Let $A$ be a nonzero commutative ring with $1$, and $n$ a positive integer. If $b_1,\ldots,b_n\in A^n$ generate $A^n$ as an $A$-module, is it true that $b_1,\ldots,b_n$ are linearly independent over $A$ (hence a basis)? I suspect this is true (It is true if $n=1$), but I can't seem to prove it.  I guess an equivalent way of asking this question is: if a linear map $A^n\to A^n$ is surjective, is it necessarily injective?  Any help is welcome!
123741
Let $L$ be a finitely generated free module over a commutative ring $A$. Set $n=\operatorname{rank} L$. Let $x_1,\dots,x_m$ be generators of $L$. Then $m \ge n$? If $m = n$, then is $x_1,\dots,x_m$ a basis of $L$?
22458
[Theorem] Let $V$ be an inner product space, and let $S$ be an orthogonal subset of $V$ consisting of nonzero vectors. Then $S$ is linearly independent.     Also, orthogonal set and linearly independent set both generate the same subspace.  (Is that right?)    Then orthogonal $\rightarrow$ linearly independent but orthogonal $\nleftarrow$ linearly independent is that right?    One more question. For T/F, Every orthogonal set is linearly independent (F) Every orthonormal set is linearly independent (T) Why?
17366
Let $R$ be a commutative ring, $P \in M_n(R)$ and $\det(P)$ is a zero divisor of $R$. Must $P$ be a zero divisor of $M_n(R)$?   Here rings mean unital rings, $M_n(R)$ denotes the ring of square matrices over $R$ of order $n$, and zero divisor is understood to be nonzero. The difficulty lies in that the adjugate matrix of $P$ may well be $0$.
20470
In the $n$-space $\Bbb R^n$ determine whether or not the set $\{e_1-e_2,e_2-e_3, \dots, e_{n}-e_{1} \}$ is Linearly Independent.  I tried solving it using the definition of a LI set, taking $e_1, e_2$ etc to be the standard basis of $\Bbb R^n$. Finally equating it to the zero of $\Bbb R^n$, I get all the scalars $a_1=a_2=\cdots=a_n$  So should I conclude that they are not necessarily zero and hence the given set is Linearly Dependent?
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The following is exercise number 10 in Hungerford's Algebra, page 190:  Let $R$ be a ring with no zero divisors such that for all $r, s \in R$ there exist $a, b \in R$, not both zero, such that $ar + bs = 0$.  a) If $R = K \oplus L$ (module direct sum) then $K = 0$ or $L = 0$  b) If $R$ has an identity, then $R$ has the invariant dimension property.  I have proved part a), but I'm having difficulty with b). I realize that it suffices to show that $R^m \cong R^n$ implies $m = n$, but I can't seem to use part a) to conclude this.
29051
Let $A$ be a commutative ring with $1$. Let $f_1,\dots,f_r$ be elements of $A$. Suppose $A = (f_1,\dots,f_r)$. Let $n &gt; 1$ be an integer. Can we prove that $A = (f_1^n,\dots,f_r^n)$ without using axiom of choice?  EDIT It is easy to prove $A = (f_1^n,\dots,f_r^n)$ if we use axiom of choice. Suppose $A≠(f_1^n,…,f_r^n)$. By Zorn's lemma(which is equivalent to axiom of choice), there exists a prime ideal $P$ such that $(f_1^n,…,f_r^n) \subset P$. Since $f_i^n \in P$, $f_i \in P$ for all $i$. Hence $A = (f_1, \dots,f_n) \subset P$. This is a contradiction.
105159
Prove that every vector space has a basis. I am going to use Zorn's lemma for this also here is a necessary definition regarding totally ordered subsets: one element will be contained in the other.   proof: A basis for $V$ is a maximally linearly independent set. Let $X$ be equal to all linearly independent subsets of $V$. On, $X$ define $A\leq B$ means $A\subset B$. Let $S$ be a totally ordered subset of $X$, let $T = \bigcup_{A_{s}\in S} A_s$ then $A_s\subset T$ for all $s\in S$. Does $T\in S$? i.e. is $T$ a linearly independent set?. Let $$O = \alpha_1 v_1 + \alpha_2 v_2 + \ldots + \alpha_n v_n \ \ \text{where} \ \ v_i\in T = \bigcup_{A_{s}\in S} A_s$$ Then there exists $S_i$ such that for all $i\in A_{s_{i}}$ all $i$ $A_{s_{1}},A_{s_{2}},\ldots, A_{s_{n}}$ (these will have one contained in the other then we would take the larger). So, there exists $A_{s_{k}}$ such that $A_{s_{k}}\supset A_{s_{i}}$ all $i$ just compare two at a time. Therefore, $V_i \in A_{s_{k}}$ all $i$, but $A_{s_{k}}$ is a linearly independent set. So, $\alpha_i = 0$ all $i$. Hence, $T\in X$ so $T$ is an upper bound for $S$. Hence every totally ordered subset of $X$ has an upper bound, hence by Zorn's lemma, there exists a maximal element and this would be the basis.  I am not sure if my proof is right, I apologize if it is a bit messy. If it is right, any suggestions on cleaning it up would be greatly appreciated.
5969
If $[L:K]$ is finite, then $[L:K]$ is algebraic.  The author's proof is to simply state that given $[L:K] = n$, then ${1, x, x^2, ... , x^n}$, for any $x$ in $L$, would contain $n+1$ elements, so it must be linearly dependent over $K$.  First of all, why must ${1, x, x^2, ... , x^n}$ be linearly dependent over $K$? I do undertsand that if $L=K(a_1, ... , a_m)$, and we pick an non-algebraic $a_i$, then ${1, a_i, a_i^2, ... , a_i^n}$ would be, by definition, linearly dependent over $K$. But how can we know that this works with all the elements in $L$?  Second of all, even if we find that $c_0+ c_1x+c_2x^2 + ... + c_{m-1}≠0$, for some $x$ in $L$, $c_i$ in $K$, why would this mean that $[L:K]=∞$?    I asked this question a couple days ago, though I didn't explain my doubt in such detail as in this post. Anyway, the replies asked if I understood the meaning of being 'linearly dependent over a field $F$'. My understanding is that a set ${b_1, ... , b_n}$ is linearly independent over $F$ if each element in $F$ can be written as $f_1b_1+...+f_nb_n$ (so that  the different combinations of $b_i$ form the whole $F$) and that the only way $f_1b_1+...+f_nb_n=0$ is for all the $b_i$ to be $0$ (so that we make sure all the elements in the set are needed to map create $F$).  I guess they asked me that question because is rather obvious why the author's proof works. Yet I don't know why, but it isn't that obvious to me.    I would really appreciate any help/thoughts.
121297
Suppose $R\neq 0$ is a commutative ring with $1$. Let $M$ be a free $R$-module. I would like to prove that $M$ is a flat $R$-module. Everywhere I have looked (mostly online) this is proved by first proving that every free module is projective, and then proving that every projective module is flat. Unfortunately, Atiyah &amp; Macdonald's "Introduction to Commutative Algebra" (Chapter 2) does not discuss projective modules. But the result that every free module is flat comes very handy in the exercises.   So my question is,     Is it possible to prove that every free module is flat just by   definitions and without appealing to projective modules?   Thanks!
98076
Let $T: V \to V$ with the minimal polynomial $z^n$. Prove that there's a vector $v$ such that $v, Tv, T^2v, ..., T^{n-1}v$ are linearly independent.   The way I did it orginally was not allowed. No Jordan Forms are allowed because we did not cover them this semester. I asked my professor for a hint and all he said was we need to use induction. I wish I had something to work with, but I'm so  lost. It's my last problem. Thanks!
21427
Let $R$ and $R'$ be rings (with 1 but no further assumptions) and $n \in \mathbb{N}$. Does the following implication hold?     If $M_n(R) \simeq M_n(R')$ then $R \simeq R'$.   If the rings are commutative then it follows from considering the centers of $M_n(R)$ and $M_n(R')$. If the rings are division it also holds and even stronger, it holds even if the matrices are not of the same size.  I thought about considering the embeddings of $R$ in $M_n(R)$ as diagonal matrices, but since I do not want to asume anything about the isomorphism I was not able to conclude that these subrings of matrices in $M_n(R)$ and $M_n(R')$ are isomorphic. I hope someone can help me or with a proof, an idea or a counterexample. Thank you.
117424
Let $K$ be a field and $A$ be a $K$-algebra.  I know, if $A$ is artinain algebra, then by Krull-Schmidt Theorem  $A$ , as a left regular module, can be written as a direct sum of indecomposable $A$-modules, that is   $A=\oplus_{i=1}^n S_i$ where each $S_i$ is indecomposable $A$-module  Moreover, each $S_i$ contains only one maximal submodule, which is given by $J_i= J(A)S_i$, and every simple $A$-module is isomorphic to some $A/J_i$.  My question is that, can you please tell me an example of a non simisimple algebra, or a ring, such that it has a simple module which does not occur in the regular module.  By occur I mean it has to be isomorphic to a simple submodule of a regular module
36949
Consider square matrices over a field $K$. I don't think additional assumptions about $K$ like algebraically closed or characteristic $0$ are pertinent, but feel free to make them for comfort. For any such matrix $A$, the set $K[A]$ of polynomials in $A$ is a commutative subalgebra of $M_n(K)$; the question is whether for any pair of commuting matrices $X,Y$ at least one such commutative subalgebra can be found that contains both $X$ and $Y$.  I was asking myself this in connection with frequently recurring requests to completely characterise commuting pairs of matrices, like . While providing a useful characterisation seems impossible, a positive anwer to the current question would at least provide some answer.  Note that in many rather likely situations one can in fact take $A$ to be one of the matrices $X,Y$, for instance when one of the matrices , or more generally if its  (so coincides with the characteristic polynomial). However this is not always possible, as can be easily seen for instance for diagonal matrices $X=\operatorname{diag}(0,0,1)$ and $Y=\operatorname{diag}(0,1,1)$. However in that case both will be polynomials in $A=\operatorname{diag}(x,y,z)$ for any distinct values $x,y,z$ (then $K[A]$ consists of all diagonal matrices); although in the example in  the matrices are not both diagonalisable, an appropriate $A$ can be found there as well.   I thought for some time that any maximal commutative subalgebra of $M_n(K)$ was of the form $K[A]$ (which would imply a positive answer) for some $A$ with minimal polynomial of degree$~n$, and that a positive answer to my question was in fact instrumental in proving this. However I was wrong on both counts: there exist (for $n\geq 4$) commutative subalgebras of dimension${}&gt;n$ (whereas $\dim_KK[A]\leq n$ for all $A\in M_n(K)$) as shown in , and I was forced to correct  in the light of this; however it seems (at least in the cases I looked at) that many (all?) pairs of matrices $X,Y$ in such a subalgebra still admit a matrix $A$ (which in general is not in the subalgebra) such that $X,Y\in K[A]$. This indicates that a positive answer to my question would not contradict the existence of such large commutative subalgebras: it would just mean that to obtain a maximal dimensional subalgebra containing $X,Y$ one should in general avoid throwing in an $A$ with $X,Y\in K[A]$. I do think these large subalgebras easily show that my question but for three commuting matrices has a negative answer.  Finally I note that  to the cited MO question mentions a result by Gerstenhaber that the dimension of the subalgebra generated two commuting matrices in $M_n(K)$ cannot exceed$~n$. This unfortunately does not settle my question (if $X,Y$ would generate a subalgebra of dimension${}&gt;n$, it would have proved a negative answer); it just might be that the mentioned result is true because of the existence of $A$ (I don't have access to a proof right now, but given the formulation it seems unlikely that it was done this way).  OK, I've tried to build up the suspense. Honesty demands that I say that I do know the answer to my question, since a colleague of mine provided a convincing one. I will however not give this answer right away, but post it once there has been some time for gathering answers here; who knows somebody will prove a different answer than the one I have (heaven forbid), or at least give the same answer with a different justification.
66341
I was reviewing my homework and it seems I overlooked something crucial while proving some ring has no Invariant Basis Number property. This is exercise VI.1.12 in Aluffi's Algebra: Chapter 0  The setup: $V$ is a $k$-vector space and let $R = \mathrm{End}_{k}(V)$.   Prove that $\mathrm{End}_{k}(V\oplus V) \cong R^4$ as an $R$-module Prove that $R$ doesn't satisfy the IBN property if $V = k^{\oplus \mathbb N}$.   For the first, I used to the fact that $V \oplus V$ is both the product and coproduct (in $k$-Vect) of $V$ with itself to get the isomorphism. What I just realized is I only showed that the two are isomorphic as groups not $R$-modules. So what would be the $R$-module structure on $\mathrm{End}_{k}(V \oplus V)$?   For the second, I used the fact that $V = k^{\oplus \mathbb N}$ implies $V \cong V \oplus V$ which in turn implies $R = \mathrm{End}_{k}(V) \cong \mathrm{End}_{k}(V \oplus V)$. Again, I just realized that I only showed the latter two are isomorphic as groups.   It may be obvious (and maybe why my professor let it pass?) but I can't come up with a good $R$-module structure that makes the two group isomorphisms $R$-linear.  Edit:  Explicitly, these are the isomorphisms I'm dealing with. Let $\pi_j, i_j$ be the natural projection/inclusion maps of the $j$-th factor resp. and $\psi: k^{\oplus \mathbb N} \oplus k^{\oplus \mathbb N} \to k^{\oplus \mathbb N}$ the isomorphism given by $\psi(e_i, 0)=e_{2i-1}$ and $\psi(0, e_i)=e_{2i}$.  Then the first isomorphism $\mathrm{End}_k(V \oplus V)\to R^4$ is given by $\varphi \mapsto (\pi_1\varphi i_1,\pi_2\varphi i_1,\pi_1\varphi i_2,\pi_2\varphi i_2)$  The second isomorphism $R \to \mathrm{End}_k(V \oplus V)$ is given by $\alpha \mapsto \psi^{-1} \alpha \psi$  The composition doesn't seem to be $R$-linear if I use the obvious $R$-module structure on $R$ and $R^4$.
40012
Say I have a module $M$ over a ring $R$ that is   $1.$ finitely-generated, i.e. $\exists$ finite $X \subset M$ such that $M = \sum_{i=1}^{n} Rx_{i}, x_{i}\in X$ (but $X$ is not necessarily free)  $2.$ free, i.e. $\exists$ a basis $Y \subset M$ a.k.a. $Y$ is free and generates M, (but $Y$ is not necessarily finite).  Now I want to show, that there exists a finite basis for $M$. For vector spaces, this would obviously be trivial, but for some  $R$-Modul, I'm not sure how to proceed.   Furthermore, is every minimal generating subset of $M$ a basis?
42525
Let $R$ be a ring (unital, not necessarily commutative), $M$ a finitely presented left $R$ module. Suppose $m_1,\ldots,m_n\in M$ generate $M$. This determines a surjection $f:R^n\to M$.     Must the kernel of $f$ be finitely generated?   This is true by definition if the generating set $m_1,\ldots,m_n$ is the one coming from the finite presentation of $M$.
11709
I know a proof of the following theorem using determinants. For some reason, I'd like to know a proof without using them.     Theorem   Let $A$ be a commutative ring.   Let $E$ and $F$ be finite free modules of the same rank over $A$.   Let $f:E → F$ be a surjective $A$-homomorphism.   Then $f$ is an isomorphism.