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3

No, the tensor product isn't zero. Be more careful about the definition. The way tensor product is constructed is to take the free group generated by elements of $A \times B$ and what we really want is to have identities such as $(a+a',b)=(a,b)+(a',b)$ so we take the subgroup generated by those three above types of elements and then taking quotient would ...


2

$\DeclareMathOperator{\h}{Hom}$ $S^{-1}M$ is characterized by the universal property $\h_{S^{-1}A}(S^{-1}M,T) \cong \h_A(M,T_{|A})$. In order to show that $L := \lim_{s \in S} M_s$ satisfies this property there are two things to do: Each $s \in S$ acts as an automorphism on $L$ (so $L$ carries a natural $S^{-1}A$-module structure). For each $A$-Module $T$ ...


2

If $F$ is free of rank one over $R$ then $F=R\cdot1=R$ and $R$ was a field to begin with. Otherwise suppose it is free of rank greater than one, so it has at least two summands which are copies of $R$ which must be generated by some fractions, which means $R\frac{a}{b}\cap R\frac{c}{d}=0$ must hold for some values $a,b,c,d\in R$... is that possible?


2

Sure, it is always true that a free module (under this definition) will be isomorphic to $A^{(I)}$ for some index set $I$, no matter if the set $I$ is finite or infinite. Suppose you are given $M=\bigoplus_{i\in I} M_i$ with isomorphisms $\phi_i:M_i\to A$. This can be composed with the map that injects $A$ into the $i$th position in $\bigoplus_{i\in I}A$. ...


2

Hint: prove it isn't generated by a single element. Then prove that any two elements are linearly dependent. More details: You can show that $M$ is not generated by a single generator using the fact that $x$ and $y$ are irreducible in the UFD $A$. If $f$ generates $M$ then $f$ divides $x$ and also $f$ divides $y$. Since these are distinct irreducible ...


2

The problem is simple: if $A$ and $B$ are both left $R$-modules, we'd want to equate $(ra,b)$ with $(a,rb)$ in the tensor product. But then we'd have $(rsa,b)=(a,rsb)$ and $(rsa,b)=(sa,rb)=(a,srb)$! So this construction forces $R$ to act commutatively on $A\otimes B$. In general the tensor product of a right and a left $R$-module is merely an abelian group. ...


1

For the second question, note that $x^4, y^4$ is a SOP. Since $R=k[x^4, x^3y,xy^3, y^4]$ is a domain, $x^4$ is a non-zero divisor on $R$. But observe that $y^4$ is a zero divisor on $R/(x^4)$ because $y^4(x^3y)^2=x^4(xy^3)^2$. Ofcourse one needs to note that $(x^3y)^2\notin (x^4)$. Hence $R$ is not CM. You already did the first part, perhaps you can now ...


1

You're being misled by the notation. The module $F$ is not $A\times B$, but the free $R$-module with the set $A\times B$ as basis. So $F$ is the set of “formal expressions“ of the form $$ r_1(a_1,b_1)+r_2(a_2,b_2)+\dots+r_k(a_k,b_k) $$ where $k$ is any natural number, $a_i\in A$, $b_i\in B$, $r_i\in R$ (for $i=1,2,\dots,k$). If $k=0$ the formal sum is empty ...


1

Consider a field $F$ and the ring $R=F\times F$. Let $M=\{0\}\times F$ with both the ordinary right $R$-module structure $(0,m)(r,s)=(0,ms)$, and let $M'$ be the same set with another $R$-module structure given by $(0,m)(r,s)=(0,mr)$. This second structure is just given by the involution on $R$ given by $(r,s)\mapsto(s,r)$. The annihilator in $R$ of $M$ ...



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