Why $G\to G/H$ is faithfully flat?

Some questions about algebraic groups.

Let $G$ be an affine algebraic group over algebraically closed field $k$. Questions: 1. $G$ is faithfully flat since it is defined over field? 2. Let $H$ be a closed subgroup of $G$, then (as I learnt from some paper) the map $\pi\colon G\to G/H$ is faithfully flat, why? reference? When it is locally trivial, especially when $G$ is linear algebraic group?

Thank you very much!

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If $k$ is a field, then any non-zero $k$-vector space whatsoever is faithfully flat. In particular, any non-trivial $k$-algebra is faithfully flat. – Zhen Lin Jul 20 '12 at 6:52
Yes, its faithfully flat over $k$. I think this is the faithfully flat means in question 1. But it is not obvious that $k[G]$ is faithfully flat over $k[G]^H$ in question 2. Thank you for your reply! – Hoxide Jul 20 '12 at 6:55
Have a look at the Book of Springer: Linear algebraic groups. (Springer is indeed the Author, not the publisher). Your claims are prooven there. I don't have the book by my side, so I cannot give you an exact reference. – sebigu Jul 20 '12 at 13:41
@sebigu Thank you very much! Actually you can download the book from Springerlink (if your institute have brought the database). You may also get it from my dropbox link (will remove after 1day for copyright issue). But I still can not find the answer. Could you help me to find it? – Hoxide Jul 20 '12 at 16:59
Sorry, I wasn't online over the weekend. I will look it up this evening, than send you the reference. – sebigu Jul 23 '12 at 13:13