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 Jan 2 comment Show that $F(x,y)=1$ for $x+y\ge 0$ and $F(x,y)=0$ otherwise does not define a joint CDF well 1 right??? Jan 2 revised Show that $F(x,y)=1$ for $x+y\ge 0$ and $F(x,y)=0$ otherwise does not define a joint CDF deleted 16 characters in body Jan 2 asked Show that $F(x,y)=1$ for $x+y\ge 0$ and $F(x,y)=0$ otherwise does not define a joint CDF Dec 22 revised Let $R$ be a Dedekind domain. If $I$,$J$ are $R$-ideals, prove that there exists an $\alpha\in I$ such that $\gcd(\langle\alpha\rangle,IJ)=I$ added 18 characters in body Dec 22 comment Let $R$ be a Dedekind domain. If $I$,$J$ are $R$-ideals, prove that there exists an $\alpha\in I$ such that $\gcd(\langle\alpha\rangle,IJ)=I$ You mean $r_1,r_2$? They are just arbitrary elements of $R$. Dec 22 comment Let $R$ be a Dedekind domain. If $I$,$J$ are $R$-ideals, prove that there exists an $\alpha\in I$ such that $\gcd(\langle\alpha\rangle,IJ)=I$ I have not, I'm not understanding how it applies. Dec 22 revised Let $R$ be a Dedekind domain. If $I$,$J$ are $R$-ideals, prove that there exists an $\alpha\in I$ such that $\gcd(\langle\alpha\rangle,IJ)=I$ edited body Dec 22 asked Let $R$ be a Dedekind domain. If $I$,$J$ are $R$-ideals, prove that there exists an $\alpha\in I$ such that $\gcd(\langle\alpha\rangle,IJ)=I$ Dec 21 comment For $I,J$ ideals, show that $IJ\subseteq (I\cap J)(I+J)$ It's that final step: $I(I+J)\cap J(I+J)=(I\cap J)(I+J)$ that requires you be in a Dedekind domain correct? What about being in a Dedekind domain allows that? Dec 21 accepted For $I,J$ ideals, show that $IJ\subseteq (I\cap J)(I+J)$ Dec 21 comment For $I,J$ ideals, show that $IJ\subseteq (I\cap J)(I+J)$ Could you talk a bit more about how being in a Dedekind domain allows you to do what you did in the paragraph that begins "For this, note that..." Thanks. Dec 21 revised For $I,J$ ideals, show that $IJ\subseteq (I\cap J)(I+J)$ added 71 characters in body Dec 21 asked For $I,J$ ideals, show that $IJ\subseteq (I\cap J)(I+J)$ Dec 18 accepted Let $\phi$ be a Euclidean function, prove that if $a|b$ and $\phi (a) = \phi (b)$, then $a\sim b$ Dec 18 comment Let $\phi$ be a Euclidean function, prove that if $a|b$ and $\phi (a) = \phi (b)$, then $a\sim b$ Thanks, this does it. Dec 18 comment Let $\phi$ be a Euclidean function, prove that if $a|b$ and $\phi (a) = \phi (b)$, then $a\sim b$ Well in general $\phi (nm)\geq \phi (n)$, for any integral domain on which a euclidean function can be defined. Dec 18 asked Let $\phi$ be a Euclidean function, prove that if $a|b$ and $\phi (a) = \phi (b)$, then $a\sim b$ Dec 2 comment The geometric intuition behind the fact that $y-x^3=0$ in $\mathbb{P}^2(\mathbb{R})$ has a singularity at infinity @Nils Matthes: Yes looking at it in the $y=1$ chart was how I discovered its singularity at infinity. My understanding was that looking at it in a different affine chart was just a tool to gain a 'fuller' understanding of the geometry of $y-x^3$. But it appears what you're implying is that the its projective version is in some sense the 'true' curve and it just so happens that its $z=1$ portion happens to miss the singularity. Nevertheless, I can't help but feel that there's probably something interesting to say which relates the way in which $y-x^3$ diverges to its singularity at infinity. Dec 2 asked The geometric intuition behind the fact that $y-x^3=0$ in $\mathbb{P}^2(\mathbb{R})$ has a singularity at infinity Nov 26 accepted Let $H$ be a subgroup of $G$ with $[G:H]=n$. Let $N$ be the kernel of the left-multiplication action on the cosets of $H$. Show $[G:N]$ divides $n!$