# Ring Homomorphism: unit in R implies unit in R'

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:

If a has a multiplicative inverse in R, then f(a) has a multiplicative inverse in S and we have f(a−1) = (f(a))−1. Therefore, f induces a group homomorphism from the (multiplicative) group of units of R to the (multiplicative) group of units of S.
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?

jgens
Gold Member
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).

jbunniii
Homework Helper
Gold Member
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?
It need not be true if the homomorphism does not map 1 to 1'.

For example, consider the map $f : \mathbb{Z} \rightarrow M_{2 \times 2}(\mathbb{R})$ defined by
$$f(n) = \left[\begin{matrix} n & 0 \\ 0 & 0 \end{matrix}\right]$$
This is a ring homomorphism that does not map 1 to 1', and clearly the image contains no units.

Thanks for the quick response!