Ring Homomorphism: unit in R implies unit in R'

AcidRainLiTE · Oct 15, 2012

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 · Oct 15, 2012

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 · Oct 15, 2012

AcidRainLiTE said: ↑

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 [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.

AcidRainLiTE · Oct 15, 2012

Thanks for the quick response!

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Ring Homomorphism: unit in R implies unit in R'

AcidRainLiTE

jgens

jbunniii

AcidRainLiTE