Talk:Algebra homomorphism

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 Field: Algebra

My understanding of homomorphism leads me to believe that the first two conditions given in the first sections follow from the nature of algebras not from the nature of homomorphism and that the third condition is the rule that makes a homomorphism what it is. Input appreciated--Cronholm144 04:55, 14 May 2007 (UTC)

In general a homomorphism preserves all the structure of the underlying object. With an algebra a homomorphism must preserve addition and multiplication (the two binary operations of the underlying set) and scalar multiplication. Contrast this to a group which only needs preserve multiplication to be a homomorphism (technically you need the identity to map to the identity and inverse to map to inverses, but this all follows from the homomorphism respecting multiplcation). TooMuchMath 05:20, 15 May 2007 (UTC)

got it, thanks... sorry for sending you down a false trail--Cronholm144 05:41, 15 May 2007 (UTC)

Injectivity of a k-algebra homomorphism isn't enough for an isomorphism. Let k be a field and K an extension field such that k\neq K. Then viewing k and K as k-algebras, the inclusion F:k\hookrightarrow K is an injective k-algebra homomorphism which isn't an isomorphism. Steve Checkoway 04:55, 12 November 2007 (UTC)

This is clear, and your edit definitely corrected a mistake. Had they left out injectivity and required surjectivity, I would assume they were talking about simple algebras. At any rate, since they weren't mentioned in the article, I linked them in with a discussion of inner automorphisms. JackSchmidt 05:06, 12 November 2007 (UTC)

Example maybe[edit]

I am not expert on the subject, I am trying to understand what is morphism and homomorphism

a = b * c

d = e + f

I think multiplication and addition are homomorphic. Applying what is written in Group homomorphism I can think of morphism h:

h(u * v) = h(u) + h(v)

This is a kind of commutative diagram:

6  = 2 * 3
     |
     |  h
     |
     V
12 = 4 + 8

(actually you could use any numbers that are valid operations)

h(2 * 3) = h(4) + h(8)

h(2 * 3) is h(6), and image of h(6) is 12

image of h(2) is 4,

image of h(3) is 8

so the h(2) + h(3) results in 4 + 8, and that is 12

I could upload an image that explains the Idea much better. I know that Wikipedia is not a place to study mathematics together with others, but I will upload image(s) if no one objects. The image is here --Pasixxxx (talk) 16:40, 13 October 2012 (UTC)