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 Apr21 comment If $A$ is a submodule of $B$ and $B$ is a submodule of $C$, is $A$ a submodule of $C$? OK, so the problem seems to be that the hypotheses are not clearly stated. As it says "Let C be a commutative ring (with 1, if this matters)." I assumed that all the modules are unitary C-modules. If they are not then of course many things can happen Apr21 comment If $A$ is a submodule of $B$ and $B$ is a submodule of $C$, is $A$ a submodule of $C$? @ClémentGuérin: No! $B$ is an ideal of the ring $C$, so submodule of $B$ , for me, means $C$-submodule of $B$ (as $C$ is the unique ring I can see around), but then $c.A\subseteq A$ for all $c\in C$. Apr21 comment If $A$ is a submodule of $B$ and $B$ is a submodule of $C$, is $A$ a submodule of $C$? I do not see your problem, a "submodule" is a subgroup such that the restriction of the action of the ring on the big thing gives it the structure of a module. In particular, any subgroup of a subgroup is a subgroup and restricting the action of C on B and then on A is the same as restricting directly to A... Apr17 answered Definition of multiplicity Apr4 comment Definition of multiplicity See the limit formula of Lech for multiplicities Jan29 comment Verifying a Vector Space Via Given Axioms The proof of 3) is the following: we have to show that there exists a neutral element for the sum in our candidate vector space. We affirm that $\{z_n\}$, with $z_n=0$ for all $n$, is our neutral element. We have already seen that addition is commutative so it is enough to verify that $\{\alpha_n\}+\{z_n\}=\{\alpha_n\}$ for all $\{\alpha_n\}$ (otherwise one should also check that $\{z_n\}+\{\alpha_n\}=\{\alpha_n\}$). So, $\{\alpha_n\} + \{z_n\} = \{\alpha_n + z_n\} = \{\alpha_n\}$, since $z_n=0$ is the neutral element for the addition in $\mathbb K$. Jan29 comment Verifying a Vector Space Via Given Axioms The proof of 2) goes like this: $\{\alpha_n\} + \{\beta_n\}=\{\alpha_n+\beta_n\}\overset{(*)}{=}\{\beta_n+\alpha_n\} = \{\beta_n\} + \{\alpha_n\}$. Where $(*)$ holds by the commutativity of addition in $\mathbb K$. Jan28 comment Verifying a Vector Space Via Given Axioms For property 7: I do not see the utility of writing $(\lambda)(\mu)$, instead of $\lambda\mu$, they mean exactly the same. What you should use here is the associativity of multiplication in the base field, do you see how? Property 8 seems OK. Jan28 comment Verifying a Vector Space Via Given Axioms (3) Your third property is the existence of zero element... from your proof it is not clear who is the zero... of course it is the sequence all of whose elements are zero, it is not clear from what you wrote. your properties 4, 5 and 6 seem OK. Jan28 comment Verifying a Vector Space Via Given Axioms Yes, there are possible improvements. Let me list some: (1) In property 1 you are actually using, implicitly, the associativity of addition in your base field, maybe you should make it explicit using parenthesis, (2) the second property does not hold for the reason you wrote, but it holds for the commutativity of addition in your base field. In fact, it is actually false that $\{\alpha_n\}+\{\beta_n\}=0$ for all $\{\alpha_n\}\,,\{\beta_n\}$... Jan28 revised Verifying a Vector Space Via Given Axioms added 333 characters in body Jan28 answered Verifying a Vector Space Via Given Axioms Jan23 answered group homomorphisms from the real line to infinite torsion abelian groups Dec16 awarded Caucus Dec9 accepted Complete but not cocomplete category Nov15 awarded Yearling Oct23 comment Complete but not cocomplete category The question was answered on mathoverflow, giving an example of a cocomplete but not complete Abelian category. Sep11 comment Uniqueness Of Linear Transformation I can probably try to interpret your question... but I think you should try to formulate it in a precise way yourself... so... how do you define $T$? I deduce it is not an arbitrary morphism otherwise you would not ask if it is 1-1 (it could be the $0$ map for example)... Sep1 comment Is every subspace of a normed linear space which is not closed a hyperspace. let $x_1$ and $x_2\in \mathfrak {X\setminus B}$. Then $x_1-x_2\notin B$ but clearly $f(x_1-x_2)=1-1=0$... so the kernel of your map may be bigger than $B$. Where are you using that your subspaces is not closed? Aug28 comment Does every free $R$-module have a maximal proper submodule? Take the quotient $M/(\mathrm{Span}(\mathfrak{B}\setminus \{x\}))$, this will be isomorphic to $R$, while a submodule $N$ of $M$ is maximal if and only if $M/N$ is simple. Your idea works iff $R$ is simple as an $R$-module.