# Tag Info

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It's just a question of terminology and usefulness. As far as terminology, there's the standard definition of linear independence and there's your definition. We need to pick one to be called "linearly independent" and the other needs to be called something else. It just so happens that the standard definition is the one we've picked. Of course this ...

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Assuming we are additionally given that $0\le m<N$, then, yes, it is possible to reveal $m$: Just try the candidates one by one and check if $(m+r)^e\bmod B$ turns out to be $C$. Mathematically, this solves the problem. Also, we have found an explicit algorithm, albeit with exponential running time ($O(N)$, which is linear in $N$, but exponential in the ...

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Direct summands of free $R$-modules are called projective $R$-modules, so I'd be using that terminology here. Let $(A, \mathfrak{m}, k)$ be a local ring with maximal ideal $\mathfrak{m}$ and residue field $k = A/\mathfrak{m}$. Let $M$ be a finitely generated projective $A$-module. Pick a minimal generating set of cardinality $n$ for $M$ (you can do this by ...

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This isn't true. Consider a finite Galois extension $K$ of $\mathbf{Q}$ in which the prime $p$ is inert. Look at the morphism $\mathrm{Spec}(\mathscr{O}_K)\to\mathrm{Spec}(\mathbf{Z})$, which is finite. The fiber over $(p)$ is the spectrum of $\mathscr{O}_K\otimes_\mathbf{Z}\mathbf{F}_p=\mathscr{O}_K/p\mathscr{O}_K$. The dimension of ...

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We can regard $M[x]$ as an $R$-algebra rather than a $R$-module. Change the word 'module' to 'algebra' in your description, then you get the universal property of $M[x]$.

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Assume you have relations $\sum_j a_j r_{j,l} = 0$ with $a_j \in P$, $r_{j,l} \in R$, $1 \le j \le n$, and $1 \le l \le p$, as in Theorem 4.24(3). Let $e_j$ denote the $j$-th standard unit vector of $R^n$. Now let $K$ be the submodule of $R^n$ generated by $\{ \sum_{j} e_j r_{j,l} \mid 1 \le l \le p\}$. Set $M=R^n/K$ and let $\lambda \colon M \to P$ be ...

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This answer uses the equivalent characterizations of injective indecomposibles from Theorem 3.52 in T.Y. Lam, Lectures on Modules and Rings. In particular, it follows from this theorem that $M$ is uniform and that $M$ is the injective envelope of each of its nonzero submodules. Since $P$ is an associated prime of $M$, there exists a nonzero submodule $N ... 1 Yes, unless$V$is trivial. I'll help you unpack the definitions, but I'll leave the meat of the problem to you. Observation. Let$V$denote a$k$-module. Then$V$is a simple$\text{End}_k V$-module iff:$V$has at least two distinct$k$-submodules, Every non-trivial$k$-submodule of$V$that is closed under the action of$\text{End}_k V$... 1 Hint: The answer is yes (as is often the case for finite dimensional vector spaces). Note that for any$v \in V \setminus \{0\}$, we may select maps$T_1,\dots,T_n \in A$so that$\{T_j(v)\}$forms a basis (or a spanning set, if you prefer). 1 Yes, the inclusion of a factor into a direct product$M \to M \oplus N$defined by$m \mapsto (m, 0)$is always an injective homomorphism no matter what modules you choose for$M$and$N$. In particular, you can inject$M \to M \oplus R[G]^n$not just for some$n$, but in fact for any$n$. If$M$and$N$are$G\$-modules then it would be a good exercise for ...

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