# Tag Info

5

Not necessarily. If there is a nontrivial ring honomorphism $\psi: A\to A$, then composing an evaluation homomorphism with $\psi$ gives you something that is not an $A$-algebra homomorphism. For example, in the case $A=\mathbb C$, $n=1$ and $\psi$ being complex conjugation we could have  \phi(a_0 + a_1X + a_2X^2 + \cdots + a_kX^k) \mapsto \overline{a_0} ...

4

What is interesting is the pair consisting of this category and its forgetful functor to rings; this exhibits the functor $R \mapsto \text{Mod}(R)$ as a fibered category or Grothendieck fibration.

4

Since $\Bbb Q$-modules are in fact vector spaces and are therefore torsion free, and $\Bbb{Q/Z}$ is not a vector space over $\Bbb Q$ as it is all torsion. The answer is that the question does not make sense.

3

What you have described is basically the "pullback" of $N$ to the category of $R$-modules via the homomorphism $\varphi|_R : R\rightarrow S$. Also, I don't like your $\sqcup$ notation. Really you're specifying the data of two things: A homomorphism $\varphi : R\rightarrow S$, and An $R$-linear map $M\rightarrow \varphi^*N$, where $\varphi^*N$ is just $N$ ...

3

Suppose that equation holds. Choose a homogeneous basis for the graded vector space $M/A_{>0}M$ and lift it to a set $S$ of homogeneous elements of $M$. There is then a canonical map $\varphi:F(S)\to M$ of graded $A$-modules, where $F(S)$ is the free module on $S$ (with the obvious grading). That equation says exactly that $F(S)_i$ and $M_i$ have the ...

2

Very generally, any model of a finitary algebraic theory is a filtered colimit of finitely presented objects. For any such object $X$ has a presentation $\langle G \mid R\rangle$, where $G$ is some set of generators and $R$ is some set of relations (and each relation only involves finitely many generators). Now consider the directed poset $P$ consisting of ...

2

A homomorphism $R^{(S)}\to R$ is determined as soon as an arbitrary function $S\to R$ is selected. The set of functions $S\to R$ is precisely $R^S$. The isomorphism can be described explicitly: if $f\colon R^{(S)}\to R$ is a homomorphism, consider $\varphi(f)=(f(e_s))_{s\in S}$, where $e_s$ is the basis element corresponding to $s\in S$. This defines a map ...

1

I will work in a category $\mathsf{Alg}(\tau)$ of algebraic structures of a given type $\tau$ (in the sense of universal algebra). Let $A$ be a finitely presented algebra. Choose some surjective homomorphism $\phi : \langle x_1,\dotsc,x_n \rangle \to A$ such that the kernel is a finitely generated congruence. Let $y_1,\dotsc,y_m$ be another generating set ...

1

Not necessarily. The kernel of the map $f$ is exactly the annihilator of $x\in M$. This can be non trivial. An example of this is if you take $\mathbb{Z}/3\mathbb{Z}$ as a $\mathbb{Z}$-module. Then the annihilator of $1$ is the the ideal $3\mathbb{Z}$ and so the map $\mathbb{Z}\to \mathbb{Z}/3\mathbb{Z}$ given by $a \mapsto a\cdot 1$ is not injective.

Only top voted, non community-wiki answers of a minimum length are eligible