Is the decomposition of an algebra into irreducible components essentially unique? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T19:49:41Z http://mathoverflow.net/feeds/question/45668 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/45668/is-the-decomposition-of-an-algebra-into-irreducible-components-essentially-unique Is the decomposition of an algebra into irreducible components essentially unique? Andrej Bauer 2010-11-11T08:59:13Z 2010-12-12T00:42:27Z <p>We consider <em>finite</em> algebras for a given signature, in the sense of universal algebra (for example, they might be groups, rings, or lattices). An algebra $A$ is <em>irreducible</em> when $A \cong B \times C$ implies that $B$ or $C$ is the one-point algebra.</p> <p>Is it the case that a $\Sigma$-algebra can be expressed as a cartesian product of irreducible algebras in an essentially unique way, i.e., unique up to permutation of factors? I suspect that this is either a theorem with somebody's name attached to it, or there is a counterexample in groups.</p> http://mathoverflow.net/questions/45668/is-the-decomposition-of-an-algebra-into-irreducible-components-essentially-unique/46302#46302 Answer by Steve Lack for Is the decomposition of an algebra into irreducible components essentially unique? Steve Lack 2010-11-17T00:06:18Z 2010-11-17T00:06:18Z <p>This is not true as stated. If you take the empty signature, or any signature with no constants, then the empty set is an algebra, and this will mess things up.</p> <p>This issue was raised in the paper </p> <blockquote> <p>M. Barr, The point of the empty set, Cahiers 13:1-12, 1972.</p> </blockquote> <p>I think that this is all that can go wrong. If you restrict to non-empty signatures, or to non-empty algebras then things are ok; but if you are also considering multisorted theories, then you need to make sure that each sort is non-empty. This issues was also discussed in the Barr paper. See also </p> <blockquote> <p>G.M. Kelly and A. Pultr, On algebraic recognition of direct-product decompositions, Journal of Pure and Applied Algebra 12:207--224, 1978.</p> </blockquote> http://mathoverflow.net/questions/45668/is-the-decomposition-of-an-algebra-into-irreducible-components-essentially-unique/46310#46310 Answer by Gerald Edgar for Is the decomposition of an algebra into irreducible components essentially unique? Gerald Edgar 2010-11-17T01:00:48Z 2010-11-17T01:09:13Z <p>Let $A, B$ be the algebras with two elements $0,1$ under addition mod $2$ and unary operation $x'=x$ in $A$ and $x'=1-x$ in $B$. Then $A \times B \cong B \times B$, though $A$ and $B$ are not isomorphic.<br> B. Jonsson </p> <p>Construct a 12-element commutative semigroup which does not have the unique factorization property.<br> R. McKinsey </p> <p>(exercises on p. 170 of Birkhoff, <em>Lattice Theory</em> (3rd edition)) </p>