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4

The ideal $R$ is always generated by the element $1$ as an ideal, for any ring. This is no surprise and has nothing to do with the noetherian property. Even if $R$ was a finitely generated $k$-algebra where $k$ is a field, it is not guaranteed that the number of generators for $R$ as a $k$-algebra is an upper bound on the number of generators of an ideal. ...


4

Let $R$ be the ring of smooth functions on $S^2$. Then $R^3 \cong \operatorname{Vect}(S^2)\oplus R$ where $\operatorname{Vect}(S^2)$ is the $R$-module of smooth vector fields on $S^2$. $\operatorname{Vect}(S^2)$ is not isomorphic to $R^2$ by the hairy-ball theorem, and the standard smooth embedding $S^2\rightarrow \Bbb R^3$ gives the first isomorphism. ...


4

According to Definition 7.1.3 in Borceux's book, the left Kan extension of a bimodule $N : A \to C$ along a bimodule $M : A \to B$ is an initial bimodule $L : B \to C$ equipped with a bimodule map $\alpha : N \to L \circ M$, where $L \circ M : A \to C$ is the bimodule $M \otimes_B L$. In other words, this is a representation of the functor $\hom(N,M ...


4

The described method where you interleave the rows gives a well-defined bijection, because a column has only finitely many nonzero entries exactly when it has both only finitely many nonzero odd entries and only finitely many nonzero even entries. The issue is that this bijection does not respect the action of $R$, assuming that the action is on the left. ...


3

Use Baer's criterion: Ideals in $R/aR$ correspond to divisors of $a$, so you have to show the following: For any $d \in R$ with $d|a$ any $R/aR$-homomorphism $f:dR/aR \to R/aR$ extends to a $R/aR$-homomorphism $R/aR \to R/aR$. A homomorphism $f:dR/aR \to R/aR$ is given by the image of $d$ and since we have $\frac{a}{d}d=0$ in $dR/aR$, the image $f(d)$ must ...


3

If $M$ is a simple module over $\mathbb{Z}$ then it is generated by an element. In particular, $M$ is a cyclic group. The modules over $\mathbb{Z}$ are precisely abelian groups, and every non-cyclic abelian group has a non-trivial subgroup (which are therefore submodules). Thus, the semisimple modules are direct sum of prime cyclics.


3

Without using any topology, you can take $R$ the ring of endomorphisms of an infinite-dimensional vector space $V$, say over $\mathbb{R}$. Then as a (left) module over itself, $R=R\oplus 0$ is isomorphic to $R\oplus R$.


3

It might be ambiguous, but I would usually parse it as a set of generators as an $R$-algebra, i.e. that every element of $A$ can be written as a polynomial in the elements of $X$ with coefficients in $R$. For instance, $\{x\}$ generates $\Bbb Z[x]$ as a $\Bbb Z$-algebra, whereas you would need something like $\{1, x, x^2, \ldots \}$ to get a set of ...


3

This already fails for $I$ a singleton: the dimension of $k[x]$ as a vector space over $k$ is countable, while the dual $$ \operatorname{Hom}_{k\text{-Mod}}(k[x],k) $$ has uncountable dimension. See Slick proof? A vector space has the same dimension as its dual if and only if it is finite dimensional on MathOverflow. So the result above is general for any ...


3

Zorn's lemma says that given any partially ordered set $S$ such that every chain in $S$ is bounded above in $S$, $S$ has a maximal element. Your $S$ is $\{B_i\}$, and there's no reason the chain $S$ should have an upper bound in $S$: claiming $A$ as such a bound is claiming $A\in S$, which is the conclusion you're trying to prove.


3

The point is that in contrast to a short exact sequence, a split short exact sequence can be viewed as a certain kind of diagram with additive commutativity relations: Definition. A sequence $A\xrightarrow{i} B\xrightarrow{\pi} C$ is split short exact if there exist $B\xrightarrow{r} A$, $C\xrightarrow{\sigma} B$ such that $$ri=\text{id}_A,\quad ...


3

More generally, for any PID $R$ and every non-zero element $e \in R$, the ring $R/(e)$ is self-injective: Baer's criterion implies that, if $S$ is a commutative ring in which every ideal is principal, an $S$-module $M$ is injective if and only if for all $a \in S$, $m \in M$ with $\mathrm{Ann}(a) \subseteq \mathrm{Ann}(m)$ we have $m \in aM$. This can be ...


3

All modules involved are cyclic (generated by one element), so it is sufficient to define maps on generators and extend them by linearity. Let $f: (p^i)/(p^n) \longrightarrow R/(p^n)$ be $R/(p^n)$-linear, with $1 \leq i \leq n$. Note that (the following is $\mod{(p^i)}$) $$0=f(0) = f(p^n)=f(p^{n-i}p^i)=p^{n-i}f(p^i)$$ this means that $f(p^i) = ap^i$ for ...


3

Suppose $R=I\oplus J$, where $I$ and $J$ are ideals of $R$, with $I\cap J=\{0\}$. Then $$ 1=x+y $$ with $x\in I$ and $y\in J$. It follows that $x=x(x+y)=x^2+xy$. Since $xy\in J$ and $xy=x^2-x\in I$, we have $xy=0$. Since $R$ is a domain, we have either $x=0$ or $y=0$. In the first case $y=1$ and $J=R$, in the second case $x=1$ and $I=R$. A finite direct ...


2

Suppose that $R=M\oplus M'$ with $M\neq 0\neq M'$. Then $(m,0)\cdot(0,m')=0$ for $m\neq 0\neq m'$ which is a contradiction to $R$ being an integral domain. Let $e\in R$ be idempotent, that is to say $e^2=e$ and $e\neq 1$. (For example $(1,0)\in K^2$ for any field $K$). Then $R=R(1-e)\oplus Re$ (easy exercise). In our example this leads to $R=K\oplus K$ as ...


2

The only way to get an exact sequence there is to have $g$ = reduction mod $2$. "Multiplication by 2" is the description of $f$, not $g$.


2

Since the submodule is torsion free, the mapping from $R$ to $\langle h \rangle$ by $r \mapsto rh$ is an isomorphism.


2

A counterexample is $\mathbb{Q}$ considered as a $\mathbb{Z}$-module. Definitely torsion-free but not free.


2

Just do it with two simultaneous inequalities: $$ ||x-2| - 3| > 1 \Leftrightarrow |x-2|-3 > 1 \text{ or } |x-2|-3 < -1 $$ $$ \Leftrightarrow |x-2| > 4 \text{ or } |x-2| < 2 $$ $$ \Leftrightarrow x-2 > 4 \text{ or } x-2 < -4 \text{ or } -2 < x-2 < 2 $$ $$ \Leftrightarrow x>6 \text{ or } x < -2 \text{ or } 0 < x < 4 $$ So ...


2

Let $R = \mathbb F_p[x, y]$ and let $A$ and $B$ be $R/x$ and $R/y$ respectively.


2

It depends on the ring. For instance, if $R=\mathbb{Z}_{(p)}$ is the localization of $\mathbb{Z}$ at the prime ideal $p$, then the maximal ideal is principal, so isomorphic as a module to the ring itself, hence projective. In case the ring is $\mathbb{Z}/4\mathbb{Z}$, the maximal ideal is not projective. Added from comment. If your aim is to discuss ...


2

Over a local commutative ring, projective modules coincide with free modules, so the question is whether $\mathfrak m$ is free. If it is free it must be of rank one, because two elements of $a,b\in A$ are necessarily $A$-linearly dependent: $a\cdot b-b\cdot a=0$ (duh!) Freeness of dimension one means that for some $m\in \mathfrak m$ the $A$-linear map ...


2

In general, no. For instance, let $A$ be a field, let $$M = N = \langle v_1, v_2, v_3, v_4 \rangle$$ be a $4$-dimensional vector space over $A$, and let $$ P = (M \otimes N) / \langle v_1 \otimes v_2 + v_3 \otimes v_4\rangle, $$ with $f$ the natural map. No non-zero decomposable tensor lies in the kernel $$\ker(f) = \langle v_1 \otimes v_2 + v_3 \otimes ...


2

Only if part: Suppose $N=f(M)\bigoplus N'$. Any vector $n\in N$ has a unique decomposition $\,n=f(m)+n'$; $m$ is unique because $f$ is injective. Define $\alpha(n)=m$; as the decomposition and $m$ are unique, it is easy to check $\alpha$ is linear. More over, for the same reason, $\alpha(f(m))=m$. If part: Suppose such an $\alpha$ exists. Then ...


2

Let $B$ be an $A$-algebra and $M$, $N$ be $B$-modules. Here is an example in which $M \otimes_A N$ and $N \otimes_A M$ are isomorphic as $A$-modules, but not as $B$-modules. $\mathbb Z[x]$ is a $\mathbb Z$-module via the inclusion $\mathbb Z \hookrightarrow \mathbb Z[x]$. Let $M = \mathbb Z[x]$ be a $\mathbb Z[x]$-module via the identity. Let $N = \mathbb ...


2

The ring $B\otimes _A B$ has a canonical $A$-algebra structure and TWO different structures of $B$-algebra, which I'll call $(B\otimes _A B)_l$ and $(B\otimes _A B)_r$, according as multiplication by elements of $B$ happens on the left or on the right. These $B$-algebra structures are in general different since if we denote by $\bullet $ and $\circ$ the ...


2

Consider the $A$-module (or $R$-module) epimorphism $$I\oplus J \rightarrow I+J=A, \;\; (i, j)\mapsto i+j \in A,$$ compute its kernel and use the first isomorphism theorem. Now, if you know that $A$ is a projective $A$-module and know what properties projective modules have, this should give you the isomorphism you are looking for. If not, the isomorphism ...


2

The issue here is that you need to know what you mean by "unique". You can show that such an object $L$ exists ; just take it to be the submodule of $M$ of elements mapped to $0$ under $f$, and check the two conditions. For "unicity", if you have two modules $L_1,L_2$ that satisfy property (i) and (ii), you cannot hope to show that $L_1 = L_2$, but you can ...


2

You have to prove that if $r\equiv s \mod I$, then $rm=sm$ for any $m\in $M$. That is equivalent to $(r-s)m=0$, which is by definition since $r\equiv s\mod I\iff r-s\in I\subseteq\operatorname{Ann}_AM$.


2

This is kind of a silly example: $\mathbb{Z} \subseteq \mathbb{Q}$, but the Krull dimension of $\mathbb{Z}$ is one while the Krull dimension of $\mathbb{Q}$ is 0.



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