Steven-Owen
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 Feb 25 accepted Subspace of Noetherian space still Noetherian Feb 25 comment Subspace of Noetherian space still Noetherian Taking the closure might cause stabilization in the chain. Feb 25 comment Subspace of Noetherian space still Noetherian No, the def of Noetherian Space is a DCC on closed sets Feb 25 comment Subspace of Noetherian space still Noetherian True, but if $Y$ is not closed then it is not a descending chain of closed sets. Feb 25 accepted The closure of an irreducible subset of an irreducible space is irreducible. Feb 25 asked Subspace of Noetherian space still Noetherian Feb 25 comment The closure of an irreducible subset of an irreducible space is irreducible. O, I see. That's confusing though going back and forth between X's topology and the induced subspace topology. Thanks for your help! Feb 25 comment The closure of an irreducible subset of an irreducible space is irreducible. That's what I originally was going for, but I thought the definition of irreducible is a decomposition into proper closed (not necessarily disjoint) subsets, not just a covering by closed sets. And from here, it is not as if we can just intersect $Y$ with $S$ and $T$, because $Y\cap S$ and $Y\cap T$ are not necessarily closed. Feb 25 asked The closure of an irreducible subset of an irreducible space is irreducible. Feb 25 accepted Radical Ideals: Show that $\sqrt{\sqrt{I}+\sqrt{J}}=\sqrt{I+J}$ Feb 25 comment Radical Ideals: Show that $\sqrt{\sqrt{I}+\sqrt{J}}=\sqrt{I+J}$ Woops, I am making typos all over the place. Thanks for your help, Thomas! Feb 25 comment Radical Ideals: Show that $\sqrt{\sqrt{I}+\sqrt{J}}=\sqrt{I+J}$ Sorry, it was supposed to say $t=p+q$. That works, right? Feb 25 revised Radical Ideals: Show that $\sqrt{\sqrt{I}+\sqrt{J}}=\sqrt{I+J}$ added 2 characters in body Feb 25 asked Radical Ideals: Show that $\sqrt{\sqrt{I}+\sqrt{J}}=\sqrt{I+J}$ Feb 25 accepted Equivalence of Affine Varieties defined as an image and as vanishing set. Feb 10 comment Equivalence of Affine Varieties defined as an image and as vanishing set. Is the first equation $st=v^2$ redundant? Starting with $t^2=s^2v$ we get $v=t^2/s^2$ which we can with $v$ and it becomes $v^2=ttv/s^2$ which by us of the second equation becomes $v^2=s^3t/s^2=st$ which is the first equation. Are these the kind of manipulations to be doing? Feb 10 comment Equivalence of Affine Varieties defined as an image and as vanishing set. Sorry, but I may be being silly here, but is the choice for $x$ just $v/s$ since we know $s=x^3$ and $v=x^4$. So know we choose our $x$ as $s/v$. I am we have $s=v^3/s^3$ $v=v^4/s^4$ and $t=v^5/s^5$. It just seems like that doesn't do anything different. Or am I supposed to start with the equations $st=v^2, s^3=vt, t^2=s^2v$ and deduce those? I am not sure why this is what we want? Sorry! Feb 10 asked Equivalence of Affine Varieties defined as an image and as vanishing set. Feb 10 comment Question about Irreducible Polynomials Being Relatively Prime. Follow up: So, if two polynomials are irreducible they cannot share a non-constant component. Suppose they did, then if that component were of smaller degree than the degree of the polynomials they could not be irreducible. This forces the two polynomials in question to be of the same degree (in $y$) and that this common component is as well. Call this common component $h\in K[y]$, then it follows that $g=\alpha h$ and also that $f=\beta h$ where $\alpha, \beta \in K=k(x)$. But then $g=\alpha/\beta f$. I am not sure how to get a contradiction. It seems reasonable that they share a component. Feb 10 accepted Question about Irreducible Polynomials Being Relatively Prime.