Our second definition of an algebra was this one:

A ring is a -algebra if there is a ring homomorphism such that and .

1. I said that the above hypothesis imply that is injective, so . Why is that?

2. Someone asked me if the condition is really necessary. Can you think of an example of and which satisfy the other conditions, but not that one?

3. Someone also asked me if the condition is really necessary. Can you think of an example of and which satisfy the other conditions, but not that one?

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With regards to 1, I feel this might be an argument:

Assume not, so there is a non-zero element such that . Then, recalling that is actually a field, we have

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(since $u$ is a homomorphism)

,

a contradiction.

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So it follows that $u$ must be injective. By the First Isomorphism Theorem of Rings, we further have that

$$.

With regards to 3, there are several examples one can cook up:

(1) Most trivial, case: Consider the trivial homomorphism from to .

(2) Let , and be defined by .

(3) Let , and be defined by , where is a matrix whose only non-zero entries are those on the diagonal (with the exemption of the leftmost one, which is also zero).

I’m not sure I understand your example (1) correctly. Are you mapping the field {0,1} to the integers, and your homomorphism doesn’t have ?

Does example number (3) require that consist only of diagonal matrices in order for ?

Again, my second post is nonsense. I don’t know what kind of tea I was drinking last night.

What the homomorphism in (1) does is that it sends everything to 0, the additive identity in .

On the other hand, here is something.

Let be a field, and be an integral domain. Let be a ring homomorphism. Since $A$ is an integral domain, we automatically have that . However, must be the trivial homomorphism if : Let $x\in\mathbb{K}$ be given. Then

Since is an integral domain,

However, I am unable to pin down a particular example using this idea. Does someone have a specific example using the above?