For any module $M$, the functor $-\otimes M $ is right exact. I want to show that if $F$ is a free module then $-\otimes F$ is a exact.   I could not see how to adapt the proof for the case of $M$ not being free to  this case.
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!
61393
I have a question which should be pretty basic commutative algebra, but I can't find a reference and I'm stuck on proving a result myself, so here it goes:  Let $\varphi: S \to R$ be a morphism of commutative rings and let $M$ be a finitely generated $S$-module. Then, we can "extend the scalars of $M$" by considering $M \otimes_S R$, which is naturally an $R$-module. I would like to get a description of $\text{Supp}(M \otimes_S R)$ (as an $R$-module) involving $\text{Supp}(M)$ and $\varphi$.  Maybe this helps: by Eisenbud's Commutative algebra with a view toward algebraic geometry, Corollary 2.7, the problem can probably be solved by finding a description of $\text{Ann}(M \otimes_S R)$ and it is not difficult to see that $\varphi(\text{Ann}(M))R \subset \text{Ann}(M \otimes_S R)$, but I fail to prove the other inclusion.
14302
Suppose we have an exact sequence of sheaves of abelian groups,  $$ 0 \longrightarrow \mathcal{F}' \stackrel{\phi'}{\longrightarrow} \mathcal{F} \stackrel{\phi}{\longrightarrow} \mathcal{F}''. $$  I want to show that the sequence of abelian groups $$0 \longrightarrow \mathcal{F}'(U) \stackrel{\phi'_{U}}{\longrightarrow} \mathcal{F}(U) \stackrel{\phi_{U}}{\longrightarrow} \mathcal{F}''(U) $$ is exact for all open sets $U$. So far I have that exactness at $\mathcal{F}'(U)$ is immediate since injectivity for morphisms of sheaves can be checked on injectivity of morphisms of sections. However I am stuck at exactness at $\mathcal{F}(U)$. I feel as though I am very close. I have determined that exactness at $\mathcal{F}(U)$ is equivalent to the presheaf image of $\phi '$ being equal to the kernel sheaf of $\phi$.   From exactness of the original sequence, I have that $\text{im} \phi ' = \ker \phi  $, but how do I then show that this is equal to the image presheaf of $\phi '$? In other words, the image presheaf of $\phi'$ was already a sheaf?  This is related to another question asked here , where user Future suggests that we can use the fact that the image sheaf can be identified with a subsheaf of the target sheaf, but I am still not sure how to do it.  I realize there are other ways to prove left exactness of the section functor, but I am very curious how to make this particular approach work.   Any help is appreciated.
111813
If $A\to B$ is such that $B$ is a free $A$-module, is it true that $Spec(B)\to Spec(A)$ is surjective? I suspect it is true that there is a projection $B\to A$ so that the composition $A\to B\to A$ is the identity map (of rings), but I'm not sure.
73900
Let $A$ be a commutative noetherian ring. Let $M, N$ be $A$-modules, and assume that $M$ is finite over $A$. Let $P$ be a flat $A$-module.  Is it true that there is an isomorphism $\operatorname{Hom}_A(M,N)\otimes_A P \cong \operatorname{Hom}_A(M,N\otimes_A P)$ ?  It appears that it holds at least locally (that is, when $P$ is free). Is it true in general? If so, is there a reference for this?  Thank you!
113148
I am trying to understand the proof of Theorem 3.4.3 from Weibel's book Introduction to homological algebra.  The statement is the following.     Let $R$ be a ring. Given $R$-modules $A$ and $B$, an extension $\xi$ of $A$ by $B$ is an exact sequence $0\to B\to X\to A\to0$. Then the connecting homomorphism $$\delta:\text{Hom}(A,A)\to\text{Ext}^1(A,B) \tag{1}$$ induces the 1-1 correspondence $\theta$ between equivalence classes of extensions of $A$ by $B$ and $\text{Ext}^1(A,B)$, which is given by $\theta:\xi\mapsto\delta(\text{id}_A)$.   In the beginning of the proof it is written the following     Fix an exact sequence $0\stackrel{j}\to M\to P\to A\to0$ with $P$ projective. Applying $\text{Hom}(-,B)$ yields an exact sequence $$\text{Hom}(P,B)\to\text{Hom}(M,B)\stackrel{\delta}\to\text{Ext}^1(A, B)\to0.\tag{2}$$   I don't understand why the connecting homomorphism in the sequence (2) is denoted by the same letter as in (1). Are they really the same?
16154
Question. Let $R$ be a ring, $\mathfrak{p}$ a prime, $M$ a finitely-generated $R$-module, and $N$ any $R$-module. Is the natural map $$\textrm{Hom}_R(M, N)_\mathfrak{p} \to \textrm{Hom}_{R_\mathfrak{p}}(M_\mathfrak{p}, N_\mathfrak{p})$$ an isomorphism (of $R_\mathfrak{p}$-modules)?  I can prove this in the case when $M$ is finitely presented: indeed, let $S$ be any flat $R$-algebra, and let $$R^m \to R^n \to M \to 0$$ be a right-exact sequence; tensoring with $S$ gives another right-exact sequence $$S^m \to S^n \to M \mathbin{\otimes_R} S \to 0$$ and applying hom functors, we get left-exact sequences $$0 \to \textrm{Hom}_R(M, N) \to N^n \to N^m$$ $$0 \to \textrm{Hom}_S(M \mathbin{\otimes_R} S, N \mathbin{\otimes_R} S) \to (N \mathbin{\otimes_R} S)^n \to (N \mathbin{\otimes_R} S)^m$$ and $S$ is flat, so tensoring the first sequence yields $$0 \to \textrm{Hom}_R(M, N) \mathbin{\otimes_R} S \to (N \mathbin{\otimes_R} S)^n \to (N \mathbin{\otimes_R} S)^m$$ but extending the sequences by $0$ to the left, and putting in vertical maps between the last two, we conclude that $\textrm{Hom}_R(M, N) \mathbin{\otimes_R} S \cong \textrm{Hom}_S(M \mathbin{\otimes_R} S, N \mathbin{\otimes_R} S)$ by the five lemma (modulo checking commutativity of diagrams). This is essentially the proof Eisenbud gives [Commutative Algebra, Prop. 2.10].  The trouble with extending it to a proof for finitely-generated modules is that we have to replace $R^m$ with a potentially arbitrary submodule $K$ of $R^n$, and that may not be free or even finitely-generated without some additional assumptions on $R$. I can't see an abstract nonsense proof of the claim, but I admit I haven't tried a bare-hands proof. However, is the claim even true?
29870
Let $M$ be an A-module and $f \in \text{End}(M)$ an idempotent endomorphism. Show that $M = \ker(f) \oplus \ker (f-\text{id}_M)$. I don't know how to approach this exercise, using only properties of modules and direct sums, without projections. Can you help me? Thanks in advance.
78204
Let $S$ be a scheme over a finite field $\mathbb{F}_q$ and $\mathcal{L}$ an invertible sheaf on $S$.   In Pink's article (, see section 1.4 on page 13), he states that:  For any section $\ell\in \mathcal{L}(S)$ and any point $s\in S$ we let $\ell(s)\in \mathcal{L}\otimes_{\mathcal{O}_S}k(s)$ denote the value of $\ell$ over the residue field $k(s)$ of $s$.  My question is: what is the definition of $\mathcal{L}\otimes_{\mathcal{O}_S}k(s)$?  We know we can define the tensor product of two $A$-modules or of two $\mathcal{O}_X$-modules, but what is the tensor product of an $\mathcal{O}_X$-module and a field?
75434
Let $K\subseteq F$ be number fields with ring of integers $\mathcal{O}_K\le \mathcal{O}_F$.   Question: Is $\mathcal{O}_F$ a free $\mathcal{O}_K$-module ?   By the integral basis theorem this is true when $K=\mathbb{Q}$ but I don't know about the general case.
77779
While reading, there is an isomorphism that I'm having trouble fulling seeing.      If you have two algebras $A$ and $B$ over a commutative ring $R$, with $I$ and $J$ two sided ideals in $A$ and $B$, then you should have an isomorphism   $$ (A/I)\otimes (B/J)\cong (A\otimes B)/(I\otimes 1+1\otimes J). $$   Now there are bilinear maps from $A/I\times B/J\to (A\times B)/(I+J)$ and so the universal property of the tensor product gives unique maps $(A/I)\otimes (B/J)\to (A\times B)/(I+J)$. Does this somehow get to the isomorphism above, or am I completely off track? What is the quick way to see this? Thanks!
39294
Let $R$ be a commutative unital ring and let $M$ be an $R$-module and let $S$ be a multiplicative subset of $R$.  Today I proved both of the following: $$ S^{-1} R\otimes_R S^{-1}M \cong S^{-1} M$$ and $$ S^{-1} R \otimes M \cong S^{-1} M$$  Now I'm slightly confused.      Either my proofs are wrong or $C \otimes A \cong C \otimes B$ does not imply $A \cong B$. But I can't come up with an example. Can someone give me an example? (Or tell me that my proofs are wrong.)
46016
Ok, I'm reading some thesis of some former students, and come up with this proof, but it doesn't really look good to me. So I guess it should be wrong somewhere. So, here it goes:     Let $R$ be a unitary commutative ring, and $I$ be an ideal, and $N$ is an $R$-module. I'll now show that $$I \otimes_R N \cong IN\,.$$   First, since $1 \in R$, and $I$ is an ideal of $R$, we must have that $I = RI$. So then $$I \otimes_R N = RI \otimes_R N = R \otimes_R IN \cong IN\,.$$  But this must be so wrong, since if this proof is true, I could prove that any $R$-module is flat. So how come this proof is wrong?
63833
Let $R$ be a commutative ring with identity and $I$ and $J$ be ideals in $R$.     Show that $R/I\otimes_R R/J\cong R/(I+J)$ as $R$-modules.   This is what I tried.  Define a map $f:R/I\times R/J\to R/(I+J)$ as $f(\bar r, \bar s)=rs+(I+J)$. This is a well-defined bilinear map.  By the universal property of tensor products, we get an $R$-linear map $F:R/I\otimes_R R/J\to R/(I+J)$ which sends $\bar r\otimes \bar s$ to $rs+(I+J)$.  This map is surjective.  So all we need to show is that the kernel of $F$ is $0$.  But I am unable to do this.
68559
I'm aware that over a commutative ring, two free finitely generated modules of finite rank are isomorphic if and only if they have the same rank.   I'm trying to understand a curious example of why this does not hold over a noncommutative ring.      Let's take $M=\oplus_{i\in\omega}\mathbb{Z}$. Then let $R=\text{End}(M)$, which is not commutative. Then $R\cong R^2$ as left $R$-modules.   It seems like one could possibly make a basis of $R$ using only endomorphisms which act on the even or odd coordinates of elements of $M$, which might suggest such an isomorphism. Is there a clear proof of why $R\cong R^2$?
44108
Let $M$, $L$, $N$ be $A$-modules and $M=N\oplus L$. If $M$ and $N$ are free, is $L$ necessarily free?
28249
I was reading about the construction of a tensor product of any right $R$-module $M$ and left $R$-module $N$ over a ring $R$.  Why is it required that $M$ and $N$ be right and left modules, respectively? How does the construction not work otherwise?