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?

  • $\begingroup$ "$rl \notin N$ 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

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 $Y\to L$.

  • $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.