# Tag Info

9

Let $M, N$ be $R-$modules. Then the following holds. If $M$ and $N$ is flat, then so is $M\otimes_{R}N$: see related question here. If $M$ and $N$ are projective, then so is $M\otimes_{R} N$. Indeed, writing $M\oplus M'=F,\ N\oplus N'=F'$, for free $R-$modules $F,\ F'$, one has that $$F'':=F\otimes_{R}F'$$ is free (tensor product of free modules) and ...

8

Let $0_M$ and ${\rm id}_M$ denote the zero map and identity map on an $A$-module $M$. We have $$M=0\iff 0_M={\rm id}_M.$$ Since $F$ is a functor, $F({\rm id}_M)={\rm id}_{FM}$. Since it's also additive, $F(0_M)=0_{FM}$.

6

To expand on Zhen Lin's comment: Let $R$ be an integral domain, $A$ a finitely presented $R$-algebra, $Q$ the fraction field of $R$. Then: to say that $A$ is smooth over $R$ can be reinterpreted geometrically as saying that the morphism of affine schemes $\operatorname{Spec} A \rightarrow \operatorname{Spec }R$ has nonsingular schemes as fibres over ...

6

Hint: An endomorphism of $\mathbb Q$ is determined by the image of $1$. Here are some details:

5

Algebra objects or monoid objects can be defined in any monoidal category. When $R$ is a commutative ring, then the category of left $R$-modules has a monoidal structure given by $\otimes_R$, and algebras in that category coincide with $R$-algebras. But the category of left $R$-modules has no "natural" monoidal structure when $R$ is not commutative - this is ...

5

Definition: A ring is Jacobson if every prime ideal is an intersection of maximal ideals. Theorem (Nullstellensatz): Let $R$ be a Jacobson ring. Then every finitely generated $R$-algebra is Jacobson. Since $\mathbb{Z}$ is Jacobson (as every nonzero prime is maximal, and $0$ is the intersection of any infinite set of maximal ideals), this shows that any ...

5

This is true with added generality, that is in the non commutative world. So, assume $\phi\colon R\to S$ is a ring homomorphism between not necessarily commutative rings. If $P_{R}$ is a projective right $R-$module, then, if $P_{R}$ is a direct summand of the free right module $R^{(X)}$ for some set $X$, we get that $$P_{R}\otimes_{R} S$$ is a direct ...

4

If $M$ does not have finite projective dimension: let $R = k[[x,y]]/(xy)$, $M = R/(y) \cong k[[x]]$. Then $x$ is a zerodivisor in $R$, but acts as a nonzerodivisor on $M$. If $R$ is not local: let $R = k \times k$, $k$ a field, $M = k \times 0 \subseteq R$, $x = (1,0) \in R$.

4

A basic fact in homological algebra is that the functor $$-\otimes_R P:_R\mathsf{Mod}\to_R\mathsf{Mod}$$ is always right exact. So, applying $-\otimes_RP$ to an exact sequence $$M\xrightarrow{f}N\to Q\to 0\tag{1}$$ gives an exact sequence $$M\otimes_R P\xrightarrow{f\otimes_R\DeclareMathOperator{id}{id}\id_P}N\otimes_RP\to Q\otimes_R P\to 0\tag{2}$$ ...

4

As ${\cal O}_K[G]\subseteq{\cal A}_{L/K}$ and ${\cal O}_K[G]$ has rank $|G|[K:\Bbb Q]=[L:\Bbb Q]$ as a $\Bbb Z$-module, ${\cal A}_{L/K}$ has at least this rank also as a $\Bbb Z$-module. If ${\cal O}_L$ were a free ${\cal A}_{L/K}$-module of rank $\ge2$, it would necessarily have rank $\ge2[L:\Bbb Q]$ as a $\Bbb Z$-module, but we know ${\cal O}_L$ has rank ...

4

What follows does not literally answer the OP's question (this is done in the other answer), but may help clear up some confusion: The OP states in comments that they are confused about the expression "finitely generated as an $R$-module". This phrasing is used in contexts when an $R$-module $M$ might also be being considered with various subsidiary ...

4

$\mathbb{Z} \to \mathbb{Z}[\sqrt{2}]$ is not smooth (for example since $\Omega^1_{\mathbb{Z}[\sqrt{2}]/\mathbb{Z}} \cong \mathbb{Z}[\sqrt{2}]/(2)$ is torsion), but $\mathbb{Q} \to \mathbb{Z}[\sqrt{2}] \otimes_\mathbb{Z} \mathbb{Q} = \mathbb{Q}(\sqrt{2})$ is smooth (in fact, étale).

3

Try using the snake lemma. The second row is short exact, and the first row is almost short exact; you can make it short exact on the nose just by replacing $K \otimes M'$ with its image in $K \otimes M$. (Note that by short exactness of the second row, the map from $K \otimes M'$ to $F \otimes M'$ factors through this image.) Now you have a map between ...

3

It means that there is a finite set $\{m_1, \ldots m_k \} \subseteq M$ such that $$M=\{r_1m_1+\cdots + r_km_k| r_1, \ldots ,r_k \in R\}$$

3

Let $R=\prod k$ (infinitely many copies) be considered a module over itself, $k$ a field. Then $\bigoplus k$ is a submodule of $R$. Suppose that $N$ is any other nontrivial submodule (equivalently, ideal) of $R$. Show that $N$ and $\bigoplus k$ intersect nontrivially: take an arbitrary nonzero element of the former, and multiply it by an appropriately ...

3

There is a more general fact at work here. Let $A\to B$ be a ring map, $M$ an $A$-module, and $N$ a $B$-module. Then there is a unique $B$-module isomorphism $M\otimes_AN\simeq(M\otimes_AB)\otimes_BN$ sending $m\otimes n$ to $(m\otimes 1)\otimes n$ (the $B$-module structure on the source is via the $B$-module structure of $N$, whereas in the target, there is ...

3

From linear algebra we know this when $R$ is a field. It follows more generally when $R$ is an integral domain, since every maximal linearly independent subset of the $R$-module $M$ induces a maximal linearly independent subset of the $Q(R)$-module $(R \setminus \{0\})^{-1} M$ (this can be checked by a direct calculation, for instance). There are many more ...

3

First remark: in general it is false that $Ae\simeq Af$ if $e=xy$ and $f=yx$. Take for example $A$ to be the ring of $2\times 2$ matrices, then $$x=\begin{pmatrix} 0 & 1\\ 0 & 0\end{pmatrix},y=\begin{pmatrix} 0 & 0\\ 0 & 1\end{pmatrix}.$$ This gives $e=xy=x\not=0$ and $f=yx=0$. However, in your title you wrote "idempotents" so I presume that ...

3

$\DeclareMathOperator{\coker}{coker}$$\DeclareMathOperator{\im}{im}Let f : M \to N. Since tensoring preserves cokernels (by right exactness), (\coker f) \otimes_A A/a \cong \coker(f \otimes A/a) = \coker(M/aM \to N/aN) = 0, so \coker f = a(\coker f). Since \coker f is f.g. (being a quotient of N which is f.g.), Nakayama's Lemma gives \coker f = ... 3 The product of free modules need not be free. Even the infinite product of free abelian groups may not be free, although this is not easy. See this answer on MO. 3 For every unital ring R, the free R-module generated by n elements is R^n. In particular, for n=1, we get that R is free (generated by one element), hence it is also projective, as each R-morphisms f:R\to M is uniquely determined by the arbitrary element f(1)\in M. 3 R is a cyclic R-module, and so if every submodule, i.e ideal, is cyclic R is a PID. Viceversa if R is a PID and M a cyclic R-module, M = <m> then$$M \cong R/Ann(m) $$so every submodule of M corresponds to a principal ideal I \subseteq R with Ann(m) \subseteq I and so is cyclic. 3 There is a short exact sequence M\xrightarrow{f} N\xrightarrow{g}\operatorname{coker} f\to 0. Applying the functor (\mathord-)\otimes_RP, which is right exact, we get a new exact sequence M\otimes_RP\xrightarrow{f\otimes1_P} N\otimes_RP\xrightarrow{g\otimes1_P}(\operatorname{coker} f)\otimes_RP\to 0 3 There is a minimal submodule K such that M/K is of exponent p, and this is clearly K = p M. So the question reduces to counting the submodules N/K of M/K such that (M/K)/(N/K) \cong M/N is isomorphic to \mathbb{Z}/p \mathbb{Z}. Since M / K \cong \mathbb{Z}/p \mathbb{Z} \oplus \mathbb{Z}/p \mathbb{Z}, a vector space of dimension two over ... 3 Here is a real life example: FIFA wants its association to last a long time. In their early history they thought about the ball being a cube. Notice that during that period of time only four colours of dye existed. Since obviously they couldn't repeat the same ball in two cups they where wondering how many world cups they could have using cube balls. To do ... 3 Tunococ's comment provides half the answer: Flatness of S implies that \{P_i\otimes_RS\} is a resolution of M\otimes_RS. The other half is to observe that this resolution consists of projective S-modules. Indeed, each P_i is a projective R-module, hence a summand of a free R-module. That direct sum decomposition is preserved by tensoring with ... 3 The reason is, basically, that rf defined this way would usually fail to be a homomorphism of R-modules. Assume that for every r \in R, g_r=rf is a homomorphism of (left) R-modules as well. Then for every r,s \in R and every x \in M, we have$$(rs)\cdot f(x)=r\cdot(s\cdot f(x))=r\cdot f(sx)=[rf](sx)=g_r(sx)=s\cdot g_r(x)=\\=s \cdot ... 2 Apply primary decomposition to these modules. Clearly, the primary decomposition of$M\oplus M$is just the primary decomposition of$M$"doubled" in the sense that all factors appear twice as many times in the decomposition of$M\oplus M$as they do in$M$. We can be certain of this because the "doubled" decomposition of$M$clearly provides a ... 2 Another way of putting it is to consider that the isomorphism class of a finitely-generated module$M$over a PID$R$is uniquely determined by the sequence of its invariant factors $$(a_{1}) \supseteq (a_{2}) \supseteq \dots \supseteq (a_{k}),$$ with all$a_{i}$not units, and $$M \cong \bigoplus_{i=1}^{k} \frac{R}{(a_{i})}.$$ Clearly the sequence of ... 2 Your action 2 is not even well-defined if$\mathfrak{h}$is not commutative: Let$i<j$such that$[x_i,x_j]\neq0\$. Then $$(x_ix_j)\cdot1=x_ix_j=(x_jx_i)\cdot1,$$ hence: $$0\neq[x_i,x_j]=[x_i,x_j]\cdot1=(x_ix_j-x_jx_i)\cdot1=0,$$ which is impossible.

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