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## Hot answers tagged modules

7

Please see the following paper by Bhargava and Satriano here, which contains the details of this computation and other related results. M. Bhargava and M. Satriano, On a notion of “Galois closure” for extensions of rings, Journal of the European Mathematical Society 16, 1881-1913 (2014). The aim of the authors is to define for commutative ring ...

3

Let $\tilde g\colon R\longrightarrow A$ be an extension of $g\colon L\longrightarrow A$ to $R$ and let $r\in L$. We have $$g(r)=\tilde g(r) = \tilde g(r\cdot 1)=r\tilde g(1),$$ so just set $a=\tilde g(1)$.

3

What's an element of $R[x]$? It's a finite sum that looks like this: $$a_0+a_1x + \cdots + a_n x^n,$$ where the $a_i \in R$. If we forget about the variable $x$ and note that all that really matters in this description is the coefficients $a_i$ and the order they appear in, we realize that this corresponds in a bijective fashion to an ordered tuple (with ...

2

A module is an abelian group, and so every subgroup (and therefore any submodule) is a normal subgroup. Further, it's easy to show that scalar multiplication is well-defined in the quotient module, and so we don't need any added conditions. For your second question, that condition (that only finitely many $x_i$ are nonzero) is basically saying it contains ...

2

You ask: wouldn't it make the most sense to just use the monoidal product from the "mother category"? but this question does not make sense, really. To judge if something makes more sense or less sense than something else, you have to spell out what you are trying to achive. Sometimes, the direct sum is the correct operation to turn modules into a ...

2

Every object in an additive category is a group and a cogroup, canonically, with respect to the monoidal structure of direct sum. This reflects the fact that vector spaces, and more generally, modules, are actually groups. So there's nothing there. Hopf algebras weren't invented to cause pain, much less because of the desire to decorate the monoidal ...

2

First of all, the quotient $M/\ker f$ may not be free -- consider $\mathbb{Z} \to \mathbb{Z}/2\mathbb{Z}$. The criterion relating injectivity of $f$ to linear independence of the columns of $A$ works in the case where both source and target are free, but not necessarily otherwise. If the quotient $M/\ker f$ is free, the matrix will not necessarily be the ...

2

Your question is how to compute the matrix exponential $\exp(a)$ where $a$ is an element of a finite-dimensional real associative algebra $A$. (Your phrasing is not more general: you're implicitly working in the algebra $M_n(A)$.) This reduces to the case of real matrices, because the left regular representation $$A \ni a \mapsto \left( L_a : b \mapsto ab \... 2 2a + 3b\in 6 \mathbb Z \subseteq 3 \mathbb Z implies a \in 3 \mathbb Z. 2a + 3b\in 6 \mathbb Z \subseteq 2 \mathbb Z implies b \in 2 \mathbb Z. Thus, 2a + 3b\in 6 \mathbb Z  iff a \in 3 \mathbb Z and b \in 2 \mathbb Z and the kernel is  3 \mathbb Z \times 2 \mathbb Z. 2 The construction you describe is that of a free R-module. It can be made precise. Let X be any set (it need not be finite, but then the following finiteness condition is superflous): Let the elements of M be functions f : X \to R (families (f_x)_{x\in X}) , such that \{f_x\neq 0 : x\in X\} is finite. Then define the module structure in the ... 1 In the second case,you can say that a simple module is a finite extension of k. 1 When b is odd 3b will also be odd and so adding an even number 2a won't make 2a+3b a multiple of 6. When b is even 3b will be a multiple of 6 and hence, for 2a+3b to be multiple of 6 we need 2a to be a multiple of 6 which is the same as a being a multiple of 3. So pairs of numbers of the form (a,b)=(3x, 2y) form the kernel ... 1 If you have a basis S for P, we know that \operatorname{Hom}(P,C)\simeq C^S, and similarly \operatorname{Hom}(P,B)\simeq B^S. Now if for each s\in S, you choose an element b_s\in B such that \;f(b_s)=g(s) (we're using the axiom of choice here if S is not finite), the family \;(b_s)_{s\in S}\in B^S defines a homomorphism h from P to ... 1 We show the implications (a) \Rightarrow (b) \Rightarrow (c) \Rightarrow (a). Assume (a) and let e be the unit of R. Let S be any ring such that R\subseteq S is an ideal. For any s\in S, we have se, es\in R and in fact they are equal:$$ es = e(es) = (es)e = e(se) = (se)e = se,  where for the second and fourth equality we have used that $... 1 It is infact never true that$\mathfrak{m}[x]\subseteq R[x]$! All of these questions become simple when you consider the following fact. An ideal$\mathfrak{p}\subseteq R$is prime if and only if$R/\mathfrak{p}$is an integral domain. Moreover$\mathfrak{p}$is maximal if and only if$R/\mathfrak{p}$is a field. The first statement is simple, the second ... 1 Additionally, you will see in category theory (if you go down that road) that direct products and direct sums satisfy different universal mapping properties: https://en.wikipedia.org/wiki/Direct_sum_of_modules As you notice in the Wiki page, the direct sum of modules is a$coproduct$, meaning it satisfies the universal mapping property$opposite\$ of that of ...

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