Direct product is cancellative for finite algebras in any variety with zero
Contents
Statement
Suppose is a variety of algebras that is a variety with zero. Suppose
are finite algebras in
. Further, suppose that the algebras
and
are isomorphic as algebras in
. Then,
is isomorphic to
.
Related facts
- Direct product is cancellative for finite groups: Essentially the same proof, tailored to groups (the key specific detail for groups is that the congruences can be described using the normal subgroups that function as their kernels).
Facts used
- Homomorphism set to direct product is Cartesian product of homomorphism sets: If
are algebras, then there is a natural bijection:
-
.
- The bijection is defined as:
.
-
- Homomorphism set is disjoint union of injective homomorphism sets: For algebras
and
, let
denotes the set of homomorphisms from
to
, and
denote the set of injective homomorphisms from
to
. Then we have:
.
Here varies over the set of all possible congruences on the algebra
.
Proof
Given: Finite algebras such that
.
To prove: .
Proof: Let be an arbitrary finite algebra in
. Note that the trivial homomorphism refers to the map that sends every element to the zero element.
Step no. | Assertion/construction | Facts used | Given data used | Previous steps used | Explanation |
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1 | ![]() |
Fact (1) | ![]() |
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2 | ![]() |
Fact (1) | ![]() |
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3 | ![]() |
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The number of homomorphisms to an algebra depends only on its isomorphism type. | ||
4 | ![]() |
Steps (1),(2),(3) | [SHOW MORE] | ||
5 | For any finite algebra ![]() ![]() ![]() ![]() ![]() ![]() ![]() |
Fact (2) | Step (4) | [SHOW MORE] | |
6 | ![]() ![]() ![]() ![]() |
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Step (5) | [SHOW MORE] | |
7 | ![]() ![]() |
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Step (6) | [SHOW MORE] |