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Apr
21
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
Apr
21
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 $c\in C$.
Apr
21
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...
Apr
17
answered Definition of multiplicity
Apr
4
comment Definition of multiplicity
See the limit formula of Lech for multiplicities
Jan
29
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 $\mathbb K$.
Jan
29
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 $\mathbb K$.
Jan
28
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.
Jan
28
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.
Jan
28
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 $\{\alpha_n\}\,,\{\beta_n\}$...
Jan
28
revised Verifying a Vector Space Via Given Axioms
added 333 characters in body
Jan
28
answered Verifying a Vector Space Via Given Axioms
Jan
23
answered group homomorphisms from the real line to infinite torsion abelian groups
Dec
16
awarded  Caucus
Dec
9
accepted Complete but not cocomplete category
Nov
15
awarded  Yearling
Oct
23
comment Complete but not cocomplete category
The question was answered on mathoverflow, giving an example of a cocomplete but not complete Abelian category.
Sep
11
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)...
Sep
1
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?
Aug
28
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.