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## Hot answers tagged modules

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

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

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

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

The first element in our basis should be such that $kv_1 = (6,9)$, and such that the coordinates are reduced to lowest terms: so $(2,3)$ is the natural choice. Now we want to expand this to a basis: e.g. we want to find $a,b$ such that the matrix $$\begin{pmatrix} 2 & a \\ 3 & b \end{pmatrix}$$ has determinant 1 ( or -1, we just want the determinant ...

2

Since $_SN$ is projective, there is $N'$ such that $N\oplus N'\cong S^{(I)}$ is a free module. Then we can tensor: $$M\otimes_SS^{(I)}\cong (M\otimes_SN)\oplus(M\otimes_SN')$$ as $R$-modules. Now $$M\otimes_SS^{(I)}\cong M^{(I)}$$ as $R$-modules. Since $M$ is projective, every direct power of it is projective as well. So $M\otimes_SN$ is a direct summand ...

2

Your edit is basically the right idea but notationally there's a problem. You can't say every $m \in M$ can be written as $m' + u$ with $m' \in M/U$ and $u \in U$ because $M/U$ is not a submodule of $M$. What you want to say is that $(v_1 + U, \ldots, v_m + U)$ are generators for $M/U$ so given $m \in M$ you first push it $M/U$ to get $m + U$. Then write ...

2

Let $\{0\}=N_0<N_1<\cdots<N_n=N$ be a composition series for $N$. In particular, $N_i/N_{i-1}$ is a cyclic module, so $N$ can be generated by (at most) $n$ elements. Then there is a surjection $R^n\to N$, and tensorizing by $M$ we get $M\otimes_R R^n\to M\otimes_RN\to 0$. But $M\otimes_R R^n\simeq M^n$, so $M\otimes_R R^n$ has finite length which is ...

1

We have $$\mathbb{Z}^{3}/M \mathbb{Z}^{3}\simeq(\mathbb{Z}\oplus\mathbb Z\oplus\mathbb Z)/(\mathbb Z\oplus\mathbb Z\oplus 32\mathbb Z)\simeq\mathbb Z/32\mathbb Z,$$ so your guess is correct.

1

Each $x\in X_1$ can be written as a $K$-linear combination of elements of $X_2$. If we multiply away denominator, we see that $cx\in X_2$ for suitable $0\ne c\in A$. If $x_1,\ldots,x_n$ generate $X_1$ and lead to corresponding factors $c_1,\ldots, c_n$ with $c_ix_i\in X_2$, then let $c=c_1\cdots c_n$ and find that $cX_1\subseteq X_2$. Then $X_1/X_2$ is a ...

1

Suppose $\mathrm{Ann}(r_2rm)\not\subseteq Q$. Then there is $a\in \mathrm{Ann}(r_2rm)$, $a\notin Q$. Now $\mathrm{Ann}(arm)\subseteq Q$: $b\in\mathrm{Ann}(arm)\Rightarrow b(arm)=0\Rightarrow (ba)rm=0\Rightarrow ba\in P\Rightarrow ba\in Q\Rightarrow b\in Q$. Since $P\subseteq\mathrm{Ann}(arm)$ and $P$ is maximal with some properties that $\mathrm{Ann}(arm)$ ...

1

$am=0\Rightarrow an_1+an_2=0\Rightarrow an_1=an_2=0$ (why?). So $\mathrm{Ann}(m)=\mathrm{Ann}(n_1)\cap\mathrm{Ann}(n_2)$. Moreover, $N_i=Rn_i$ for $i=1,2$. We have $R/\mathrm{Ann}(m)\simeq Rm$, $Rm=N_1\dotplus N_2$, and $N_1\dotplus N_2\simeq R/\mathrm{Ann}(n_1)\oplus R/\mathrm{Ann}(n_2)$, therefore $$R/\mathrm{Ann}(n_1)\cap\mathrm{Ann}(n_2)\simeq ... 1 An answer to your last question: If \mathbb Z \to N is onto, then any homomorphism f:\mathbb Q \to N is easily shown to be trivial. In particular we can always chose h = 0 (the only homomorphism \mathbb Q \to \mathbb Z anyway) to satisfy g \circ h = f. This shows (a boring fact, since the involved maps are trivial, so nothing happens here) that ... 1 R is obviously generated by 1 as an R-module. The ideal (submodule) (x_1,x_2,\ldots) of R is not finitely generated. For otherwise (x_1,x_2,\ldots)=(f_1,\ldots,f_k) for some f_1,\ldots,f_k\in R, and every f_i involves only finitely many indeterminates. 1 Since 3^2\equiv -1\pmod {10}, one has$$\begin{align}5^{31}\cdot 2^{789}-23^{23}&\equiv0-3^{23}\\&\equiv -(3^2)^{11}\cdot 3\\&\equiv -(-1)^{11}\cdot 3\\&\equiv -(-1)\cdot 3\\&\equiv 3.\end{align}

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Hint: $A\oplus B$ is equipped with inclusions $\iota_A,\iota_B\colon A,B\to A\oplus B$ and projections $\pi_A,\pi_B\colon A\oplus B\to A,B$ such that $\pi_A\circ\iota_A=\operatorname{id}_A$, $\pi_B\circ\iota_B=\operatorname{id}_B$.

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Since the map $\phi_m$ is surjective we have $M=Rm$ for all $m\in M-\{0\}$. Suppose $\mathrm{Ann}(m)$ is not left maximal for some $m\in M-\{0\}$. Then there is $\mathrm{Ann}(m)\subsetneq I\subsetneq R$ a left ideal. Let $a\in I-\mathrm{Ann}(m)$. Then $am\ne 0$, and $M=R(am)$. Since $M=Rm$ we get $m\in R(am)$, so there is $r\in R$ such that $m=ram$, that is, ...

1

Choose generators of $P$ to get a surjection from a free module $F$ onto $P$. This gives you a split (by your definition!) exact sequence $0 \to K \to F \to P \to 0$, showing $F = K \oplus P$.

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