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When I was reading this post, it was mentioned that a field $k$ is an initial object in the category of $k$-algebras. But if I understand things correctly, this seems to rely on some (rather reasonable) convention.

A $k$-algebra is simply a ring morphism with domain $k$, so saying that $k$ is an $k$-algebra amounts to an endomorphism $k \to k$ (aka the structure map). But it seems to me that in the category of $k$-algebras (i.e. the coslice category $k/\mathbf{Ring}$), $k$ need not be initial if the $k$-algebra structure map on $k$ is not an automorphism. So when regarding $k$ as an $k$-algebra without any mention of the algebra structure, is it a common convention to take the identity map for the structure map?

The question might seem trivial to someone, but I'm still an absolute beginner in $R$-algebra.

Ray
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  • Since you would like a map $k\to k$ to be a $k$-algebra map, it is determined by the image of 1, which has to be 1, which forces the $k$-agebra map $k\to k$ to be the identity. – Evans Gambit Jul 26 '21 at 07:44

2 Answers2

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A $k$-algebra is a ring morphism $f\colon k\to A$, where the image of $f$ is contained in the center of $A$.

Then the identity morphism $\iota\colon k\to k$ is an initial object in the category of $k$-algebras. Not $k$: your definition requires specifying a ring morphism!

What's a morphism $\varphi\colon(k\xrightarrow{f}A)\to(k\xrightarrow{g}B)$?

You need a ring morphism $\varphi\colon A\to B$ such that $g=\varphi f$. Now, suppose you have $\varphi\colon(k\xrightarrow{\iota}k)\to(k\xrightarrow{f}A)$: this requires $f=\phi\iota=\phi$, so there is a unique $k$-algebra morphism from $k\xrightarrow{\iota}k$ to any $k$-algebra.

egreg
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  • That's correct. That's the point for the coslice category. Saying $k$ is an $k$-algebra means there is a underlying structure map (ring map) $k \to k$. The point of my post is that people just omit the structure map when declaring $k$ as an $k$-algebra (probably because they use the module definition), but in reality $k$ can be an $k$-algebra in some nontrivial way (via non-identity structure map). If the structure map is an automorphism, then $k$ will be an initial object because it is invertible. If the structure map is not nvertible, then there might be no $k$-alg homo $k \to A$ at all. – Ray Jul 26 '21 at 10:30
  • @Ray Yes, there can be proper embeddings $k\to k$, but the initial object has to be declared as an object of the category. Since the identity is “obvious”, it is commonly omitted. – egreg Jul 26 '21 at 10:36
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You are correct that this is potentially ambiguous and your guess is correct: If nothing else is stated, then one understands $k$ as a $k$-algebra via the identity map $\mathrm{id}_k$. As you have realized in your categorical argument, this is (up the unique isomorphism) the only natural choice to make it $k$ into the initial object in $k/\mathbf{Ring}$.

Qi Zhu
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