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seen Jun 5 '13 at 0:57

Math! :)


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.
Feb
10
comment Question about Irreducible Polynomials Being Relatively Prime.
So the idea here is that since $k[x,y]$ is the same as $k[x][y]$ we think of $f,g$ as irreducible in $k[x][y]$ which implies irreducible in $k(x)[y]$ where $k[x]$ is the ring playing the role of the integers and $k(x)$ is its field of fractions playing the role of the rationals?
Feb
10
comment Question about Irreducible Polynomials Being Relatively Prime.
Is it somehow equivalent to irreducibility lemma on en.wikipedia.org/wiki/Gauss%27s_lemma_(polynomial) If so, I'm struggling a bit to see the connection.
Feb
10
comment Question about Irreducible Polynomials Being Relatively Prime.
Could you point me in the direction of a reference for this particular lemma of Gauss? He seem's to have quite a few ;)
Feb
10
comment Question about Irreducible Polynomials Being Relatively Prime.
$k(x)$ is the field of rational expressions/functions.
Feb
10
asked Question about Irreducible Polynomials Being Relatively Prime.
Jan
21
answered Finite group and subgroups
Dec
10
accepted Find three integers $a,\, b,\,$ and $c$ such that $\sqrt{a^2+b^2}$, $\sqrt{a^2+c^2}$, $\sqrt{c^2+b^2}$, and $\sqrt{a^2+b^2+c^2}$ are all integers.
Dec
10
comment Find three integers $a,\, b,\,$ and $c$ such that $\sqrt{a^2+b^2}$, $\sqrt{a^2+c^2}$, $\sqrt{c^2+b^2}$, and $\sqrt{a^2+b^2+c^2}$ are all integers.
Sorry, I meant nonzero solutions. All three integers should be nonzero