# Tag Info

10

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 $a ... 8 If$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]$. We can also see this isomorphism more directly by ... 7$\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 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$S=\mathbb Z-\{0\}$. The isomorphism is given by$\dfrac ab\otimes x\mapsto\dfrac{ax}{b}$. This way$1\otimes(1,1,\dots)$corresponds to the fraction ... 7 No, if the ring is commutative, yes, if the ring is noncommutative. The easiest counterexample is the endomorphism ring$R$of$V=F^{(\omega)}$, where$F$is a field and$F^{(\omega)}$denotes a direct sum of countably many copies of$F$(as vector space). Let homomorphisms act on the left, so$V$becomes a left$R$-module. Then$R\cong R^2$as left ... 7 Yes, it may even happen that$N'=0$, i.e. that$M$is isomorphic to$M/N$but$N \neq 0$. Take$M=R \oplus R \oplus \dotsc$and$N = R \oplus 0 \oplus 0 \oplus \dotsc$. 6 Here's one (class of) example(s). Let us restrict our attention to$\mathbb Z$-modules, i.e. abelian groups. Note that two abelian groups of order$p$are always isomorphic, thus it is sufficient to find a group$P$of order$p^i$(for any$i$) with two subgroups$A$and$B$of order$p^{i-1}$such that$A\not\cong B$. For instance, if$P = \mathbf C_4 ...

6

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 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

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

$M$ is module isomorphic to $R/I$, but that doesn't mean it has suddenly become a field. Now, you can treat the module $M$ as a ring by transferring the structure of $R/I$ back to $M$ through the bijection, but until you do that, $M$ does not have any binary multiplication operation, and hence it does not make any sense to call it a field (or a ring.)

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

Yes, there is such a finite subset. Because $X$ is a generating set for the module, given any $m \in M$ there exist $x_1,\dots, x_k \in X$ and scalars $r_1,\dots,r_k \in R$ such that $m = r_1 x_1 + \cdots + r_k x_k$. Now suppose $m_1,\dots, m_n \in M$ is a finite generating set. Then for each $m_i$ there exist $x^{(i)}_1, \dots, x^{(i)}_{k(i)} \in X$ such ...

6

In general , there is no one big theorem about monoidal closed categories that justifies or illuminates them. The internal hom does what it says on the tin: it lets you represent your homsets as objects of your category. This allows you to state questions and theorems that simply wouldn't make sense in a non-closed setting. My advice would simply be to keep ...

6

Your idea is right: If $x$ has finite order and your group homomorphism is $\phi$, then $\phi(x)$ has an order that is less than or equal to that of $x$. This is true regardless of whether $\phi$ is 1-1 or not. Thus you can't map onto elements of infinite order. When I read the original post, I misread that the indexing was over $p$! But of course, if you ...

6

The answer is negative since $A\subset B$ flat and $B$ regular implies $A$ regular; see Bruns and Herzog, Theorem 2.2.12. But in this case $A\simeq k[a,b,c]/(ac-b^2)$, so $A$ is not regular. Edit. A simpler approach: let $I=(x^2,xy)$ and $A/I\to A/I$ be the multiplication by $y^2$. Since $A/I\simeq k[y^2]$ this is injective, but on $A/I\otimes_AB\to ... 6 The argument for modules generalizes trivially, once you phrase it properly. For any object$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 It is not true that if$A$is not finitely generated, you can find$\mathbb Z$-independent elements. For example,$\mathbb Q$is a non finitely generated$\mathbb Z$-module, but, clearly, you cannot find two$\mathbb Z$-independent rationals$\frac ab$,$\frac cd$, for$ac = bc\times \frac ab = ad\frac cd$. When we prove that$\mathcal O_K$is finitely ... 6 For the non-free part: Take any two nonzero elements$x, y ∈ ℚ$and show they satisfy$λx + μy = 0$for some nonzero$λ, μ ∈ ℤ$, hence any two elements are linearly dependent. Thus, since$ℚ$is not cyclic, it cannot have a basis. For the non-finitely-generated part: If$ℚ$was finitely generated then without loss of generality (by finding the common ... 5 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 ... 5 If$m = n$, then$0 \to \mathbb{Z}/n\mathbb{Z} \xrightarrow{\operatorname{id}} \mathbb{Z}/n\mathbb{Z} \to 0$is a free resolution. If$m < n$, it will be useful to write$n = km$, where$k > 1$. The first step of the resolution looks like$F_0 \xrightarrow{\epsilon} \mathbb{Z}/m\mathbb{Z} \to 0$for some free$\mathbb{Z}/n\mathbb{Z}$-module$F_0$. ... 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. 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 Let$R$be ring for which$F$is a free module. Then we have isomorphisms$R^n \rightarrow F$and$F \rightarrow R^{n+1}$which gives us an isomorphism$R^n \rightarrow R^{n+1}$. Thus for any$m \geq n$we have an isomorphism$R^m \cong R^n \oplus R^{m-n} \cong R^{n+1} \oplus R^{m-n} \cong R^{m+1}$. Composing these isomorphisms we get an isomorphism$R^n ...

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 eigenvalues of a matrix $A$ over a field $k$ are the roots of the characteristic polynomial $\chi(A)=\det(\lambda I-A)$. But what is this thing we're taking the determinant of? $\lambda$ is not an element of $k$, but an indeterminate, so to properly describe $\lambda I-A$ we ought to say its coefficients, e.g. $2-\lambda$, are elements of the polynomial ...

5

In fact all submodules of free $\mathbb{Z}$-modules are projective, even free, so the experimentation you're doing won't succeed. (This is because $\mathbb{Z}$ is a principal ideal domain.) Taking $(x,y)$ a submodule of $k[x,y]$ for $k$ some field will work better. $(x,y)$ is not projective, which we can show by showing it's not flat, since projective ...

5

It is still true because: Any $A$-module is the direct limit of its finitely generated submodules (for any ring $A$). Tensor products commute with direct limits. Direct limit is an exact functor. Btw, a submodule of an $A$-module with this property (the morphism $M'\hookrightarrow M$ is universally exact) is called a pure submodule of $M$, and the ...

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