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0

Since the question now has two answers, I'll just add an illustrative example. The $\mathbb Z$-module $M=\mathbb Z/2\mathbb Z$ has non-zero localization $M_{2\mathbb Z}\neq 0$ at the prime $2\mathbb Z=(2)\subset \mathbb Z$ . But all its other localizations are zero: $M_{\mathfrak p}=0$ for all primes $\mathfrak p\neq (2)$ namely for $\mathfrak ...


0

Suppose $M$ is nonzero. And let $I=\{a\in R\mid aM=0\}$. Show that $I\neq A$ hence there is a maximal ideal $p$ containing $I$. Now $M_p\neq0$.


1

$M_p=0$ if and only if $x/s=0/1$ for all $x\in M$ and $s\in R-p$ if and only if for all $x\in M$ there is $t_p\in R-p$ such that $t_px=0$. If all localizations $M_p$ are $0$, then $M=0$. Pick an element $x\in M$. Since $x/s=0/1$ in $M_p$ there is $t_p\in R-p$ such that $t_px=0$. Now notice that the ideal generated by all $t_p$, when $p$ runs through ...


1

In general, if you define a $\mathbb{Z}$-module homomorphism from a base to anything in $\mathbb{Z}$, it extends to an homomorphism from the whole module. In your case, the homomorphism extends to $$ \varphi(a_0+a_1x+a_2x^2+\dots+a_{n-1}x^{n-1})=a_1+2a_2+\dots+(n-1)a_{n-1} $$ This construction is forced, and now $\varphi:M\to \mathbb{Z}$ is a module ...


1

You should look at a much more general statement: For any ideal $I\subset R$, and any homomorphism $f: N\to N'$ of $R$-modules, $f(IN) \subset IN'$. This statement is immediate from the definition of a module homomorphism. If $f$ is an isomorphism, then we can apply the same claim to $f^{-1}$, and see that $f$ restricts to an isomorphism $IN\to IN'$. ...


0

Let $f:R^n\rightarrow R^m$ be an isomorphism. Consider $\overline{f}:(R/M)^n\rightarrow (R/N)^m$ given by $\overline{f}([r])=f(r)$ (the map is clearly well-defined). Proceeding in an analogue way we can construct $\overline{f^{-1}}$. It is clear that $\overline{f^{-1}}$ is the inverse of $\overline{f}$.


1

In the unlikely case by semisimple you mean "has zero radical" (this is sometimes called "Jacobson semisimple" ) then yes, it's always true that $\mathrm{rad}(V/\mathrm{rad}(V))=\{0\}$. But if you mean "is a direct sum of simple modules" then (as PavelC has already pointed out) there are many counterexamples. The polynomial ring is nice, but I'll try to ...


2

For non-finite-dimensional algebras certainly not. Take $A=k[X]$ and $V=k[X],$ the regular module. Then $\mathrm{rad} V=0,$ basically since $k[X]$ is a PID, but $V$ is not semisimple (because, for example, all simple $A$-modules are easily seen to be finitely dimensional. However, $A$ is not, and if $A$ was semisimple, it would be a sum of finitely many ...


1

Any valuation domain $R$ satisfies the condition "P". If $I$ is a proper ideal of a valuation domain $R$, then $\sqrt I$ is a prime ideal. Let $a,b\in R$ such that $ab\in\sqrt I$. Then there is $n\ge 1$ such that $(ab)^n\in I$. If $a\mid b$ write $b=ax$ and note that $b^{2n}\in I$, so $b\in\sqrt I$. The conclusion: it is not possible to bound the ...


1

Associativity may or may not be an axiom, depending on your context. Requiring associativity results in a strictly smaller class of objects. While associative algebras are common, the study of nonassociative algebras is also full of important topics (Lie theory is a good example.) A basic example to keep in mind is $\Bbb R^3$ with the cross product. That ...


2

Yes, the trivial homomorphism. But apart from that, there is no natural homomorphism. Proof. Suppose that $\eta_{A,B} : A \otimes B \to A \times B$ is a natural homomorphism, meaning that $\eta_{A,B}$ defined for all $R$-modules $A,B$ and $\eta = (\eta_{A,B})_{A,B}$ is a natural transformation. Let $(r_1,r_2) := \eta_{R,R}(1 \otimes 1) \in R \times R$. If ...


3

You're right to be skeptical. As you've already noted at your other question, Artinian rings are ruled out from consideration since the Krull-Schmidt theorem can be used to prove they have the aforementioned property. Happily, I can demonstrate a ring without the property that is even von Neumann regular. I was inspired by a theorem in Goodearl's book von ...


1

Let $I = \langle(2,1+\sqrt{-5}),(1-\sqrt{-5},2)\rangle$. As you've probably seen, $2 \times 2 - (1+\sqrt{-5})(1-\sqrt{-5}) = -2 \neq 0$ so $I$ is a free $R$-module of rank $2$, and we even have an $R$-basis of $I$. More importantly this calculation also shows that $(0,2)$ and $(2,0)$ are in $I$ (just do the combination that cancels each component in the ...


2

Since then $D^i$ would be a finite dimensional $\Bbb R$ division algebra, the Frobenius theorem says $D^i$ has to be $\Bbb R$, $\Bbb C$ or $\Bbb H$. But $\Bbb R$ and $\Bbb H$ are not $\Bbb C$ algebras because their centers are both $\Bbb R$, and so neither center can contain a copy of $\Bbb C$. So $D^i=\Bbb C$ for every $i$. That means each matrix ring has ...


1

It is not the case that: $$ \int |g(x)| \, dx = \left|\int g(x) \, dx\right| $$ Instead, to get rid of the absolute values, we use the fact that: $$ |x - 1| = \begin{cases} x - 1 &\text{if } x \geq 1 \\ 1 - x &\text{if } x < 1 \end{cases} $$ We thus split the integral into two separate ones: $$ \int_{-5}^1 [(\tfrac{-1}{5}x + 7) - (2 + (1 - x))] \, ...


1

Hints: For (i): consider any ring $R$ with nonzero zero divisors. (In other words, the example you gave was fine. It sounds like you are forgetting to specify which ring the module is over, and that will be critical in deciding if it is free or not.) For (ii): consider any ring $R$ without zero divisors and with a nonprojective ideal $I$. $I$ is your ...


1

By the Artin-Wedderburn theorem, a semisimple ring is a direct product of simple rings, i.e. $R=\prod_{i=1}^n R_i$. The maximal ideals are of the form $M_j=\prod_{i=1}^n I_i$ where $I_i=R_i$ for $i\neq j$ and $I_j=\{0\}$ for a particular $j$, $1\leq j\leq n$. Thus the only semisimple rings with a unique maximal ideal are the simple ones.


2

The Krull-Schmidt theorem literally says: If you decompose a module with finite composition length into a direct sum of indecomposables in two ways, then the two decompositions are the same length, and there's some permutation that pairs up factors from each decomposition into isomorphic pairs. So the idea is that you take the two decompositions ...


1

Here's a more elementary explanation (from Exercise 19.7 in Eisenbud): Proposition: Let $(R,P)$ be an Artinian local ring, not a field. Then no submodule of $PR^n$ is free. Proof: $P$ is nilpotent in $R$, say $P^k = 0$, $P^{k-1} \ne 0$ for some $k > 1$, so any submodule of $PR^n$ is annihilated by $P^{k-1}$, hence has nontrivial annihilator, and thus ...


1

As you say, take any non-free module, for example $\mathbb{Z}/2$ over $\mathbb{Z}$. But even for free modules the claims (i) and (ii) are wrong. (i) $\{2,3\}$ is a generating set of the $\mathbb{Z}$-module $\mathbb{Z}$, but it does not contain a basis. (ii) $\{2\}$ is a linearly independent subset of the $\mathbb{Z}$-module $\mathbb{Z}$, but it cannot be ...


3

$R=k[x]/(x^n)$ is an artinian local ring. Let $M$ be an $R$-module having a finite free resolution, or equivalently $\operatorname{pd}_RM<\infty$. Now we can apply the Auslander-Buchsbaum formula and get $\operatorname{pd}_RM=0$, that is, $M$ is projective, hence free.


0

There is a simpler definition which is equivalent to the category of monoid objects in the category of $R$-bimodules. nLab calls them rings over $R$, and describes them with the "coslice category" or "under category" $R/\operatorname{Ring}$. An object in this category is a ring homomorphism $h: R \to A$ for some $A$, and a morphism from $h$ to $h': R \to ...


1

To start with, let me correct your claim: suppose $I$ is finitely generated, then there is a subset $S \subset I$ with $S=\{f_{1},...,f_{n}\}$ such that $\langle S\rangle=I$, where $f_i$ are polynomials in $I$. These polynomials need only finitely many generators, say $x_1,\dots,x_N$. That is, we can write $f_i=x_1g_{i1}+\cdots+x_Ng_{iN}$ for all ...


0

A finite subset of $A$ contains words, which are made of a finite number of the $x_i$s. Given such a finite set $S$, there is $i$ such that $x_i$ does not appear in any element of $S$. It follows that $x_i\not\in\langle S\rangle$, unless $S$ contains some non-zero element of $K$, in which case $\langle S\rangle=A$.


3

Let me show how a family of isomorphisms $\mathcal{C}(X,Z)\cong\mathcal{C}(Y,Z)$ natural in $Z$ gives us an isomorphism $Y\cong X$. I will assume that you are familiar with functors, in particular the $\operatorname{Hom}$-funcotrs in question and natural transformations, because otherwise discussing implications of Yoneda's Lemma won't be very fruitful. For ...


1

This is always the case. The obvious isomorphism of $\mathbb{Z}$-modules $$i^{*}(M \otimes_{\mathbb{Z}} N) \rightarrow i^{*}(M) \otimes_{\mathbb{Z}} i^{*}(N)$$ defined by $i^{*}(m \otimes n) \mapsto i^{*}(m) \otimes i^{*}(n)$ (really the identity map as $\mathbb{Z}$-modules) is always a map of $\mathbb{Z}[H]$-modules. This is true for modules over Hopf ...


1

Your proof is basically correct. But $\{x\}$ is not always minimal (look at $M=0$). Here is a quick proof: We prove by induction on $n$ that every module generated by $n$ elements has a minimal generating set taken from these elements. The case $n=0$ is clear, the minimal generating set is $\emptyset$. Now let $x_1,\dotsc,x_n$ be a generating set. If it is ...


1

The reason we are interested in algebras is that lots of the rings we are interested in are algebras! About your first question: yes. In fact, the condition that the morphism has image in the center is precisely that the left and right module structures coincide, and is in fact phrased in thay often.


0

(2nd question) Naively, a $k$-Algebra is a generalization of a (not necessarily commutative) polynomial ring with coefficients in a field, modulo relations.


3

your morphism is correct. For the surjectivity, if you take any $(m,s)\in M\oplus S$, take $f(m)+s'-f\circ\psi(s')$ where $s'$ is any preimage of $s$ under $g$. For injectivity, if $(\psi(n),g(n))=(0,0)$, then $n$ is in the image of $f$ because the sequence is exact, i.e. $n=f(m)$, but then $0=\psi (n)=\psi\circ f (m)=m$ and so $n=0$


0

I've put the question on MO, where Steven Landsburg provided a counter example: http://mathoverflow.net/questions/184387/lifting-a-direct-summand-of-a-free-module/184399#184399


0

Note that if $\mathbb Z/2\mathbb Z$ was free, it could only be generated by its only nonzero element $1$. However, it is not.


3

Clearly, the only candidate for a basis is $\{1\}$. However, is $\{1\}$ a linearly independent set? That is, is it true that for all $n \in \Bbb Z$, $n\cdot 1 = 0 \iff n = 0$?


0

Consider the short exact sequence of $\mathbb Z$-modules $$ 0 \to \mathbb Q \hookrightarrow \mathbb R \xrightarrow{\varphi} \mathbb R / \mathbb Q \to 0. $$ Since $\mathbb Q$ is injective, the sequence splits. Thus, there exists a homomorphism $\lambda: \mathbb R / \mathbb Q \to \mathbb R$ such that $\varphi \circ \lambda = \operatorname{id}_{\mathbb R / ...


2

Essentially you are asking how to write down a homomorphism of abelian groups $f : \mathbb{R} \to \mathbb{Q}$ such that $f|_{\mathbb{Q}}=\mathrm{id}_{\mathbb{Q}}$. Because then $\mathbb{R} = \mathbb{Q} \oplus \ker(f)$. Notice that $f$ is automatically $\mathbb{Q}$-linear. Existence of $f$ follows directly from linear algebra when one chooses a ...


1

HINT: Consider the $M_m(k)$ module $k^m$. Any $M_m(k)$ module is a direct sum of copies of $k^m$. The same is true for $M_n(k)$ and $k^n$. Under the morphism $M_n(k) \to M_m(k)$ $\ \ k^m$ becomes an $M_n(k)$ module. Hence it's a sum of copies of $k^n$. Now take $\dim_k$.


3

The definition of the tensor product is its universal property, which is quite simple. What you are struggling with is the construction of the tensor product - this is something different. If you want to see a construction of the tensor product which avoids free modules at all, see here. $K$ is by definition a submodule, since it is defined as the submodule ...


3

If $R$ is any ring and $M$ is a left $R$-module, then there is always an isomorphism of abelian groups $R \otimes_R M \cong M$ given by $r \otimes m \mapsto rm$ and $m \mapsto m \otimes 1$ in the other direction. In fact, $(r,m) \mapsto rm$ is $R$-balanced, hence induces a homomorphism $R \otimes_R M \to M$ of abelian groups. Clearly, $m \mapsto m \otimes ...


1

If $R$ has a unit, this is still true. $R$ is an $(R,R)$-bimodule, $M$ is a left $R$-module, therefore $R \otimes_R M$ is an $R$-left module with the action given by $r \cdot (x \otimes m) = rx \otimes m$. Define $f : M \to R \otimes_R M$ by $f(m) = 1 \otimes m$. Then $f$ is a morphism of $R$-left modules (it's obviously additive): $$f(r \cdot m) = 1 \otimes ...


1

You could of course make that definition, but you should call it something other than rank: if $P$ is any projective module that is not free, then it is not a direct summand of $R^{\mathrm{rank}(P)}$ (rank is additive and its complement would therefore be rank $0$ projective, hence $0$). For an explicit example, you have to start with $R$ something other ...


1

Let $A,B$ be $R$-modules. The direct sum $A\oplus B= \{(a,b) | a\in A, b\in B \}$ is a module under component wise operations: $(a_1,b_1)+(a_2,b_2)=(a_1+a_2,b_1+b_2)$ and $r(a,b)=(ra,rb)$. This extends to a direct sum of finitely many $R$-modules. However, for a direct sum of infinitely many $R$-modules, there is a further requirement that elements have ...


1

In the second definition, the $\oplus$ in the set is just a symbol - $a\oplus b$ is just another way of writing the pair $(a,b)$, which should tell you why this definition is the same as the one in your lecture notes. The first definition is slightly different because it's telling you how to define the direct sum of any (possibly infinite) set of modules. If ...


1

For finitely many summands everything agrees. Then $A\oplus B$ is what you wrote, and this holds in the same way for finitely many summands. For infinitely many summands there are differences, and we need to distinguish between $\oplus M_i$ and $\prod M_i$. This is expalined in books on rings and modules. A good example is the free $\mathbb{Z}$-module ...


2

Here's a basic argument without using any structure theorem (though I shall use that $K[x]$ is a PID). First off, $K[x]$ does not have any simple modules that are infinite dimensional over $K$. Any element$~v\neq0$ of a simple module generates the whole module, and if $P.v=0$ for some nonzero $P\in K[x]$, the generated module would be of dimension $\deg ...


2

If $V$ is finite dimensional then $V$ is a finitely generated torsion $K[x]$-module. Now apply the structure theorem and find that $V$ is a direct sum of cyclic $K[x]$-modules. Since $\chi_T$ is the product of all invariant factors and $m_T$ is the last of them, we can conclude that $V$ has only one invariant factor equal to $\chi_T=m_T$, so $V\simeq ...


1

This holds for any $A$-module: $$ M\otimes_A k(\mathfrak{m})\simeq M\otimes_A (A_{\mathfrak m}\otimes_{A_{\mathfrak m}}k(\mathfrak{m}))\simeq (M\otimes_A A_{\mathfrak m})\otimes_{A_{\mathfrak m}}k(\mathfrak{m})\simeq$$ $$M_{\mathfrak m}\otimes_{A_{\mathfrak m}}k(\mathfrak{m})\simeq M_\mathfrak{m}/\mathfrak{m}M_\mathfrak{m} $$ The first isomorphism follows ...


2

If $\mathfrak{p}$ is any prime ideal of $A$, we have $$ A_\mathfrak{p}/\mathfrak{p}A_\mathfrak{p} \cong \text{Frac}(A/\mathfrak{p}).$$ Let's assume this for now. In general, we have $$M_\mathfrak{p}/\mathfrak{p}M_\mathfrak{p} = (A_\mathfrak{p} \otimes_A M) / (\mathfrak{p} A_\mathfrak{p} \otimes_A M) = (A_\mathfrak{p} / \mathfrak{p}A_\mathfrak{p})\otimes_A ...


0

I found the following "definition" in Hoffman's Linear Algebra (2ed.): "If the set $S$ contains only finitely many vectors $\alpha_1,\ldots,\alpha_n$, we sometimes say that $\alpha_1,\ldots,\alpha_n$ are linearly independent instead of saying that $S$ is linearly independent." To be extremely pedantic, this definition (or convention) still has an issue ...


1

Denote $I=\ker f$. $R/I=M$ is simple, hence the only submodule (ideal) of $R/I$ is $0$ and itself. This show that $R/I$ is a field, so $I$ is maximal. If $I$ is maximal, then $M=R/I$ is a field hence simple.


1

It is an abuse of language. It means that the set $\{v, w\}$ is linearly independent. Since I took my time to answer here, might as well add something: there is a difference between sets and sequences. Notice that $\{v, v, w\} = \{v, w\} = \{w, v\}$, but $(v,w)$ has just one meaning. The set $\{v, v, w\}$ is linearly independent. The sequence $(v,v,w)$ is ...



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