ANT 3.1

Disc and Linearly Independent.The involution monoid isomorphism

Disc and Linearly Independent.

For readers with sufficient background, the result follows directly from the fact that $-{\otimes }_K\bar{K}$$-\otimes_K \overline K$ is faithful and exact. Hence it preserves and reflects linearly independent property. This completes the proof.

Let $L/K$$L/K$ be a finite separable extension with $[L:K]=n$$[L:K]=n$.

Then by the Primitive Element Theorem we have $L=K[\theta ]\cong K[x]/(f(x))$$L=K[\theta]\cong K[x]/(f(x))$. Hence we have

$$L{\otimes }_K\bar{K}\cong \bar{K}[x]/(f(x))\cong \bar{K}[x]/(x-{\theta }_1)\times ...\times \bar{K}[x]/(x-{\theta }_n)\cong {\bar{K}}^n$$

By

$$\begin{array}{c}\alpha \otimes 1⟼\left(\begin{array}{c}{\sigma }_1(\alpha )\\ .\\ .\\ .\\ {\sigma }_n(\alpha )\end{array}\right)\end{array}$$

Notice that we are working in ${\mathrm{Vect}}_{\mathrm{K}}$$\mathrm{Vect_K}$. Every space is free, hence projective and flat. Thus

$$-{\otimes }_K\bar{K}:{\mathrm{Vect}}_K\to {\mathrm{Vect}}_{\bar{K}}$$

is exact.

Recall that $\mathfrak{v}=\{v_1,...,v_n\}\in V$$\mathfrak v=\{v_1,...,v_n\}\in V$ is linearly independent iff $f^{\mathfrak{v}}:K^n\to V,f^{\mathfrak{v}}(c_1,...,c_n)=c_1{v}_1+...+c_n{v}_n$$f^{\mathfrak v}:K^n\to V,f^{\mathfrak v}(c_1,...,c_n)=c_1v_1+...+c_nv_n$ is injective.

Hence we have $\mathfrak{A}=\{{\alpha }_1,...,{\alpha }_n\}\in L$$\mathfrak{A}=\{\alpha_1,...,\alpha_n\}\in L$ is linearly independent iff the following sequence is exact.

$$0⟶K^n\stackrel{f^{\mathfrak{A}}}{\to }L$$

And

$$\begin{array}{c}M=\left(\begin{array}{ccc}{\sigma }_1({\alpha }_1)& ...& {\sigma }_1({\alpha }_n)\\ .\\ .\\ .\\ {\sigma }_n({\alpha }_1)& ...& {\sigma }_n({\alpha }_n)\end{array}\right)\end{array}$$

the column vectors are linearly independent iff $M$$M$ is injective.

The fact that the functor $-{\otimes }_K\bar{K}$$-\otimes_K\overline K$ is exact and faithful implies

$$0⟶K^n{\otimes }_K\bar{K}\stackrel{M=f^{\mathfrak{A}}\otimes 1}{\to }{\bar{K}}^n\text{ is exact}⟺0⟶K^n\stackrel{f^{\mathfrak{A}}}{\to }L\text{ is exact}$$

Hence we have

$$\mathrm{disc}({\alpha }_1,...,{\alpha }_n)=\det M^2\ne 0⟺{\alpha }_1,...,{\alpha }_n\text{ are linearly independent.}$$

The involution monoid isomorphism

Let $R[x]$$R[x]$ be a polynomial ring and define

$$f(p(x))=x^{\mathrm{deg}(p)}p\left(\frac{1}{x}\right),a_n{x}^n+a_{n-1}{x}^{n-1}+...+a_{1}x+a_0⟼a_n+a_{n-1}x+...+a_1{x}^{n-1}+a_0{x}^n$$

This is a involution, and a monoid isomorphism.

$$f(pq(x))=x^{\mathrm{deg}(p)+\mathrm{deg}(q)}pq\left(\frac{1}{x}\right)=x^{\mathrm{deg}(p)}p\left(\frac{1}{x}\right)x^{\mathrm{deg}q}q\left(\frac{1}{x}\right)=f(p(x))f(q(x))$$

Hence it preserves and reflect irreducible property.

Remark. Involution is really interesting and important. Here is some introduction and application.

https://marco-yuze-zheng.blogspot.com/2024/03/introduction-to-involution.html

https://marco-yuze-zheng.blogspot.com/2024/11/a-new-proof-of-kings-rule-using.html