Let $G$ be an affine, connected, reductive group and $X$ a $G$-module. Choose a maximal torus $T\subseteq G$, a Borel $B\subseteq G$ containing $T$ and let $U$ be the unipotent radical of $B$. Denote by the character group of $T$. Let $\Lambda\subseteq\mathbb{X}$ be the set of dominant weights of $G$ with respect to these choices. We can decompose the graded coordinate ring $\mathbb{C}[X]=\bigoplus_{\lambda\in\Lambda} V_{(\lambda)}$ into its isotypic components $V_{(\lambda)}$ of weight $\lambda$. Let \[ \Lambda_X=\{ \lambda\in\Lambda \mid V_{(\lambda)}\ne\{0\}\} \] be the set of weights that occur in $\mathbb{C}[X]$. Let $V_{(\lambda)}\cong V_\lambda^{\oplus n_\lambda}$, where $V_\lambda$ is the irreducible module of highest weight $\lambda$. Each $V_\lambda$ has a highest weight vector which is unique up to scaling - let $f_{\lambda 1}, \ldots, f_{\lambda n_\lambda}\in V_{(\lambda)}$ be linearly independent highest weight vectors. Claim. *For each $1\le k\le n_\lambda$, if the function $f:=f_{\lambda k}\in\mathbb{C}[X]$ is reducible, then there exist weights $\lambda_1,\ldots,\lambda_r\in\Lambda_X$ such that $\lambda$ is an $\mathbb N$-linear combination of the $\lambda_i$.* Indeed, about a year ago this statement was completely unclear to me. However, it's actually not that hard to see and I felt like sharing my proof. (more…)

I asked a question on mathoverflow, and while researching I stumbled upon a theorem which I did not know before. I consider it quite powerful and the proof is short and sweet, I copied it from this expository paper by Bryden Cais: Theorem. Let $X$ be a normal, separated, Noetherian scheme and $U\subseteq X$ a nonempty affine open subset. Then, $X\setminus U$ is pure of codimension one. Proof. We need to show that for each generic point $y\in Y=X\setminus U$, the local ring $\mathcal{O}_{X,y}$ has dimension one. Let $y$ be the generic point of a component of $Y$. Denote by $\mathfrak{m}$ the maximal ideal of the local ring $A:=\mathcal{O}_{X,y}$. Let $S:=\mathrm{Spec}(A)$. The inclusion morphism $f:S\to X$ is affine because $X$ is separated (Lemma 1). Thus, $V:=f^{-1}(U)=S\setminus\{ \mathfrak{m} \}$ is an affine, open subscheme of $S$. The morphism on global sections $A=\mathcal{O}_S(S)\to\mathcal{O}_{V}(V)$, corresponding to the inclusion $V\to S$, is therefore injective. It can not be surjective because $V\ne S$. Pick some $a\in\mathcal{O}_{V}(V)\setminus A$. If we now had $\dim(A)>1$, then the prime ideals of height one $\mathfrak{p}\subseteq A$ would satisfy $\mathfrak{p}\ne\mathfrak{m}$, i.e. they would correspond to points contained in $V$. Consequently, we would have $a\in \mathcal{O}_{V,\mathfrak{p}} = A_{\mathfrak{p}}$. Because $A$ is a normal Noetherian domain, it is the intersection of all localizations at prime ideals of height one - this is Corollary 11.4 in Eisenbud's book. This yields the contradition that $a\in A$. Lemma 1. If $f:X\to Y$ is a morphism of schemes with $Y$ separated and $X$ affine, then $f$ is an affine morphism. Proof. This is exercise 3.3.6 in Qing Liu's book. Let $V\subseteq Y$ be an open affine subset. Proposition 3.9(f) in the same book tells you that there is a closed immersion $f^{-1}(V)\cong X\times_Y V\to X\times V$, so $f^{-1}(V)$ is isomorphic to a closed subscheme of an affine scheme, hence affine.