isomorphism theorem.
As a corollary we find things such as .
We finish with an obvious but important theorem:
Theorem: Let and be two sets of rings with then .
Proof: Let be one of the guaranteed isomorphism then we now prove that
is an isomorphism. We already know that it’s a morphism, and so it suffices to show that it’s injective and surjective. Surjectivity is obvious since if is in then we know for each there exists some such that and so if we define in the domain by then evidently and so our map is surjective. It’s evidently injective since if for in the domain then for some and so we know and so evidently and so our map is injective. The conclusion follows.
We thus have from previous theorem that:
Theorem: If and be two sets of rings. Then,
1. Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 2004. Print.
2. Bhattacharya, P. B., S. K. Jain, and S. R. Nagpaul. Basic Abstract Algebra. Cambridge [Cambridgeshire: Cambridge UP, 1986. Print.