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Let $A$ and $B$ be commutative rings, and suppose (the underlying group of) $B$ has a structure of $A$-module. "Obviously", that doesn't imply that $B$ gets a structure as an $A$-algebra, but I can't come up with a simple example.

Surely there must be some relatively down-to-earth example, though. Can you help me?

user26857
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Ketil Tveiten
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    $A=\mathbb{Z}[x]$, with $x$ acting on $B$ by any additive non-homomorphism? –  Sep 23 '14 at 15:56
  • @PiotrAchinger you mean abelian group homomorphism but not right $B$-module homomorphism, don't you? – Fernando Muro Sep 23 '14 at 15:57
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    Take $B$ to be polynomial ring $k[t]$. Take $A$ to be the endomorphism ring of $B$ as a $k$-vector space. Surely $B$ is an $A$ module, but not an algebra (where would you send the derivative map?). –  Sep 23 '14 at 15:58
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    I'm not sure why you say $\mathbb{R}$ is a $\mathbb{C}$-vector space but not a $\mathbb{C}$-algebra; any embedding of $\mathbb{C}$ in $\mathbb{R}$ makes $\mathbb{R}$ a $\mathbb{C}$-algebra. In your situation, $B$ will be an $A$-algebra iff the action of $A$ on $B$ commutes with the action of $B$ on itself. Equivalently, $B$ is an $A$-algebra iff the action of $a\in A$ coincides with multiplication by $a\cdot 1_B$ in the ring structure of $B$. – Eric Wofsey Sep 23 '14 at 16:05
  • Uh, yes, that's true. Edited to remove that non-example. – Ketil Tveiten Sep 24 '14 at 08:07
  • Also, the endomorphism ring of a vector space isn't commutative, so that's also not an example. – Ketil Tveiten Sep 24 '14 at 08:09
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    @EricWofsey, there are no $\mathbb C$-algebra structures on $\mathbb R$: such a thing is a map of rings $\mathbb C\to\mathbb R$, which would have to map $i$ to something squaring to $-1$. – Mariano Suárez-Álvarez Sep 24 '14 at 08:18
  • @MarianoSuárez-Alvarez: Oh, oops, of course, I wasn't thinking straight in that first sentence. – Eric Wofsey Sep 24 '14 at 22:33
  • @EricWofsey I think you are right, though not in the way you wanted to be: The isomorphism of the additive groups of $\mathbb{C}$ and $\mathbb{R}$ carries along the field structure, but the resulting ring structure is not the usual ring structure on $\mathbb{R}$. – Ketil Tveiten Sep 25 '14 at 08:27

2 Answers2

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Let $B=\def\QQ{\mathbb Q}\QQ(t)$, which is among other things a rational vector space of countable dimension. In particular, it has a structure of a $\QQ(t_1,t_2)$-module of rank one. But $B$ is not an $\QQ(t_1,t_2)$-algebra in any way, as there are no ring maps $\QQ(t_1,t_2)\to B$.

Mariano Suárez-Álvarez
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    @user26857, I have no idea. All I know is that $\mathbb Q(t_1,t_2)$ has countable dimension as a $\mathbb Q$-vector space, so that there exists an isomorphism $\mathbb Q(t_1,t_2)\to B$ of $\mathbb Q$-vector spaces, and then there is a unique way of turning $B$ into a $\mathbb Q(t_1,t_2)$-module so that that map is an isomorphism of $\mathbb Q(t_1,t_2)$-modules. – Mariano Suárez-Álvarez Jun 30 '15 at 16:38
  • I am not being ironic. I don't know what the module structure is, as it depends on the choice of $\mathbb Q$-bases on $\mathbb Q(t_1,t_2)$ and on $\mathbb Q(t)$. My comment above is the explanation you wanted. – Mariano Suárez-Álvarez Jun 30 '15 at 16:44
  • WHen someone asks for extra details on my answer, I provide them. You asked for details and I provided them. The person who asked this question, on the other hand, did not, probably because he did not need them — he even accepted the answer, as you can see. – Mariano Suárez-Álvarez Jun 30 '15 at 16:48
  • I do not know what you are asking for, really. Are you asking how to turn $\mathbb Q(t)$ into a $\mathbb Q(t_1,t_2)$-modules so that a givem fixed isomorphism of $\mathbb Q$-vector spaces $\mathbb Q(t)\to\mathbb Q(t_1,t_2)$ becomes an isomorphism of $\mathbb Q(t_1,t_2)$-modules? If so, I suggest you ask a separate question. – Mariano Suárez-Álvarez Jun 30 '15 at 16:48
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    @user26857, I kindly request that if you intend to communicate with me you do not remove your comments after I have replied to them. – Mariano Suárez-Álvarez Jun 30 '15 at 16:55
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    One way to avoid that urge of yours to remove comments in which you attribute others intentions imagined by you is, simply, not to make them in the first place. – Mariano Suárez-Álvarez Jun 30 '15 at 17:03
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I came up with a counterexample myself: let $A=k[\partial_t]$, and $B=k[t]$, with $A$ acting by differentiation; clearly there is no ring map $\phi:A\to B$ compatible with that action.

Ketil Tveiten
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