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July 30

group isomorphism

I'm a bit uncertain about the definition for group isomorphism. The article says "a bijective group homomorphism". I think this nullifies statements (and their proofs) such as "a group homomorphism is an isomorphism iff it is bijective", and therefore should be changed to something that requires the existence of a sibling homomorphism from the codomain group to the domain one. Unfortunately, i've forgotten the details, but at least that's what i remember from my Algebra 101 class of 9 years ago. -- Jokes Free4Me (talk) 12:28, 30 July 2009 (UTC)[reply]

You are right; according to the general definition in category theory, a homomorphism h:G→G' is an isomorphism if and only if there is a homomorphism k:G'→G such that hk=1_G' and kh=1_G. But the equivalence is quite immediate and the various definitions live together... --pma (talk) 12:57, 30 July 2009 (UTC)[reply]

Thanks. I was actually thinking about a general isomorphism, and in that base being bijective is not enough. The inverse must also be a morphism. -- Jokes Free4Me (talk) 13:40, 30 July 2009 (UTC)[reply]

Indeed, bijectivity is a useful criterion for isomorphisms in special categories; in general, morphisms need not even be functions, in which case bijectivity for them is just meaningless. --pma (talk) 13:58, 30 July 2009 (UTC)[reply]

I've heard people distinguish between an "isomorphism into" another group and an "isomorphism onto" another group. With the former, the "other group" is not isomorphic to the domain, but of course it has a subgroup that is. Michael Hardy (talk) 16:44, 30 July 2009 (UTC)[reply]

Wouldn't an "isomorphism into" a group just be a monomorphism (i.e. an injective homomorphism)? Maelin (Talk | Contribs) 04:16, 2 August 2009 (UTC)[reply]

irreducibility

Eisenstein's criterion mentions that "Then f(x) is irreducible over F[x], where F is the field of fractions of D. When f(x) is primitive [...], it is also irreducible over D[x]." I believe Irred(F[x]) implies Irred(D[x]), since if something is reducible over D[x] it must clearly be immediately reducible over F[x], using the same decomposition. -- Jokes Free4Me (talk) 13:40, 30 July 2009 (UTC)[reply]

You are wrong. The polynomial 2x is reducible over the integers but irreducible over the rationals. Algebraist 13:49, 30 July 2009 (UTC)[reply]

Oh, right. But then, irreducible polynomial is a bit incomplete and misleading: it only defines irreducibility over fields, and then looks at the ring Z. Maybe such an example as 2X could be added to it, after expanding the definition? Thanks. -- Jokes Free4Me (talk) 14:16, 30 July 2009 (UTC)[reply]

The more general definition is given further down the page, at irreducible polynomial#Generalization. That is a rather silly order for things to be in. Algebraist 14:19, 30 July 2009 (UTC)[reply]

PS Speaking of the incompleteness of that article, am i right in assuming that 6=2*3 is reducible in Z[X]? -- Jokes Free4Me (talk) 14:16, 30 July 2009 (UTC)[reply]

Yes. Algebraist 14:19, 30 July 2009 (UTC)[reply]