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1d
revised Groebner basis and prime ideals.
added 2 characters in body
1d
comment Groebner basis and prime ideals.
That is a typo, thank you. I missed.
1d
asked Groebner basis and prime ideals.
2d
comment Quotient Ideal, quick equivalence.
I didn't say it wasn't correct. I just asked if it was what i wrote, because I wasn't sure; but thanks. i would've given you an up vote at least to say thanks.
2d
comment Quotient Ideal, quick equivalence.
Was I right? My ideas are pretty loose, maybe you can check it and correct it.
2d
comment Quotient Ideal, quick equivalence.
I mean $r<y> \subset <x>$
2d
comment Polynomial ring ideal. Show that $(f) \cap R_N = (0) \iff f \notin R_N$
so if you multiply $f$ by any nonzero element of $R$, you get a mixed polynomial, so you end up in $R$ again.
2d
revised Polynomial ring ideal. Show that $(f) \cap R_N = (0) \iff f \notin R_N$
edited body
2d
revised Quotient Ideal, quick equivalence.
added 33 characters in body
2d
asked Polynomial ring ideal. Show that $(f) \cap R_N = (0) \iff f \notin R_N$
2d
reviewed Reject Quotient Ideal, quick equivalence.
2d
asked Quotient Ideal, quick equivalence.
Feb
2
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Feb
1
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Jan
29
comment What does “associate” mean in a relation?
Ah or $a = ub$ for $a,b \in R$ in a ring with $u$ being a unit.
Jan
29
asked What does “associate” mean in a relation?
Jan
29
accepted kernel of action.
Jan
29
comment Error analysis for Runge Kutta, how to take Big O of 2 variables?
@David, No you do, but you know that it is not defined at $t_0$. Example (1 equation): $y' = 1/y$, with $y(0) = 0$ but $y'(0) = 1/y(0)$ is not defined, so the idea is to approximate the solution near $0$ via series recursion. Here we modify it, say, by $y(0 + 0.1) = y(0.1) = \text{something nonzero}$, now it is okay to use the Runge Kutta Alg start at $t=0.1$ because $y'(0.1) = 1/y(0.1)$ has no problems.
Jan
29
comment Looking for some insight or explanation in Normal group and similar matrix
Group action by conjugation I believe.
Jan
28
comment Group Action: Group $(\mathbb{Z}, +)$ acting on $\mathbb{R}$
@AlexanderMcFarlane, can you give an example of your arbitrary action? It needs to satisfy the group action axioms.