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This comes from an earlier question of mine that wasn't asked in the best way. Let $R$ be some ring (can be left and right Noetherian to make things easier), and let $M$ be a finitely generated $R$-module and $N\leq M$ an $R$-submodule of $M$. Now let $L$ be a non-zero cyclic submodule of the quotient $M/N$ generated by a single element $l\in M$ such that $l\not\in N$ and $rl\not\in N$ for any $r\in R$. Can I construct a surjective homomorphism $f:M/N\to L$? I was thinking of perhaps defining $f$ by mapping $l+N\mapsto l$. This would obviously be surjective, but does it need to be a homomorphism?

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  • $\begingroup$ "$rl \notin N$ for any $r \in R$" cannot hold ($r=0$). The answer assumes you actually want $L\cap N = \{0\}$, is that correct? $\endgroup$ – Torsten Schoeneberg Dec 24 '17 at 22:05
  • $\begingroup$ It would be helpful to give a link to that "earlier question" so that people can see what has already been discussed. $\endgroup$ – Torsten Schoeneberg Dec 24 '17 at 23:39
  • $\begingroup$ @TorstenSchoeneberg It seems to be the case in the earlier question. It does not make sense to say that $L\leq M/N$ is generated by an element of $M$. In that case, it would mean $L$ is the image of a cyclic submodule $Z$ of $M$ under the canonical projection $M\to M/N$ such that $Z$ contains $N$. If $Z$ is generated by $l$, then $Z\geq N$ implies that $rl\in N$ for some $r$. In particular, $r=0$ works. $\endgroup$ – Batominovski Dec 25 '17 at 2:13
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This may be an overkilling answer. However, I only have a counterexample from representation theory at this moment. It would be an interesting question if $M$ is also required to be indecomposable. In my counterexample, $M$ is decomposable.

Let $\mathfrak{g}$ be the special linear Lie algebra $\mathfrak{sl}_2(\mathbb{C})$. Write $h$ for $\begin{bmatrix}1&0\\0&-1\end{bmatrix}$. Set $\mathfrak{h}:=\mathbb{C}h$, which is a Cartan subalgebra of $\mathfrak{g}$. Take $\mathfrak{n}^+$ to be the subalgebra of $\mathfrak{g}$ consisting of strictly upper triangular matrices, and $\mathfrak{n}^-$ the subalgebra consisting of strictly lower triangular matrices. Denote by $R$ the universal enveloping algebra $\mathfrak{U}(\mathfrak{g})$ of $\mathfrak{g}$. Note that $R$ is Noetherian.

For each $\lambda\in\mathbb{C}$, the cyclic $R$-module $\mathfrak{M}(\lambda)$ is generated by $1_\lambda$ under the condition that $h\cdot 1_\lambda=\lambda\,1_\lambda$ and $\mathfrak{n}^+\cdot 1_\lambda=\{0\}$. The module $\mathfrak{M}(\lambda)$ is known as the Verma module over $\mathfrak{g}$ with highest weight $\lambda$ (with respect to the Borel subalgebra $\mathfrak{b}+\mathfrak{n}^+$).

Now, let $M:=\mathfrak{M}(0)\oplus\mathfrak{M}(0)$, $N:=\mathfrak{M}(0)\oplus\{0\}\leq M$, and $L:=\{0\}\oplus\mathfrak{M}(-2)\leq M$. Then, $M/N\cong\mathfrak{M}(0)$, which contains $\mathfrak{M}(-2)\cong L$ as a cyclic submodule. However, there does not exist a surjective $R$-module homomorphism $\mathfrak{M}(0)\to\mathfrak{M}(-2)$. Indeed, any $R$-module homomorphism $\mathfrak{M}(0)\to\mathfrak{M}(-2)$ is zero.


P.S. I found a counterexample in which $M$ is indecomposable. If $\mathfrak{g}:=\mathfrak{so}_3(\mathbb{C})$ instead, then one can pick $M$ to be the quotient by its unique simple submodule of a Verma module with regular integral dominant highest weight (with respect to a fixed Borel subalgebra of $\mathfrak{g}$). The module $M$ will be indecomposable with two trivially intersecting simple submodules $N$ and $L$. Again, the only homomorphism from $M/N$ to $L$ is zero.

Nonetheless, if $R$ is a principal ideal domain and $M$ is required to be a (finitely generated) unitary $R$-module, then the answer is positive. The condition that $N\cap L=\{0\}$ implies that there exist submodules $X$ and $Y$ of $M$ such that $M=X\oplus Y$ with $N\leq X$ and $L\leq Y$. It can be easily proven that $L$ is also isomorphic to a quotient of $Y$. Thus, there exists a surjective $R$-module homomorphism $M/N\to L$, which can be defined by composing a surjective homomorphism $M/N\to M/X\cong Y$ with a surjective homomorphism $Y\to L$.

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  • $\begingroup$ Why can't you just say $M = \mathfrak{M}(0), N= 0, L = \mathfrak{M}(-2)$ in your first (counter)example? And more generally, if we're looking for counterexamples, can we not wlog assume $N=0$? $\endgroup$ – Torsten Schoeneberg Dec 24 '17 at 23:37
  • $\begingroup$ I was assuming that the OP does not want an example with zero submodule. But yes, what you suggest works. $\endgroup$ – Batominovski Dec 25 '17 at 2:07
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Positive results:

  • The assertion is true, almost by definition, if $M/N$ is semisimple. In particular, for all modules $N\subseteq M$ over a semisimple ring $R$.

  • As Batominovski points out, it is true for finitely generated $M$ over a principal ideal domain $R$. One can e.g. derive this from the fundamental theorem about such modules.

  • I think from that one can conclude that it also holds if the ring $R$ is a quotient of a PID; and further, that might extend to any (commutative) principal ideal ring, but I am not totally sure about this.

However, as far as commutative domains go, PIDs are the only ones that work, because there is ...

A big class of counterexamples:

Let $R$ be a commutative domain. If $I$ is a non-principal ideal in $R$, and $l\in I$ is any non-zero element, $L$ the cyclic submodule generated by $l$, then there is no surjective homomorphism $I \rightarrow L$. So, for finitely generated $I$, one can take $M = I, N= 0$ as counterexample. If you want it more concrete, $R = k[x,y], M=I = (x,y)$, and e.g. $l = x$. If you want $N$ to be-non trivial, take whatever $N$ you like and set $M = I \oplus N$ and $0 \neq l \in I \oplus 0$.)

To see this, note first that since $R$ is a domain, $L \simeq R$ as $R$-module. On the other hand, we have

Lemma: Let $R$ be a commutative domain, $K$ its field of fractions, $I \subseteq R$ any ideal. Every $R$-module homomorphism $I\rightarrow R$ is given by multiplication with some $x_{\phi} \in K$. In particular, every non-zero homomorphism $I \rightarrow R$ is injective.
Proof: Let $\phi \in Hom_R(I,R)$ and assume there are $a,b \in I \setminus \{0\}$. Then $$ab-ba=0 \implies a\phi(b)-b\phi(a)= 0 \implies \frac{\phi(a)}{a} = \frac{\phi(b)}{b} =: x_{\phi} \in K.$$

Corollary: If $I$ is not a principal ideal, there is no surjective $\phi:I \rightarrow R$. For if there were, it were bijective, and hence $I \simeq R$ would be a principal ideal.

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