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Nov
21
awarded  Nice Answer
Nov
1
comment Direct limits of injective modules
To be clearer: the answer is yes for all direct limits of injective module if and only if the ring is noetherian.
Nov
1
comment Direct limits of injective modules
See page 81 Theorem 3.46 Lectures on Modules and Rings by Lam. This has exactly the theorem refered to in the comment below and is a nice textbook.
Nov
1
comment Pure Subgroups of $\mathbb Z\times\mathbb Z$
The subgroup $\langle (1,2),(3,2)\rangle$ contains $(2,0)$. It does not contain $(1,0)$ since if it did there would exist $n,m\in\mathbb{N}$ such that $1=n+3m$ and $0=2n+2m$. But this would mean $1=2m$. You don't need any tools to do this question.
Oct
24
comment Pure Subgroups of $\mathbb Z\times\mathbb Z$
Is $(2,0)\in\langle (1,2),(3,2)\rangle$? Is $(1,0)\in \langle (1,2),(3,2)\rangle$?
Oct
24
comment Pure Subgroups of $\mathbb Z\times\mathbb Z$
Yes, or you could just prove that it is not pure. Tho' it is probably a good exercise to "guess" what the quotient is isomorphic to and then exhibit an isomorphism.
Oct
22
comment Pure Subgroups of $\mathbb Z\times\mathbb Z$
let us continue this discussion in chat
Oct
22
comment Pure Subgroups of $\mathbb Z\times\mathbb Z$
@BabakSorouh So a subgroup $S$ of $G$ is pure if all elements of $S$ which are $n$-divisible in $G$ are $n$-divisible in $S$.
Oct
22
comment Pure Subgroups of $\mathbb Z\times\mathbb Z$
@BabakSorouh So perhaps you could first find some examples of elements that generate pure-subgroups.
Oct
22
comment Pure Subgroups of $\mathbb Z\times\mathbb Z$
@BabakSorouh An element $a\in G$ a group is $n$-divisible if $a=b^n$ for some $b\in G$.
Oct
22
comment Pure Subgroups of $\mathbb Z\times\mathbb Z$
Why don't you think about which elements of $\mathbb{Z}\times\mathbb{Z}$ are $n$-divisible. For instance, are $(1,2)$ or $(3,2)$ $2$-divisible?
Oct
19
comment Can we make $\mathbb{R}^{2}$ an ordered field?
Do you want the underlying abelian group structure of $\mathbb{R}^2$ to be preserved?
Oct
19
awarded  Editor
Oct
19
revised $\mathbb R^3$ is not a field
improved content
Oct
19
comment $\mathbb R^3$ is not a field
Yes - but I have to pop out for a few hours. I promise to edit it to make it more clear when I get back. Sorry!
Oct
19
answered $\mathbb R^3$ is not a field
Oct
19
accepted Transcendental extensions of $\mathbb{Q}$ containing algebraic elements.
Oct
19
comment Transcendental extensions of $\mathbb{Q}$ containing algebraic elements.
I think maybe I get it now. If we take $f\in\mathbb{Q}[x]$ to be the minimal polynomial of the primitive element of $\mathbb{Q}(a,b)$ then $f$ does not become reducible over $F$ but does become reducible over $K(v)$. Thus, $[K(v):F]$ is the degree of $f$ and so is $[K(v):\mathbb{Q}(v)]$. Thus $[F:\mathbb{Q}(v)]=1$.
Oct
19
comment Transcendental extensions of $\mathbb{Q}$ containing algebraic elements.
I don't understand the line "Then by (1) $\[F:\mathbb{Q}(v)\]=1$" i.e. $F$ might contain no elements algebraic over $\mathbb{Q}$ but still contain elements algebraic over $\mathbb{Q}(v)$. Could you explain this? Thanks for the answer. I think I can construct my own proof (which is heavily inspired by your proof) but I'd like to understand this step in your proof (and then accept your answer!).
Oct
18
comment Complete DVR with finite residue field is compact?
I'm only guessing but since it is complete, can you mimic the proof that the ring of p-adic integers is compact?