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From Springer, Linear Algebraic Groups, first page of Chapter 3.

Let $G$ be a commuative algebraic group. If we regard $G$ as a closed subgroup of $GL_{n}$, then we can identify its semisimple part $G_{s}$ and unipotent part $G_{u}$ with subgroups of upper-triangular matrices $T_{n}$. We have $G_{s}=G\cap D_{n}$. This can be done via general eigenspace decomposition and the fact that $G$ is commutative.

My question is why the projection map $G\rightarrow G_{s}$ is a morphism of algebriac varieties, or a homomorphism $k[G_{s}]\rightarrow k[G]$ of coordiniate rings. The author commented this is because "the map sending $x\in G$ to its semi-simple part $x_{s}$ is also a morphism, since it maps $x$ to a set of its matrix elements". I cannot really follow this argument, though I can construct the obvious map $k[G_{s}]\rightarrow k[G]$ which is a homomorphism induced by injection. Can someone give me a hint what does "its matrix elements" mean at here? Sorry this question is really trivial.

Bombyx mori
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When $x_s$ is diagonal and $x_u$ is triangular, the map $x=(t_{ij})_{i, j}\mapsto x_s$ is just $(t_{ij})_{i,j}\mapsto \mathrm{diag}(t_{ii})_i$ and this is clearly an algebraic morphism.