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Let $K$ be a field and $GL_n(K)$ the set of all invertible $n$ by $n$ matrices over $K$. Let $m: GL_n(K)\times GL_n(K) \to GL_n(K)$ be the usual multiplication of matrices. Why the map $m$ is regular? Thank you very much.

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    $\begingroup$ The maps are just polynomials in the entries? In particular, if you identify $\text{GL}_n(K)$ as the set of pairs $(A,B)$ in $K^{n^2}$ such that $AB=1$ (where you use regular matrix multiplication), then this is obvious. $\endgroup$ – Alex Youcis Sep 8 '13 at 5:02
  • $\begingroup$ @AlexYoucis, thank you very much. But what is the product of $(A, B)$ and $(C, D)$ in $K^{n^2}$? $\endgroup$ – LJR Sep 8 '13 at 5:07
  • $\begingroup$ I am not saying you should. If $R$ is an algebraic ring, then you show that $R^\times$ is a variety by identifying it with the set $(x,y)$ in $R^2$ with $xy=1$. For example, $k^\times$ is an affine $k$-variety, isomorphic to the set $xy=1$ in $\mathbb{A}^2$. $\endgroup$ – Alex Youcis Sep 8 '13 at 5:12
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First forget $GL_n(K)$ and work in $M_n(K)$. The multiplication map $$ M_n(K)\times M_n(K)\to M_n(K)$$ is polynomial in the entries: $$((x_{ij})_{ij}, (y_{kl})_{kl})\mapsto (\sum_{r} x_{ir}y_{rl})_{il}, $$ so it is a regular map. When you restrict to $GL_n(K)$, you get a regular map $$ GL_n(K)\times GL_n(K)\to M_n(K).$$ As the multiplication lands in $GL_n(K)$, you get the statement you want to prove.

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  • $\begingroup$ thank you very much. But the multiplication is the multiplication of matrices. Why it is a polynomial? $\endgroup$ – LJR Sep 8 '13 at 11:13
  • $\begingroup$ @IJR: a matrix $(x_{ij})_{ij}$ is viewed as element of $K^{n^2}$. $\endgroup$ – Cantlog Sep 8 '13 at 11:41

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