# heat death

less info
reputation
312
bio website reddit.com/r/… location meatspace age member for 2 years, 1 month seen 2 hours ago profile views 1,037

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

# 589 Actions

 Jul4 accepted Trying to sort the coefficients of the polynomial $(z-a)(z-b)(z-c)…(z-n)$ into a vector Jul4 accepted Pairwise non-integral numbers Jul4 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. Jul4 revised A property equivalent to $K$ being a finite-dimensional, normal extension field of $F$ added 17 characters in body Jul4 revised A property equivalent to $K$ being a finite-dimensional, normal extension field of $F$ added 12 characters in body Jul4 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 $K[x]$ Jul4 asked A property equivalent to $K$ being a finite-dimensional, normal extension field of $F$ Jul3 comment Finding a splitting field of $x^3 + x +1$ over $\mathbb{Z}_2$ @Dylan Moreland: Ok cool, thanks for the help. =] Jul3 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? Jul3 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? Jul3 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? Jul3 revised Finding a splitting field of $x^3 + x +1$ over $\mathbb{Z}_2$ added 123 characters in body Jul3 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! Jul3 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 $X$ Jul3 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. =] Jul3 asked Finding a splitting field of $x^3 + x +1$ over $\mathbb{Z}_2$ Jul2 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. Jul2 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. Jul1 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 $X$ I certainly believe you, I just wonder if maybe there is an error in the book and parts (a) and (b) were switched. Anyways thanks for the help, I should hopefully be able to figure it out from here. Jul1 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 $X$ This is part (b) of the problem almost word for word, my problem is part (a), I'm not sure what to make of that.. Except that they use min instead of max.