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7

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 ...


5

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 ...


4

This is false. Taking these two facts: tensoring $- ⊗_R M$ is exact iff M is a flat module a module over a PID is flat iff it is torsion free we see that $- ⊗ \mathbb Q : \mathrm{Ab} → \mathrm{Ab}$ is an exact functor, but it doesn't preserve free objects: $\mathbb Z ⊗ \mathbb Q = \mathbb Q$, and $\mathbb Q$ is not free over $\mathbb Z$ (any two elements ...


3

If $R$ is any ring and $n \geq 1$, then the evident map $M_n(R) \to \bigoplus_{i=1}^{n} R^n$ which decomposes a matrix into its columns is an isomorphism of left $M_n(R)$-modules. Hence, $R^n$ is a projective left $M_n(R)$-module. It is usually not free. If $R$ is commutative and non-zero (more generally, when $R$ has IBN), then free $M_n(R)$-modules of rank ...


3

$M_2(\mathbb{C})$ is a semisimple ring, so every module over this ring is projective and injective. Moreover, $\mathbb C^2$ is a simple $M_2(\mathbb{C})$-module, and if it is free it must be isomorphic to $M_2(\mathbb{C})$. In fact, $M_2(\mathbb{C})\simeq \mathbb C^2\oplus\mathbb C^2$, and therefore by taking the ranks we get a contradiction.


2

It's not an isomorphism, it's equality! The module $\mathfrak{a}(M/N)$ is generated by the elements of the form $a(x+N)$, for $a\in\mathfrak{a}$ and $x\in M$. Clearly $a(x+N)=ax+N\in (\mathfrak{a}M+N)/N$, so one inclusion is settled up. Conversely, an element in $(\mathfrak{a}M+N)/N$ is of the form $$ y+z+N $$ for some $y\in\mathfrak{a}M$ and $z\in N$. ...


2

Let us try this : $$\mathfrak{a}M+N=M\Rightarrow(\mathfrak{a}M+N)/N=M/N $$ Now, we always have $\mathfrak{a}(M/N)\subseteq M/N$ so the only thing to show to get the equality is to show the reverse inclusion, take $m_0\in M$ then there exists $a\in\mathfrak{a}$, $m\in M$ and $n\in N$ such that : $$m_0=am+n$$ Now modulo $N$ we have $[m_0]=[am]$ hence ...


1

It is because $A\simeq\operatorname{Im}f$ and tensoring an isomorphism always gives an isomorphism. Hence it's equivalent to prove that tensoring the inclusion morphism or tensoring the original injective morphism results in an injective morphism. Btw, any rings of fractions is a flat $A$-module.


1

$aL=0$ and $bN=0$ implies $(ab)x=0$ for all $x\in M$:


1

Since $I$ is nonzero, choose $0 \neq y \in I$. Then $y \cdot \bar x$ is a nontrivial linear combination of $\bar x$, so should be nonzero because $\{\bar x\} \subset R/I$ is independent. But this is absurd, because $yx \in I$ so $y \cdot \bar x = \overline{yx} = 0 \in R/I$.


1

Linear combinations are themselves by definition so that almost all summands are zero. For an algebraist the sum of infinitely many nonzero summands is not really a sum. Also, the infinite variant $R^X$ is certainly not free over the given set. As such it would have to be true that for any other module $A$ and map $f\colon X\to A$ there exists a unique ...


1

Consider $R = \mathbb{Z}$ and $A=C= \mathbb{Z}/(p)$ and $B=\mathbb{Z}/(p^2)$ and the short exact sequence $$0 \to \mathbb{Z}/(p) \to \mathbb{Z}/(p^2) \to \mathbb{Z}/(p) \to 0$$ where the first morphism is given by $1\mapsto p$, the second is the projection mod$(p)$. Clearly for $p$ prime, this does not split, since $\mathbb{Z}/(p^2)$ is not isomorphic to ...


1

As $M$ is cyclic, there is an epimorphism $R \rightarrow M$. Because $M$ is assumed to be projective, it is thus isomorphic to a direct summand of $R$. Note that every direct summand of $R$ is of the form $Re$ where $e \in R$ is an idempotent. Now we have $(1-e)e = 0$ and as $R$ was assumed to be an integral domain, it follows that $e = 1$ or $e = 0$, that ...


1

This is true because a projective module over an integral domain is torsion-free (direct summand of a free module), hence its annihilator is $0$.


1

I'm afraid not. The invariant basis number or IBN property is not satisfied by all rings. It is trueu for the so-called stably free rings, for which, given two $n\times n$ matrices, $AB=I_n\Rightarrow BA=I_n$. Among stably free rings are commutative rings, noetherian rings, artinian rings and semi-local rings. An example of a ring that does not satisfy ...



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