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I apologize if this question is trivial, but I just cant figure it out. Let $K$ be a field and let $K\longrightarrow A$ be an epimorphism of rings. Is it necessary that $A=K$?

stupidq75
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    Yes, it is necessary, unless $A=0$. Not appropriate for this site. – Alex Degtyarev Apr 12 '15 at 17:28
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    The typical way to approach this is to consider the pair of maps $i_1, i_2: A \to A \otimes_K A$ where $i_1(a) = a \otimes 1$ and $i_2(a) = 1 \otimes a$, show their restrictions along $K \to A$ agree, and show $i_1, i_2$ disagree if the dimension of $A$ as a vector space over $K$ is greater than 1. – Todd Trimble Apr 12 '15 at 17:31
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    If this is so trivial it ought to have a rigorous proof. Why shouldn't the dimension of $A$ over $K$ be infinite? – Paul Taylor Apr 12 '15 at 18:25
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    I guess that you mean "epimorphism" in the categorical sense (some people use it for "surjective homomorphism", which is stronger since the ring homomorphism $\mathbf{Z}\to\mathbf{Q}$ is a non-surjective epimorphism). An ambiguity is on what you call "ring": associative? commutative? there are people on mathOverflow using various conventions. – YCor Apr 12 '15 at 20:50

2 Answers2

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Assume $A$ is not just the zero ring (consisting of just one element). Via the given map $i: K \to A$, we may regard $A$ as a $K$-module or in other words a vector space over $K$. Now let us consider the two ring maps $i_1, i_2: A \to A \otimes_K A$ where $i_1(a) = a \otimes 1$ and $i_2(a) = 1 \otimes a$. Observe that for any $k \in K$, we have

$$(i_1 \circ i)(k) = i(k) \otimes 1 = (1 \cdot k) \otimes 1 = 1 \otimes (k \cdot 1) = 1 \otimes i(k) = (i_2 \circ i)(k)$$

so $i_1 \circ i = i_2 \circ i$, whence $i_1 = i_2$ by epimorphicity of $i$. But now $i_1 = i_2$ forces $\dim_K A = 1$. Indeed, if $\dim_K A > 1$, we may pick an element $a$ that is linearly independent of the identity $1$. Then we can restrict $i_1, i_2$ to the subspace $V$ spanned by basis elements $e_1 = a, e_2 = 1$, and we just observe that $i_1(a) = e_1 \otimes e_2 \in V \otimes V$, $i_2(a) = e_2 \otimes e_1$, which are clearly linearly independent ($e_i \otimes e_j$ are basis elements of $V \otimes V$). Thus $i_1, i_2$ disagree at $a$, contradicting $i_1 = i_2$.

Todd Trimble
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  • So it still requires axiom of choice? – Fan Zheng Apr 12 '15 at 19:20
  • @FanZheng Presumably not; I'm using just basic notions of independence etc. and cutting down to a finite-dimensional subspace $V$ to finish the argument. I didn't need a basis for all of $A$. Can you say more explicitly what you have in mind? – Todd Trimble Apr 12 '15 at 19:23
  • Is it clear that without a basis of $A$ that the map $V\otimes_KV\to A\otimes_KA$ is injective, which you are implicitly using in the argument? – Emil Jeřábek Apr 12 '15 at 19:28
  • @Emil Yeah, I was just wondering that myself. Need that all modules over a field are flat for that part. I'm not sure. – Todd Trimble Apr 12 '15 at 19:33
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    I convinced myself that it should work without choice. One can construct $A\otimes_KA$ as the quotient of the space with basis ${u\otimes v:u,v\in A}$ over a subspace generated by $uk\otimes v-u\otimes kv$ and friends. If this subspace contained $i_1(a)-i_2(a)$, this would be witnessed by a finite linear combination involving only finitely many elements of $A$, hence it would already show up for a finite-dimensional subspace of $A$, where we can do the algebra as usual. – Emil Jeřábek Apr 12 '15 at 20:09
  • @EmilJeřábek Yep! – Todd Trimble Apr 12 '15 at 20:13
  • @Todd Trimble: Over a field each module is flat. For, a module $A$ is flat, iff $Tor_1(A,-)=0$. But $A$ is the direct limit of finitely generated modules $A_i$ and $Tor$ commutes with direct limits. But f.g. modules over a field a free and thus flat. Hence $Tor_1(A_i,-)=0$ showing $Tor_1(A,-)=0$. – tj_ Apr 12 '15 at 20:27
  • @tj_ I expect what you mean is that $Tor$ commutes with filtered or directed colimits, not to be confused with direct limits = colimits. But your argument still goes through. And perhaps more to the point, I take it that you meant to offer a choice-free argument (I already knew modules over fields are flat assuming choice), and the argument seems valid in that respect as well. Thanks! – Todd Trimble Apr 12 '15 at 20:31
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This is even true for $K$ a division ring. See my question and the answers here: Are epimorphisms from a division ring isomorphisms ?

Added: That epimorphisms from a field (or a division ring) are isomorphisms doesn't require the Axiom of Choice.

For, as noted in the comment to the accepted answer in the link, $\alpha: R \to S$ is epi iff $S \otimes_R S/\alpha(R)=0$. If $R$ is a division ring, the epi $\alpha$ is injective (due to the lack of non-trivial ideals). Since every module over a division ring is flat (this doesn't need AC!), tensoring the exact sequence $0 \to R \xrightarrow[]{\alpha} S$ with $S/\alpha(R)$ yields the exact sequence $0 \to S/\alpha(R)\to S \otimes_R S/\alpha(R)=0$. Thus $S/\alpha(R)=0$, i.e. $\alpha(R)=S$.

Community
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tj_
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  • Which does support considering this question as a duplicate (subsumed under a more general question). – Todd Trimble Apr 12 '15 at 19:26
  • Right! But the discussion on the need of AC is a new aspect of the question. – tj_ Apr 12 '15 at 19:38
  • Technically that's a separate question. But I'm not really fussed about closing this question (I can imagine that the software might not have revealed your question as a possible duplicate when the question was being compiled, in which case I wouldn't blame the OP). – Todd Trimble Apr 12 '15 at 19:42