Ben West
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 1d comment If $E$ is finite, show that $E=F(u)$ for some $u\in E$. By your work, it seems that $E$ is a finite field, not just a finite extension? Since $u\in F(u)$, so is any power of $u$. Thus $E^\ast=\langle u\rangle\subset F(u)$. Since $0\in F(u)$, $E\subset F(u)$. Apr 6 comment Showing isomorphism between $\langle 3\rangle / \langle 12\rangle$ and $\Bbb Z_4$ To start, a quotient of a cyclic group is cyclic since the class of a generator is still a generator, and cyclic groups of a given order are unique up to isomorphism. Apr 2 comment How to reduce to the affine case? $\phi(X)$ contains a nonempty open subset of $\overline{\phi(X)}$ I saw your other comments, and I agree, I'm dubious too now. If you unaccept, I will deleted the post for now (the site won't allow me to otherwise). I'll think about it if I have time, or hopefully someone more well-versed in AG will step in. Apr 2 comment How to reduce to the affine case? $\phi(X)$ contains a nonempty open subset of $\overline{\phi(X)}$ Digging back through my old notes on Springer I think this bothered me as well. I made a note to myself there that Proposition 2.2.13 of Algebraic Groups and Differential Galois Theory by Crespo proves this in more detail. Apr 2 comment How to reduce to the affine case? $\phi(X)$ contains a nonempty open subset of $\overline{\phi(X)}$ You're right, that's a good point. If I recall correctly, one first reduces to the case that $X,Y$ are irreducible, $Y$ is affine, and $\phi$ is dominant. To reduce down to $X$ affine, if $\bigcup_i U_i$ is an open affine cover of $X$, then the $U_i$ are dense in $X$, so their images $\phi(U_i)$ are dense in $Y$, hence the restrictions $U_i\to Y$ are dominant morphisms of irreducible affines. Then you can take the union of opens of $\overline{\phi(U_i)}$ which are in $\phi(U_i)$. Mar 27 comment Show that $\mathbb{Q}(\sqrt2, \sqrt[3]2)$ is a primitive field extension of $\mathbb{Q}$. @Mathmo123 Simple, it's less work for me. Mar 27 comment Show that $\mathbb{Q}(\sqrt2, \sqrt[3]2)$ is a primitive field extension of $\mathbb{Q}$. Do you have to do it explicitly? By the primitive element theorem, finite separable extensions are simple. This extension is easily seen to be finite, and it's separable since it's a finite extension of $\mathbb{Q}$ and $\mathbb{Q}$ is perfect. Mar 25 reviewed Approve Find the elements of $S_6$ that commute with $(1234)$. Mar 25 comment If $H$ is a closed subgroup of $G$, is the Lie algebra of $H$ contained in the Lie algebra of $G$? This is briefly discussed in the beginning of section 9.4 in Humphreys' Linear Algebraic Groups. Mar 25 answered Show that $SL(n, \mathbb{R})$ is a $(n^2 -1)$ smooth submanifold of $M(n,\mathbb{R})$ Mar 23 answered Is this the correct subgroup of order 12 I found? Mar 22 answered In a reflection group, the longest word $w_0$ contains all simple reflections Mar 16 answered How to define the Yoneda embedding Mar 16 answered Express $x^8-x$ as a product of irreducibles in $\Bbb Z_2[x]$ Mar 10 answered Show that the polynomial $(x-1)(x-2) \cdots (x-n)-1$ is irreducible on $\mathbb{Z}[x]$ for all $n \geq 1$ Mar 7 comment Show that a Z-module A is flat if and only if it is torsion free? More generally, it's a standard result that over a PID, (or a Dedekind domain even), flat modules and torsion-free modules are the same. You can find this as Corollary 3.51 in Rotman's Homological Algebra, for instance. Feb 28 comment How to prove that if $S: U\rightarrow V$ and $T: V\rightarrow W$ are isomorphisms, then $TS$ is also an isomorphism? You're practically there, $TS$ is an isomorphism if it has an inverse which is also a linear transformation. You've found an inverse, now just check it is also a linear map, which might be immediate based on the facts you have at your disposal. Feb 28 answered Abelian group in short exact sequence Feb 28 comment How do we show that it is a subset of the centralizer? $H\cap M=1$ means $H\cap M$ is the trivial group consisting of just the identity. Feb 28 answered How do we show that it is a subset of the centralizer?