on the definition of an algebra.

Our second definition of an algebra was this one:

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?

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5 comments

  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,

    a contradiction.

    \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?

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