Finitely presented group with fewer relations than generators.
2013年8月29日 · I claim that if there are fewer relations than generators of a finitely presented group, then the group has to be infinite. I would like to see is it true in general?? Is any finitely presented group with fewer relations than generators an infinite Group? Please suggest some idea to see if this is true/false?
abstract algebra - Definition: finite type vs finitely generated ...
2013年10月22日 · But "finitely generated" is also used for modules and in fact arbitrary algebraic structures (see below), so that one often prefers "finite type" in the setting of ring homomorphisms. In order to differentiate these notions even more, one says "finite" if the corresponding module is finitely generated.
Finitely presented module + flat implies projective
2017年4月11日 · Every finitely generated module is the quotient of a finitely generated projective module. 10 Non-finitely generated, non-projective flat module, over a polynomial ring
abstract algebra - Example of finite extension which is not finitely ...
I don't see a precision problem. Anyway, your current version is correct. (In my opinion it is unnecessarily complicated: the degree is finite means the extension is finitely generated as a module, which means that it is finitely generated as an algebra, which means that it is finitely generated as a field extension). $\endgroup$ –
abstract algebra - $\mathbb Z[1/2]$ is not finitely generated ...
I believe you are confusing finitely generated modules with finitely generated algebras. In a module we only have addition and scalar multiplication by the base ring, which in this case is $\mathbb Z$. $\frac1 4$ is not in the submodule generated by $1$ and $\frac 1 2$ because $\frac 1 4$ is not an integer linear combination of these.
Is any finite-dimensional extension of a field, say
Other answers provide nice proofs, here is a very short one based on the multiplicativity of the degree over field towers: If $ K/F $ is a finite extension and $ \alpha \in K $, then $ F(\alpha) $ is a subfield of $ K $, and we have a tower of fields $ F \subseteq F(\alpha) \subseteq K $.
What is a finitely generated sheaf? - Mathematics Stack Exchange
But the property of being noetherian is hereditary, so we may use [Chapter II, Exercise 1.11] to deduce that $\Gamma (U, -)$ preserves directed colimits. Thus $\mathbb{Z}_U$ is indeed finitely presentable. That said, Hartshorne speaks of subsheaves of $\mathbb{Z}_U$ as well. It is not at all clear to me that these are finitely presentable.
Equivalent definitions of purely finitely additive measures
I don't know much about purely finitely additive measures, but since no one has answered yet let me give my interpretation of the definition given by Yosida and Hewitt in their paper Finitely additive measures (1952). As you expected, a (non-zero) purely finitely additive measure is clearly not countably additive. If it were, taking $\nu = \mu ...
abstract algebra - Intuitive explanation of Nakayama's Lemma ...
Suppose your module is of finite length. Then you can consider on it the so called radical filtration, which organises the module into an onion-like thing, with elements of the maximal ideal pushing elements of the module farther in from their starting layer to one right below and, moreover, each layer obtained from the one above it in this way.
logic - some problems about axiomatizable classes resulting from ...
Show that a class K is finitely axiomatizable iff K and its complement are axiomatizable. Clearly if it's finitely axiomatizable then it's axiomatizable. For the other direction, I've been playing with it but not sure how to frame K's complement. Show that the class K of all infinite sets is axiomatizable but not finitely axiomatizable.