# on the definition of an algebra.

A ring $A$ is a $\mathbb{K}$-algebra if there is a ring homomorphism $u:\mathbb{K} \rightarrow A$ such that $u(\mathbb{K}) \subseteq Z(A)$ and $u(1_{\mathbb{K}}) = 1_A$.

1. I said that the above hypothesis imply that $u$ is injective, so $\mathbb{K} \cong u(\mathbb{K}) \subseteq Z(A)$. Why is that?

2. Someone asked me if  the condition $u(\mathbb{K}) \subseteq Z(A)$ is really necessary. Can you think of an example of $\mathbb{K}$ and $A$ which satisfy the other conditions, but not that one?

3. Someone also asked me if  the condition $u(1_{\mathbb{K}}) = 1_A$ is really necessary. Can you think of an example of $\mathbb{K}$ and $A$ which satisfy the other conditions, but not that one?

1. Duquec

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

Assume not, so there is a non-zero element $x\in\mathbb{K}$ such that $u(x) = 0$. Then, recalling that $\mathbb{K}$ is actually a field, we have

\vspace{3mm}

$u(x) = 0$

$u(x)u(x^{-1}) = 0u(x^{-1})$

$u(xx^{-1}) = 0$ (since $u$ is a homomorphism)

$u(1_\mathbb{K}) = 0$,

\vspace{3mm}

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

$$\frac{\mathbb{K}}{ker u} \cong u(\mathbb{K}) \Longrightarrow \mathbb{K} \cong u(\mathbb{K})$$.

2. Duquec

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

(1) Most trivial, case: Consider the trivial homomorphism from $Z_2$ to $\mathbb{Z}$.

(2) Let $\mathbb{K} = \mathbb{Q}$, $A = \mathbb{R}$ and $u : \mathbb{K} \longrightarrow A$ be defined by $u(x) = e^x$.

(3) Let $\mathbb{K} = \mathbb{Q}$, $A = Mat_{n\times n} (\mathbb{Q})$ and $u : \mathbb{K} \longrightarrow A$ be defined by $u(x) = X$, where $X$ is a matrix whose only non-zero entries are those on the diagonal (with the exemption of the leftmost one, which is also zero).

3. briansdumb

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 $u(1)=1$?

Does example number (3) require that $A$ consist only of diagonal matrices in order for $u(K) \subset Z(A)$?

• Duquec

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 $\mathbb{Z}$.

• Duquec

On the other hand, here is something.

Let $\mathbb{K}$ be a field, and $A$ be an integral domain. Let $u : \mathbb{K} \rightarrow A$ be a ring homomorphism. Since $A$ is an integral domain, we automatically have that $u(\mathbb{K}) \subseteq Z(A)$. However, $u$ must be the trivial homomorphism if $u(1_{\mathbb{K}}) \neq 1_A$: Let $x\in\mathbb{K}$ be given. Then

$u(x) = u(1_{\mathbb{K}}x) = u(x)$

$u(1_{\mathbb{K}})u(x) = u(x)$

$(u(1_{\mathbb{K}}) - 1_A)u(x) = 0.$

Since $A$ is an integral domain, $u(x) = 0.$

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