# Tag Info

5

It does. If $C$ is any chain complex in your additive category and $F = 0$ the zero functor, then $H_n(FC) \cong F\bigl(H_n(C)\bigr)$ for each $n \in \mathbf Z$, as both sides are zero.

5

The eigenvalues of a matrix $A$ over a field $k$ are the roots of the characteristic polynomial $\chi(A)=\det(\lambda I-A)$. But what is this thing we're taking the determinant of? $\lambda$ is not an element of $k$, but an indeterminate, so to properly describe $\lambda I-A$ we ought to say its coefficients, e.g. $2-\lambda$, are elements of the polynomial ...

4

The problem with your argument is quite subtle: you can't say $\pi f(m/1)\in N$, because the canonical homomorphism $i:N\to S^{-1}N$ may not be injective (because elements of $S$ might annihilate elements of $N$). That is, there is always some element $n\in N$ such that $i(n)=\pi f(m/1)$, but that $n$ might not be unique, and it is not clear that you can ...

3

To explain what I said on MO (and also explaining what goes on in knsam's answer): The way to think about this is that since the group permutes the given basis vectors, it fixes the sum of all the given basis vectors. This gives a $1$-dimensional invariant submodule.

3

You're making things too complicate. ;-) Let $f\colon R/I\to R/J$ be a homomorphism. Consider the composition map $g=f\circ \pi\colon R\to R/J$ and write $g(1)=x+J$. Then, for every $r\in R$, $g(r)=g(1r)=g(1)r=xr+J$. Since $g(r)=0$, for every $r\in I$, we know that $xI\subseteq J$, so that $x\in (J\mathbin:I)$. Now, ...

2

Hint. Arguably, the simplest invariant subspace would be one of dimension $1$. What would such a thing be? Do you see any such subspace in this case?

2

$(x,y)$ is a regular sequence in $R=\mathbf Q[x,y]$ , hence the Koszul complex: \begin{alignat*}{3}0\longrightarrow R&\xrightarrow{\begin{bmatrix}x\!&\!\!y\end{bmatrix}} R^2&\xrightarrow{\smash[t]{\begin{bmatrix}-y\\x\end{bmatrix}}}&R\longrightarrow R/(x,y)\longrightarrow 0 \\ t&\longmapsto \rlap{(xt, yt)}\\ &(u,v)&\longmapsto ...

2

It seems like, by $K$-module, you are really talking about vector spaces, and they are isomorphic if and only if they each have a basis of the same cardinality. The size of the basis for $K[x]$ is easily seen to be the cardinality of the natural numbers. Can the basis of $K[[x]]$ be the cardinality of the natural numbers? It's not clear why you are ...

1

A normalized vector (more commonly known as a unit vector) has norm 1. The reason why you don't have exactly one is that you're using finite precision on a computer ($\sqrt{25+49}$ isn't 8.6. It's $8.6023252670...$).

1

A normalized vector has unit norm, that's the definition of being normalized, and it also follows from your definition. Let's check (in symbols): $$\def\abs#1{\left|#1\right|}\abs{\def\n{\mathord{\rm normalized}}\n(a)} = \sqrt{\n(a).x^2 + \n(a).y^2} = \sqrt{\frac{a.x^2}{\abs a^2} + \frac{a.y^2}{\abs a^2}} = \sqrt{\frac{a.x^2 + a.y^2}{\abs a^2}} = 1$$ The ...

1

An ideal has to be a subgroup (of the additive group of the ring) to begin with. So if you prove that all subgroups are already ideals, you are done. PS Just correct the statement every sub-group is $mZ_n$ for $0<m<n$ to every subgroup is of the form $m Z_{n}$ for $m \mid n$.

1

In general, the answer is no. See here for example. There it is shown that $\mathbb {Z}^{\mathbb {N}}$ is not free as $\mathbb {Z}$-module.

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