# catamite

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210
bio website reddit.com/r/… location meatspace age member for 1 year, 11 months seen 14 hours ago profile views 1,026

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

# 575 Actions

 Jul7 comment If $E$ and $F$ are subfields of a finite field $K$ and $E\cong F$, prove that $E = F$ It seems intuitively obvious, but how do we know all the roots of $x^q - x$ are distinct? Jul7 revised If $E$ and $F$ are subfields of a finite field $K$ and $E\cong F$, prove that $E = F$ added 42 characters in body Jul7 comment If $E$ and $F$ are subfields of a finite field $K$ and $E\cong F$, prove that $E = F$ Sorry I should have mentioned, I have no Galois Theory at my disposal. Group Theory up through Sylow Theorems, and Field Theory through splitting fields and general results on the classification of finite fields. Also undergrad ring theory, but that's probably not important. Jul7 asked If $E$ and $F$ are subfields of a finite field $K$ and $E\cong F$, prove that $E = F$ Jul6 accepted A property equivalent to $K$ being a finite-dimensional, normal extension field of $F$ Jul4 accepted Finding a splitting field of $x^3 + x +1$ over $\mathbb{Z}_2$ 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 $X$