Sarastro
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 Jan 5 comment Which of the following field properties are correct? One way would be to use that $x^4-x^3+x^2-x+1=\Phi_5(-x)$, and that $\Phi_5(x)$ is irreducible. Dec 14 comment Showing that $|\text{Hom}_\mathbb{Q}(K,K)|=6$ Consider complex conjugation as a $\Bbb{Q}$-automorphism of $L$. Also, note that $\Bbb{Q}(\alpha,\beta,\bar{\beta})=\Bbb{Q}(\alpha,\beta)$, which can save you some writing. Dec 14 comment How do I find a splitting field $x^8-3$ over $\mathbb{Q}$? It is not enough that $\zeta\notin\mathbb{Q}(\sqrt[8]{3})$ for $K$ to be of dimension $4\cdot8=32$ over $\mathbb{Q}$. For example, the splitting field of $x^8-2$ over $\Bbb{Q}$ has degree $16$ even though $\zeta\notin\mathbb{Q}(\sqrt[8]{2})$. Nov 20 comment Proving 2047 is not an euler pseudoprime It's a Jacobi symbol. Nov 14 comment Subgroups of $S_n$ There's also an element of order $10$. Nov 14 comment Subgroups of $S_n$ If $n\geq 3+4+5=12$, $S_n$ contains a cyclic subgroup of order $60$: take the subgroup generated by an element of cycle type $(3,4,5)$. If $n<12$, $S_n$ has no cyclic subgroup of order $60$, since no element of $S_n$ has order $60$. May 30 comment determining a linear isomorphism so that two quadratic forms become equivalent @moran I have edited my answer. May 28 comment Using the Uniform Cauchy Criterion theorem In fact even more is true: not only do the $(f_n)$ converge uniformly, the derivative of the uniform limit is the uniform limit of the $(f_n')$. May 24 comment The Image of normal subgroup is also normal subgroup? You need $\phi$ to be surjective. May 20 comment Class of graphs with symmetric random walk That's a reversible Markov chain, and it holds if the initial distribution is the invariant distribution. The kind of random walk you describe is always reversible for a finite connected graph. May 17 comment Proving Irreducibility of $x^4-16x^3+20x^2+12$ in $\mathbb Q[x]$ Do you know Gauss' Lemma? May 16 comment In a commutative ring with identity, if $p$ is irreducible, is ($p$) a maximal ideal? I will have to think about it May 16 comment In a commutative ring with identity, if $p$ is irreducible, is ($p$) a maximal ideal? In a general commutative ring $R$, $R/I$ is a field iff $I$ is maximal in $R$. So, what you are really asking is: if $(p)$ is maximal whenever $p$ is irreducible, is $R[X]$ a PID? May 12 comment $A$ is normal matrix.If $A{A^T}$ has $n$ distinict eigenvalue,why $A$ is symmetric? $A$ normal means that $AA^T=A^TA$. But $AA^T=A^TA$ does not imply $A=A^T$ in general. Apr 22 comment Question on the uniqueness of LU decomposition If the leading $k\times k$ submatrices of $A$ (i.e. the top left $k\times k$ submatrices) are all invertible for $k=1,\cdots,n-1$, then an LU factorisation exists and is unique. Conversely if one of them is singular, the LU factorisation of $A$ is either nonexistent or not unique. Unfortunately, I can only prove it one way (not the converse). Apr 22 comment A small complex number whose total distance from other given complex numbers is large Do you mean $|z|\leq 1$? Mar 14 comment Prove that if $N(z)$ is irreducible in $\Bbb{Z}$ then $z$ is irreducible in $\Bbb{Z}[\sqrt{\alpha}]$. In an integral domain $R$, $r,s\in R$ are associates iff $r=su$ for some unit $u\in R$. If $r$ divides $z$, $rs=z$ for some $s$. By my definition, if $z$ is irreducible, either $r$ or $s$ is a unit. So $r|z$ implies $r$ is a unit or $r$ is an associate of $z$. Mar 14 comment Prove that if $N(z)$ is irreducible in $\Bbb{Z}$ then $z$ is irreducible in $\Bbb{Z}[\sqrt{\alpha}]$. For a ring $R$, $r\in R$ is a unit if $\exists s\in R$ such that $rs=sr=1$. An element in a ring is irreducible if it is not the product of two nonunits. Mar 10 comment If h is a holomorphic non-vanishing function in the complex plane then h is the exponential of another function. It should be $\exp(g(0))=h(0)$. Feb 11 comment The function given by the Rao-Blackwell theorem is a statistic No, $\mathbb E_\theta[V_n|T_n=t]=\sum_{\lbrace x:T_n(x)=t\rbrace}\mathbb P_\theta(X=x|T_n=t)V_n(x)$ follows by definition of expectation. We sum over all possibilities of $x$ given $T_n=t$. It is not true in general that $\mathbb E_\theta[V_n|T_n=t]=\sum_{\lbrace x:T_n(x)=t\rbrace}\mathbb P_\theta(V_n=V_n(x)|T_n=t)V_n(x)$