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Ring Homomorphism: unit in R implies unit in R'

  1. Oct 15, 2012 #1
    I was just looking at wikipedia's article on ring homomorphisms (http://en.wikipedia.org/wiki/Ring_homomorphism) and I am a little confused.

    If you look at the definition they give for ring homomorphism, they require only that addition and multiplication is preserved over the homomorphism (and not that it maps 1 to 1'). Then, under the 'Properties' section, the third bullet down claims:

    Don't you need to explicitly require that the homomorphism maps 1 to 1' in order for that statement to be true? Or is there someway to deduce this without specifying that requirement?
     
  2. jcsd
  3. Oct 15, 2012 #2

    jgens

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    Not all rings are unitary, so the definition of ring homomorphism given on wikipedia is for a general ring. When a ring is unitary, most people require that a homomorphism map 1 to 1, but occasionally you will find authors who do not use this convention (in this case, 1 just maps to some unit in the ring).
     
  4. Oct 15, 2012 #3

    jbunniii

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    It need not be true if the homomorphism does not map 1 to 1'.

    For example, consider the map [itex]f : \mathbb{Z} \rightarrow M_{2 \times 2}(\mathbb{R})[/itex] defined by
    [tex]f(n) = \left[\begin{matrix}
    n & 0 \\
    0 & 0
    \end{matrix}\right][/tex]
    This is a ring homomorphism that does not map 1 to 1', and clearly the image contains no units.
     
  5. Oct 15, 2012 #4
    Thanks for the quick response!
     
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