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楼主 |
业余的业余
发表于 2023-11-5 01:25
本帖最后由 业余的业余 于 2023-11-7 21:33 编辑 重新利用此楼。以下内容引自9楼所列电子书第520页。个人觉得对证明 $b$ 有必要
Proposition 9. Let $\alpha$ be algebraic over $F$. Then there is a unique monic irreducible polynomial $m_{\alpha,F}(x)\in F[x]$ which has $\alpha$ at a root. A polynomial $f(x)\in F[x]$ has $\alpha$ as a root if and only if $m_{\alpha,F}(x)$ divides $f(x)$ in $F[x]$.
Proof: Let $g(x)\in F[x]$ be a monic polynomial of minimal degree having $\alpha$ as a root. Suppose $g(x)$ were reducible in $F[x]$, say $g(x)=a(x)b(x)$ with $a(x),b(x)\in F[x]$ both of degree smaller than $\deg(g)$. Then $g(\alpha)=a(\alpha)b(\alpha)$ in $K$, and since $K$ is a field,either $a(\alpha)=0$ or $b(\alpha)=0$, contradicting the minimality of the degree of $g(x)$. It follows that $g(x)$ is a monic irreducible polynomial having $\alpha$ as root.
Suppose now that $f(x)\in F[x]$ is any polynomial having $\alpha$ as a root. By the Euclidean Algorithm in $F[x]$ there are polynomials $q(x),r(x)\in F[x]$ such that\[ f(x)=q(x)g(x)+r(x) \qquad \text{with }\deg(r)=\deg(g).\]
Then $f(\alpha)=q(\alpha)g(\alpha)+r(\alpha)$ in $K$ and since $\alpha$ is a root of both $f(x)$ and $g(x)$, we obtain $r(\alpha)=0$, which contradicts the minimality of $g(x)$ unless $r(x)=0$. Hence $g(x)$ divides any polynomial $f(x)$ in $F[x]$ having $\alpha$ as a root...$\box$
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hbghlyj
发表于 2023-11-5 00:10
