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visits member for 2 years, 6 months
seen Sep 11 at 23:52

when someone smiles at me, all I see is an ape bearing its teethe


Jul
6
accepted A property equivalent to $K$ being a finite-dimensional, normal extension field of $F$
Jul
4
accepted Finding a splitting field of $x^3 + x +1$ over $\mathbb{Z}_2$
Jul
4
accepted Trying to sort the coefficients of the polynomial $(z-a)(z-b)(z-c)…(z-n)$ into a vector
Jul
4
accepted Pairwise non-integral numbers
Jul
4
comment A property equivalent to $K$ being a finite-dimensional, normal extension field of $F$
@Arturo Magidin: you're right, thanks for pointing that out.
Jul
4
revised A property equivalent to $K$ being a finite-dimensional, normal extension field of $F$
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Jul
4
revised A property equivalent to $K$ being a finite-dimensional, normal extension field of $F$
added 12 characters in body
Jul
4
comment A property equivalent to $K$ being a finite-dimensional, normal extension field of $F$
Ah yes sorry forgot to say that, $f(x)$ is irreducible in $F[x]$. Normal means if it is irreducible in $F[x]$ and has one root in $K[x]$ then it has all roots in $K[x]$
Jul
4
asked A property equivalent to $K$ being a finite-dimensional, normal extension field of $F$
Jul
3
comment Finding a splitting field of $x^3 + x +1$ over $\mathbb{Z}_2$
@Dylan Moreland: Ok cool, thanks for the help. =]
Jul
3
comment Finding a splitting field of $x^3 + x +1$ over $\mathbb{Z}_2$
ohh nice! but we still have $\mathbb{Q}(\sqrt[3]{2})\cong \mathbb{Q}(\sqrt[3]{2}w)\cong \mathbb{Q}[x]/x^3 -2$, where $w$ is a third root of unity, correct?
Jul
3
comment Finding a splitting field of $x^3 + x +1$ over $\mathbb{Z}_2$
Ok interesting, so is there a canonical example of when the quotient field does not contain all the roots of the polynomial?
Jul
3
comment Finding a splitting field of $x^3 + x +1$ over $\mathbb{Z}_2$
I thought forming the quotient field adjoined just a single root? And that sometimes you would get lucky and the remaining roots could be formed within the quotient field as well, but that sometimes they couldn't, is this not correct?
Jul
3
revised Finding a splitting field of $x^3 + x +1$ over $\mathbb{Z}_2$
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Jul
3
comment Finding a splitting field of $x^3 + x +1$ over $\mathbb{Z}_2$
oh shoot I must have miscaculated in my quotient field, ok cool, thanks!
Jul
3
accepted Let $Y$ be an ordered set in the order topology with $f,g:X\rightarrow Y$ continuous, show that $\{x:f(x)\leq g(x)\}$ is closed in $X$
Jul
3
comment Let $Y$ be an ordered set in the order topology with $f,g:X\rightarrow Y$ continuous, show that $\{x:f(x)\leq g(x)\}$ is closed in $X$
Ahhh thank you for this. =]
Jul
3
asked Finding a splitting field of $x^3 + x +1$ over $\mathbb{Z}_2$
Jul
2
comment Let $Y$ be an ordered set in the order topology with $f,g:X\rightarrow Y$ continuous, show that $\{x:f(x)\leq g(x)\}$ is closed in $X$
Ok after going over this all yesterday I'm pretty sure my book doesn't have (a) and (b) switched. To prove your hint requires the pasting lemma which requires that my set $A$ be closed. So as far as I can tell part (a) is necessary for part (b) and not the other way around.
Jul
2
comment Let $Y$ be an ordered set in the order topology with $f,g:X\rightarrow Y$ continuous, show that $\{x:f(x)\leq g(x)\}$ is closed in $X$
On line two of your proof I think you meant "in $Y$" not "in $X$". Also could you explain how it follows that in these two disjoint open sets in $Y$ we always have $f(y) > g(z)$ for any $y\in U_1$ and $z\in U_2$? I don't see how this follows from what you have above it. Thanks.