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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 $\mathbb Z\times\mathbb Z$ 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?