most recent 30 from http://mathoverflow.net 2013-05-25T12:00:07Z http://mathoverflow.net/feeds/question/13593 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/13593/is-every-monomorphism-of-commutative-hopf-algebras-over-a-field-injective Nicolas Schmidt 2010-01-31T22:30:25Z 2010-02-05T12:19:39Z

<p>Is it true that any monomorphism of commutative Hopf algebras over a field is injective? Moreover, is it true that any epimorphism of commutative Hopf algebras over a field is surjective?</p>

http://mathoverflow.net/questions/13593/is-every-monomorphism-of-commutative-hopf-algebras-over-a-field-injective/13598#13598 Ben Webster 2010-01-31T23:22:44Z 2010-01-31T23:28:25Z

<p>No. The category of commutative Hopf algebras over a field is opposite to the category of affine group schemes, so your question is "is every epi of affine $k$-group schemes surjective?" But there are maps of finite groups that are epis, but not surjective (this happens when the image has normal closure the whole group), for example the inclusion of a transposition into $S_3$.</p> <p>I suspect every epi in commutative Hopf algebras is surjective. Certainly one can't construct an example from finite groups. I might be forgetting some funniness about group schemes.</p>

http://mathoverflow.net/questions/13593/is-every-monomorphism-of-commutative-hopf-algebras-over-a-field-injective/13639#13639 Nicolas Schmidt 2010-02-01T09:44:08Z 2010-02-01T09:44:08Z

<p>Yes, the category of commutative Hopf algebras over k is antiequivalent to the category of (pro-)affine group schemes over k, but this equivalence does not respect the obvious functors to (Set). I really meant injective as maps between algebras, not between maximal spectrum, prime spectrum or rational points.</p>

http://mathoverflow.net/questions/13593/is-every-monomorphism-of-commutative-hopf-algebras-over-a-field-injective/14234#14234 Pavel Etingof 2010-02-05T05:47:13Z 2010-02-05T12:19:39Z

<p>It seems that Ben is nevertheless right that the answer to the first question is "NO". Let $G=SL(2,\Bbb C)$, and $B$ be the subgroup of lower triangular matrices. Then the inclusion $B\to G$ is an epi, since every algebraic representation of $B$ that extends to $G$ does so uniquely (on the nose, not just up to an isomorphism!). This follows from the fact that in any finite dimensional representation $V$ of the Lie algebra $sl(2)$, the operator $e$ is determined by $f$ and $h$. Indeed, the kernel $K$ of $e$ is spanned by vectors $v$ satisfying $hv=mv$ and $f^{m+1}v=0$ for some integer $m\ge 0$, and since $V=\Bbb C[f]K$, the operator $e$ on $V$ is uniquely determined.</p> <p>So one might guess that a morphism of complex affine algebraic groups $\phi: H\to G$ is an epi if and only if $G/\phi(H)$ is connected and proper (but I did not check this). </p>

http://mathoverflow.net/questions/13593/is-every-monomorphism-of-commutative-hopf-algebras-over-a-field-injective/14250#14250 Dai Tamaki 2010-02-05T08:41:14Z 2010-02-05T08:41:14Z

<p>I haven't read through yet but the following paper by Chirvasitu seems to answer your second question and discusses closely related problems. See discussion in page 7 after Proposition 2.5.</p> <p><a href="http://arxiv.org/abs/0907.2881" rel="nofollow">http://arxiv.org/abs/0907.2881</a></p> <p>Namely an epimorphism of Hopf algebras over a field $$f : H \longrightarrow K$$ is necessarily surjective if $K$ is commutative.</p>