Consider the category of $k$-algebras with $k$-algebra homomorphisms. It's clear that the field $k$ itself is the initial object, since any $k$-algebra morphism must fix $k$.

Do terminal objects exist in this category? I looked all over and couldn't find mention of one. Any terminal object would have to contain all of $k$, but I feel there should be other $k$-algebras out there which have some independent generators over $k$, and then by choosing where to send the generators, you could construct multiple $k$-algebra homomorphisms into any other $k$-algebra, so a terminal object can't exist.

Am I over looking something?


1 Answer 1


The ring $\{ 0 \}$ admits a unique $k$-algebra structure, and it is easy to see that there is a unique $k$-algebra homomorphism $A \to \{ 0 \}$ for any $k$-algebra $A$.

(If your definition of $k$-algebra does not allow for $\{ 0 \}$, then it is wrong.)

  • 5
    $\begingroup$ +100 for "then it is wrong" since still many authors use this "definition". Have a look at Banach algebra texts, which ones require $\lVert 1 \rVert \leq 1$ (which is correct and convenient)? And which ones require $\lVert 1 \rVert = 1 $ (which is not correct and only convenient as long as you do not look at the details of certain constructions)? $\endgroup$ Apr 8, 2015 at 1:18

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .