Ring Homomorphism: unit in R implies unit in R'

[AcidRainLiTE]

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]

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]

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]

Thanks for the quick response!

 

[blue_raver22]

I can understand your confusion. The definition of a ring homomorphism does not explicitly state that the unit element must be preserved. However, this is actually implied by the definition itself.

A ring homomorphism is a function that preserves the structure of a ring, meaning that it preserves the operations of addition and multiplication. This means that the unit element in the original ring R must be mapped to the unit element in the target ring R'.

If a is a unit in R, then there exists an element a^-1 in R such that aa^-1 = 1. Since the homomorphism preserves multiplication, we can apply it to both sides of this equation, giving f(aa^-1) = f(1). By the definition of a homomorphism, this is equal to f(a)f(a^-1) = 1'. This shows that f(a) is a unit in R'.

Therefore, the statement in the 'Properties' section is true because it is implied by the definition of a ring homomorphism. It is important to note that while it is not explicitly stated, it is still a necessary requirement for a homomorphism to preserve the unit element.