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 Apr 6 revised the number of inversions in the permutation “reverse” added 6 characters in body Apr 6 answered the number of inversions in the permutation “reverse” Apr 6 comment Intuition for theorem about compact subsets of topological groups So you do know a proof of the statement, but you still want some intuition for it, right? Apr 6 revised If $F$ has characteristic $p$ and $f(x)=x^p-a\in F[x]$, then $f(x)$ is either irreducible over $F$ or $f(x)$ splits in $F$. added 1 character in body Apr 5 comment If $F$ has characteristic $p$ and $f(x)=x^p-a\in F[x]$, then $f(x)$ is either irreducible over $F$ or $f(x)$ splits in $F$. @OLP Yeah, I’ve updated it. Apr 5 revised If $F$ has characteristic $p$ and $f(x)=x^p-a\in F[x]$, then $f(x)$ is either irreducible over $F$ or $f(x)$ splits in $F$. expanded Apr 5 comment Find the dimension of $\mathbb F[T]$ over $\mathbb F$ @quid »$ℚ[\sqrt{-5}]$« is short for »$ℚ[X]/(X^2 + 5)$«. On the other hand, »$F[T]$ with $m(X)$ being the minimal polynomial of $T$« isn’t short for »$F[X]/(m)$«. Apr 4 answered Let $\alpha\in R$. Prove that $\mathbb Q(\alpha)\cong\mathbb Q(x)$ iff $\alpha$ is transcendental. Apr 4 comment Let $\alpha\in R$. Prove that $\mathbb Q(\alpha)\cong\mathbb Q(x)$ iff $\alpha$ is transcendental. $ℚ[X] ≠ ℚ(X)$ if $ℚ[X]$ denotes the ring of polynomials @OLP. By the way, I recommend using capital letters if you are refering to indeterminates for polynomials/rational functions. Apr 4 comment Prove that there is no irreducible polynomial in $\mathbb Q[x]$ which is zero at both $x=\sqrt 5$ and $x=\sqrt 7$ So you already know about Galois automorphisms and minimal polynomials and such? Apr 4 comment If $F$ has characteristic $p$ and $f(x)=x^p-a\in F[x]$, then $f(x)$ is either irreducible over $F$ or $f(x)$ splits in $F$. »Frobenius endomorphism« is just a fancy name for any map given by $x ↦ x^q$ whenever the Freshman’s Dream $(x+y)^q = x^q + y^q$ holds (so whenever the given map is additive/an endomorphism of an abelian group). Don’t worry about it, but keep it in mind because it’s very important. Apr 4 comment If $F$ has characteristic $p$ and $f(x)=x^p-a\in F[x]$, then $f(x)$ is either irreducible over $F$ or $f(x)$ splits in $F$. @OLP Don’t worry too much about the word »Frobenius homomorphism«, you already realized that the Freshman’s Dream holds and $f$ splits as $f = (X-α)^p$ (in its splitting field). I’ve added another hint. How do proper divisors of $f$ look like in $L$? Apr 4 revised If $F$ has characteristic $p$ and $f(x)=x^p-a\in F[x]$, then $f(x)$ is either irreducible over $F$ or $f(x)$ splits in $F$. added 41 characters in body Apr 4 comment Prove that every open interval in $R$ is a union of at most countable collection of disjoint segments And a segment would be …? A closed intervall possibly? Apr 4 comment If $F$ has characteristic $p$ and $f(x)=x^p-a\in F[x]$, then $f(x)$ is either irreducible over $F$ or $f(x)$ splits in $F$. @OLP No, it can’t. I gave a hint in my answer. See if that helps. Apr 4 comment Monic polynomial … which is called Gauss’s lemma (like a lot of other lemmata), by the way. Apr 4 answered If $F$ has characteristic $p$ and $f(x)=x^p-a\in F[x]$, then $f(x)$ is either irreducible over $F$ or $f(x)$ splits in $F$. Apr 4 reviewed Edit Group having exactly one non trivial proper subgroup. Apr 4 revised Group having exactly one non trivial proper subgroup. fixed latex Apr 4 revised Prove a map is closed deleted 1 character in body