Surjectivity of Invariants - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T14:33:15Z http://mathoverflow.net/feeds/question/91615 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/91615/surjectivity-of-invariants Surjectivity of Invariants mark 2012-03-19T12:56:03Z 2012-03-20T01:47:33Z <p>Suppose $V, W, U$ are $Z_p$ module over a field $F$ of characteristic $p$ and $V=W \oplus U$. Is there a degree preserving surjective map from $F[V]^{Z_p}$ to $F[W]^{Z_p}$ ? In non-modular case the Reynold's operator does the job but in this case I don't have any clue. </p> http://mathoverflow.net/questions/91615/surjectivity-of-invariants/91626#91626 Answer by Ralph for Surjectivity of Invariants Ralph 2012-03-19T15:20:37Z 2012-03-20T01:47:33Z <p>That holds more generally: If $k$ is any commutative ring with unit, $G$ any group and if $U,W$ are $k$-free $kG$-modules, $V = U \oplus W$, then there is a $k$-linear surjection $k[V]^G \to k[W]^G$: </p> <p>Let $r: V \to W$ be the projection. Then $k[V]=k[W] \otimes k[U]$ and $r$ induces a $k$-algebra hom. $r: k[V] \mapsto k[W]$. Aussume for the moment we know that $r$ is compatible with the $G$ action, i.e. $r ( g \cdot v) = g \cdot r(v)$. Then $r$ preserves the invariants: $r: k[V]^G \to k[W]^G$. </p> <p>Moreover, $r$ is surjective on the invariants. For, let $w \in k[W]^G$ and put $v := w \otimes 1$. Then $g \cdot v = (g\cdot w)\otimes (g\cdot 1) = w \otimes 1 = v$, i.e. $v \in k[V]^G$ and $r(v) = w$. </p> <p>Now let's show that $r$ is compatible with the $G$-action. Note that we have an $k$-algebra homomorphism $\epsilon: k[U] \to k$ that satisfies $\epsilon(g \cdot u) = \epsilon(u)$: If $\lbrace x_1,...,x_n\rbrace$ is a $k$-basis of $U$ then $\epsilon$ is just the augmentation $k[x_1,...,x_n] \to k$. </p> <p>Now $r$ is given by $r(w \otimes u) = \epsilon(u) \cdot w$ and we obtain $$r(g (w \otimes u)) = r(g w \otimes g u) = \epsilon(g u) (g w) =\epsilon(u) (gw) = g (\epsilon(u)w) = g\; r(w \otimes u).$$</p> <p>BTW: To my understanding, the Reynold's operator yields $k[V]^H \to k[V]^G$ for $H \le G$ and adresses another problem than yours. </p>