Hot answers tagged

12

Nakayama Lemma. Let $N$ be a finitely generated $R$-module, and $J\subseteq R$. Suppose that $J$ is closed under addition and multiplication and $JN=N$. Then there is $a\in J$ such that $(1+a)N=0$. (Here by $JN$ we denote the subset of linear combinations of $N$ with coefficients in $J$.) Cayley-Hamilton Theorem. Let $A$ be a commutative ring, $I$ an ...


10

No: For a local ring $(R,{\mathfrak m})$ with ${\mathfrak m}^2=0$ you have $\text{Hom}_R(R/{\mathfrak m},R)\cong\{x\in R\ |\ {\mathfrak m}x=0\}={\mathfrak m}$, so it suffices to choose $R$ such that ${\mathfrak m}$ is not finitely generated, e.g. $R := {\mathbb k}[x_1,x_2,\ldots]/(x_i^2, x_i x_j)$.


9

The simpler tensor product $\mathbb{Z}[i] \otimes \mathbb{Z}[i]$ is free abelian on the generators $1 \otimes 1, 1 \otimes i, i \otimes 1, i \otimes i$. As either a left or a right $\mathbb{Z}[i]$-module it is free on two generators $1 \otimes 1, i \otimes i$. This tensor product is the quotient of the above tensor product by the additional relation that $a ...


9

No, your argument isn't correct: if $1\otimes x=1\otimes y$ then not necessarily $x=y$. I'd do this as follows: if $M$ is a $\mathbb Z$-module, then $\mathbb Q\otimes_{\mathbb Z}M\simeq S^{-1}M$, where $S=\mathbb Z\setminus\{0\}$. The isomorphism is given by $\dfrac ab\otimes x\mapsto\dfrac{ax}{b}$. This way $1\otimes(1,1,\dots)$ corresponds to the fraction ...


9

A simple counterexample is the ring $\mathbb{Z}$, and the $\mathbb{Z}$-modules $M=\mathbb{Z}/2\mathbb{Z}$ and $N=\mathbb{Z}/3\mathbb{Z}$. There is no non-trivial $\mathbb{Z}$-module homomorphism from $M$ to $N$, because there is no element of $N$ that has order $2$. (One could replace $2$ and $3$ with any relatively prime integers.) Another counterexample ...


8

Sometimes it is better to look at a more general situation. In fact, this makes it easier to see what is really going on. 1) Let $V$ be a $K$-vector space and let $L/K$ be a field extension. Then $L \otimes_K V$ carries the structure of a vector space over $L$ via (linear extension of) $\alpha(\beta \otimes x) = \alpha\beta \otimes x$. 2) If ...


8

If $R=\mathbb{Z}[2i]$, then $\mathbb{Z}[i] \cong R[X]/I$, where $I=(X^2+1,2X-2i)$. So we have $\mathbb{Z}[i] \otimes_R R[X]/I \cong \mathbb{Z}[i][X]/I$. Substituting $Y=X-i$, this is $\mathbb{Z}[i][Y]/(Y^2+2iY,2Y) \cong \mathbb{Z}[i][Y]/(Y^2,2Y) \cong \mathbb{Z}[i] \oplus \mathbb{Z}[i]/2\mathbb{Z}[i]$. We can also see this isomorphism more directly by ...


8

No, for two reasons. First, $\frac{1}{q} m$ must be a $q^{th}$ root of $m$, but such a $q^{th}$ root need not exist: for example take $M = \mathbb{Z}$. Second, if $m = qn$ then $\frac{1}{q} m = \frac{1}{q} qn = n$, so $q^{th}$ roots also need to be unique, and even when they exist in some abelian group they need not be unique: for example take $M = ...


7

It is enough to prove it preserves short exact sequences: $\;0\to M\to N\to P\to 0$. As the tensor product is right-exact, and $S^{-1}M\simeq M\otimes_A S^{-1}A$, it is even enough to prove it preserves injectivity. So consider an injective morphism $\varphi\colon M\to N$ and suppose $\;S^{-1}\varphi\Bigl(\dfrac ms\Bigr)=0$ in $S^{-1}N$. This means there ...


7

Proposition: $\mathbb{Q}$ is flat as a $\mathbb{Z}$-module. (Which is to say, the functor $ \_ \otimes_{\mathbb{Z}} \mathbb{Q}$ is exact.) Pf: The easiest is to observe that tensoring with $\mathbb Q$ is the same as localizing at $\mathbb{Z} \setminus \{0\}$, and localization is exact. Fancier proof: $\mathbb{Q}$ is the filtered colimit of the free modules ...


7

$\mathbb{Z}$-modules are precisely abelian groups. As every ring is an abelian group, it is a $\mathbb{Z}$-module. It is entirely possible to be a module over more than one ring. For example, if $M$ is an $R$-module then it is also an $S$-module for any subring $S$ of $R$ (you seem to be interested in the case where $M = R$). Another example is given by ...


7

This is not true in general. For example, $\mathbb{Q}\otimes_{\mathbb{Z}}(\mathbb{Z}/p\mathbb{Z}) \cong 0 \cong 0\otimes_{\mathbb{Z}}(\mathbb{Z}/p\mathbb{Z})$ but $\mathbb{Q}$ and $0$ are not isomorphic as $\mathbb{Z}$-modules.


7

The only other examples of non IBN rings that I am aware of are built using Leavitt path algebras, and they do exactly what you want. It is possible to specify positive integers $n<m$ such that $R^i\ncong R^j$ for $(i,j)$ less than $(n,m)$ in the lexicographic order, but $R^n\cong R^m$ as modules. (The $(m,n)$ pair is called the 'module type' in the ...


7

For question 2, $F\otimes_A -$ has a left adjoint iff $F$ is finitely generated, and the left adjoint is always exact. For if $(f_i)\in F^I$ is an element of an infinite product of copies of $F$, then it is easy to see $(f_i)$ is in the image of the canonical map $F\otimes_A A^I\to F^I$ iff $\{f_i\}$ is contained in a finitely generated submodule of $F$. ...


6

For commutative rings there is a geometric way to think about these things. Every module $M$ over a commutative ring $R$ has a support, which is the set of prime ideals $P$ such that the localization $M_P$ at $P$ is nonzero. This use of "support" is analogous to the notion of support of a function: it's "where the module is nonzero." For example, when $R = ...


6

For every nonzero $x, y\in\mathbb{Q}$, there are nonzero integers $m, n$ such that $$mx=ny$$ (where $mx, ny$ are interpreted in the obvious way). Now, any homomorphism $f$ between $(\mathbb{Q}, +)$ and another structure $(G, *)$ must preserve multiplication by integers: $$f(mx)=mf(x).$$ So if $(\mathbb{Q}, +)\cong(\mathbb{Q}^n, +)$, then $(\mathbb{Q}^n, +)$ ...


6

No. Free abelian groups don't have any nonzero divisible elements, but every element of $\mathbb{R}$ is divisible. $\mathbb{R}$ is free as a $\mathbb{Q}$-module (this requires the axiom of choice), since $\mathbb{Q}$ is a field.


6

If $A$ is any ring, then $A$ and $M_n(A)$ have equivalent categories of modules, and usually $A$ and $M_n(A)$ are not isomorphic. This is the simplest example of a Morita equivalence.


6

For the second question, the finite rings are precisely those for which every finitely generated module has finitely many submodules. For a ring $R$ and $r\in R$, let $M_r$ be the submodule of $R\oplus R$ generated by $(1,r)$. Then $(1,r)$ is the only element of $M_r$ whose first coordinate is $1$, and so $M_r\neq M_s$ for $r\neq s$, and so if $R$ is ...


6

No. Let $G$ be any non-zero abelian group. Then the obvious morphism $(0,G)\to (G,G)$ is a monomorphism and epimorphism, but not an isomorphism, which can't happen in an abelian category.


6

Let $\mathbb{F}$ be a field and let $V$ be a module over $\mathbb{F}$ (i.e. a vector space over $\mathbb{F}$), then $V^*$ is also a module over $\mathbb{F}$. The tensor product $V^*\otimes_{\mathbb{F}}V$ is canonically isomorphic to $\operatorname{End}_{\mathbb{F}}V$ via the map induced by the bilinear map $V^*\times V \to ...


6

This is never true unless $\mathfrak{m}=0$, i.e. unless $A$ is a field. Let $x\in\mathfrak{m}$, and consider the short exact sequence $0\to (x)\to A\to A/(x)\to 0$. If $A/\mathfrak{m}$ is flat, we can tensor it with the short exact sequence to get a short exact sequence $0\to (x)/\mathfrak{m}(x)\to A/\mathfrak{m}\to A/\mathfrak{m}\to 0$. Exactness of this ...


6

The argument for modules generalizes trivially, once you phrase it properly. For any object $A$, there is an epimorphism $p:F\to A$ from a free object $F$. If $A$ is projective, then applying the definition of projectivity to $p$ and the identity map $1:A\to A$, there exists a map $i:A\to F$ such that $pi=1$. That is, $A$ is a retract of $F$. (In the ...


6

(I'm implicitly assuming that $R$ is a commutative ring.) An $R[x_1,\ldots,x_n]$-module is the same thing as an $R$-module equipped with $n$ mutually commuting endomorphisms. One way of putting it is that a left $S$-module $M$, where $S$ is a (not necessarily commutative) $R$-algebra, is the same thing as an $R$-module $M$ together with an $R$-algebra map ...


5

Not necessarily. If there is a nontrivial ring honomorphism $\psi: A\to A$, then composing an evaluation homomorphism with $\psi$ gives you something that is not an $A$-algebra homomorphism. For example, in the case $A=\mathbb C$, $n=1$ and $\psi$ being complex conjugation we could have $$ \phi(a_0 + a_1X + a_2X^2 + \cdots + a_kX^k) \mapsto \overline{a_0} ...


5

The abstract approach. In general, given any ring homomorphism $\phi:R\to S$, and any $s\in S$, there is a unique ring homomorphism: $\phi':R[X]\to S$ that is equal to $\phi$ on $R$ and with $\phi'(X)=s$. Given any additive abelian group, $(A,+)$, the set of homomorphisms of the group, $\mathrm{End}(A),$ becomes a ring. An $R$-module $M$ can be seen as an ...


5

As user54748 says in comments, if $A$ is a noncommutative ring then it's not clear that there's any good meaning for "$A$-linear". You say you mean that homsets are $A$-modules and composition is $A$-linear, but if you mean that $(a\theta)\circ\phi=a(\theta\circ\phi)$ and $\theta\circ(b\phi)=b(\theta\circ\phi)$, then this implies that ...


5

Your guess is right, it's indeed not true in general: Any choice of an $R$-module $X$ such that $X\otimes_R X=0$ and $f\neq\text{id}: X\to X$ gives a counterexample, for example you could take $R := {\mathbb Z}$, $X := {\mathbb Q}/{\mathbb Z}$ and $f = 2\cdot \text{id}$.


5

Hint: If one can find a pair of cardinals $\kappa$ and $\lambda$ such that $\kappa\neq\lambda$ but $2^\kappa=2^\lambda$, then $\mathbb Z_2^{(\kappa)}$ and $\mathbb Z_2^{(\lambda)}$ give a counterexample. Since ZFC theory doesn't violate the existence of such pair, I think your conjecture is not true, but I'm not sure if it's false... By the way, this ...


5

One way to proceed is to notice that $\mathbb Z_8$ is a local ring, so that its finitely generated projective modules are in fact free. Of course, this implies that finitely generated modules have at least $8$ elements.



Only top voted, non community-wiki answers of a minimum length are eligible