## 张祖锦2023年数学专业真题分类70天之第41天
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921、 (3)、 $\displaystyle x^4-9$ 在复数域上的因式分解为 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$. (云南大学2023年高等代数考研试题) [多项式 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /
\begin\{aligned\} x^4-9=&(x^2+3)(x^2-3)\\\\ =&\left(x+\sqrt\{3\}\mathrm\{ i\}\right)\left(x-\sqrt\{3\}\mathrm\{ i\}\right)\left(x+\sqrt\{3\}\right)\left(x-\sqrt\{3\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
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922、 (4)、 $\displaystyle x\_1x\_2-x\_2x\_3+x\_1x\_3$ 的正惯性指数为 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$. (云南大学2023年高等代数考研试题) [多项式 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 题中二次型的矩阵为
\begin\{aligned\} A=\left(\begin\{array\}\{cccccccccccccccccccc\}0&\frac\{1\}\{2\}&\frac\{1\}\{2\}\\\\ \frac\{1\}\{2\}&0&-\frac\{1\}\{2\}\\\\ \frac\{1\}\{2\}&-\frac\{1\}\{2\}&0\end\{array\}\right), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
它的特征值为 $\displaystyle \frac\{1\}\{2\},\frac\{1\}\{2\},-1$, 而正惯性指数为 $\displaystyle 2$.跟锦数学微信公众号. [在线资料](
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923、 1、 填空题 (每题 5 分, 共 30 分). (0-3)、 当 $\displaystyle \lambda=\underline\{\ \ \ \ \ \ \ \ \ \ \}$ 时, $\displaystyle x^2+3x+2\mid x^4+\lambda x^2-3x+2$. (长安大学2023年高等代数考研试题) [多项式 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由
\begin\{aligned\} x^4+\lambda x^2-3x+2 =(x^2-3x+\lambda+7)(x^2+3x+2)-(\lambda+6)(3x+2) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle \lambda=-6$.跟锦数学微信公众号. [在线资料](
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924、 2、 (10 分) 求多项式 $\displaystyle x^4-6x^2+8x-3$ 在有理数域上的标准分解. (长安大学2023年高等代数考研试题) [多项式 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 通过有理根判定知 $\displaystyle 1,-3$ 是题中多项式 $\displaystyle f$ 的根. 再作带余除法知 $\displaystyle f(x)=(x-1)^3(x+3)$.跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
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925、 (6)、 设 $\displaystyle f(x)$ 是首一三次多项式, 满足 $\displaystyle (x-1)^2\mid f(x)-1, (x+1)^2\mid f(x)+1$, 则 $\displaystyle f(x)=\underline\{\ \ \ \ \ \ \ \ \ \ \}$. (郑州大学2023年高等代数考研试题) [多项式 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /
\begin\{aligned\} &\quad \ (x-1)\mid f'(x), (x+1)\mid f'(x)\\\\ &\Rightarrow (x^2-1)\mid f'(x)\Rightarrow f'(x)=(x+a)(x^2-1)\\\\ &\Rightarrow f(x)=\frac\{x^4\}\{4\}+\frac\{ax^3\}\{3\}-\frac\{x^2\}\{2\}-ax\\\\ &\stackrel\{f(1)=1,f(-1)=-1\}\{\Rightarrow\}a=-\frac\{3\}\{2\}, b=\frac\{1\}\{4\}\\\\ &\Rightarrow f(x)=\frac\{x^4-2x^3-2x^2+6x+1\}\{4\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
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926、 2、 计算和证明题. (1)、 设 $\displaystyle f(x)=1+x+\cdots+x^\{n-1\}$. 证明: $\displaystyle f(x)\mid [(f(x)-x^n)^2-x^n]$. (郑州大学2023年高等代数考研试题) [多项式 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle g(x)=(f(x)-x^n)^2-x^n$, $\displaystyle \omega\_k=\mathrm\{e\}^\{\mathrm\{ i\}\frac\{2k\pi\}\{n\}\}, 1\leq k\leq n-1$, 则
\begin\{aligned\} f(x)=&\frac\{x^n-1\}\{x-1\}=\prod\_\{k=1\}^\{n-1\}(x-\omega\_k),\\\\ g(\omega\_k)=&\left\[f(\omega\_k)-1\right\]^2-1=(0-1)^2-1=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle f\mid g$.跟锦数学微信公众号. [在线资料](
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927、 1、 (1)、 设 $\displaystyle g(x),h(x)$ 互素, 求证: $\displaystyle (fh,g)=(f,g)$. (中国科学院大学2023年高等代数考研试题) [多项式 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 只要证明 $\displaystyle fh,g$ 与 $\displaystyle f,g$ 的公因式相同即可. (1)、 若 $\displaystyle d\mid f, d\mid g$, 则 $\displaystyle d$ 也是 $\displaystyle fh,g$ 的公因式. (2)、 若 $\displaystyle d\mid fh, d\mid g$, 则由
\begin\{aligned\} (g,h)=1\Rightarrow&\exists\ u,v,\mathrm\{ s.t.\} ug+vh=1 \Rightarrow vfg+vfh=f\\\\ \Rightarrow& d\mid (vfg+vfh)=f \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle d$ 也是 $\displaystyle f,g$ 的公因式.跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
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928、 2、 设 $\displaystyle f(x)=x^3+ax^2+bx+c$ 为整系数多项式. 证明: 若 $\displaystyle ac+bc$ 为奇数, 则 $\displaystyle f(x)$ 在有理数域上不可约. (中国科学院大学2023年高等代数考研试题) [多项式 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 用反证法. 若 $\displaystyle f(x)$ 在 $\displaystyle \mathbb\{Q\}$ 中可约, 则存在正数 $\displaystyle \alpha,d,e$, 使得
\begin\{aligned\} &f(x)=(x-\alpha)(x^2+dx+e)\\\\ \Rightarrow&x^3+ax^2+bx+c=x^3+(d-\alpha)x^2+(e-\alpha d)x-\alpha e\\\\ \Rightarrow&a=d-\alpha, b=e-\alpha d, c=-\alpha e. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由 $\displaystyle (a+b)c$ 是奇数知
\begin\{aligned\} \mbox\{ $\displaystyle c$ 是奇数\}&\Rightarrow \mbox\{ $\displaystyle \alpha,e$ 均是奇数\}\left(c=-\alpha e\right),\\\\ \mbox\{ $\displaystyle a+b$ 是奇数\}&\Rightarrow \left\\{\begin\{array\}\{llllllllllll\} \mbox\{ $\displaystyle a$ 奇, $\displaystyle b$ 偶\}&\Rightarrow \mbox\{ $\displaystyle d$ 偶\}\left(a=d-\alpha\right)\\\\ &\Rightarrow\mbox\{偶 $\displaystyle b=e-\alpha d$ 奇\}\Rightarrow\mbox\{矛盾\},\\\\ \mbox\{ $\displaystyle a$ 偶, $\displaystyle b$ 奇\}&\Rightarrow \mbox\{ $\displaystyle d$ 奇\}\left(a=d-\alpha\right)\\\\ &\Rightarrow \mbox\{奇 $\displaystyle b=e-\alpha d$ 偶\}\Rightarrow\mbox\{矛盾\}. \end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
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---
929、 1、 选择题. (1)、 设 $\displaystyle k,l$ 是整数, 且都大于 $\displaystyle 1$, 则 $\displaystyle x^\{3k\}+x^\{3l\}$ 除以 $\displaystyle x^2+x+1$ 的余式为 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$.
A. $\displaystyle x+1$
B. $\displaystyle 0$
C. $\displaystyle 1$
D. $\displaystyle 2$ (中国矿业大学(北京)2023年高等代数考研试题) [多项式 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / $\displaystyle D$. 设
\begin\{aligned\} x^\{3k\}+x^\{3l\}=q(x)(x^2+x+1)+ax+b, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则令 $\displaystyle x=\frac\{-1\pm \sqrt\{3\}\mathrm\{ i\}\}\{2\}$ 后得
\begin\{aligned\} 2=a\frac\{-1+\sqrt\{3\}\mathrm\{ i\}\}\{2\}+b, 2=a\frac\{-1- \sqrt\{3\}\mathrm\{ i\}\}\{2\}+b\Rightarrow a=0, b=2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
930、 2、 已知多项式 $\displaystyle f(x)$ 满足 $\displaystyle f(x)\mid f(x^n)$, 证明: $\displaystyle f(x)$ 的根只能是 $\displaystyle 0$ 或单位根. (中国矿业大学(北京)2023年高等代数考研试题) [多项式 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /
\begin\{aligned\} \mbox\{ $\displaystyle \alpha$ 是 $\displaystyle f$ 的根\}&\Rightarrow f(\alpha)=0\\\\ &\Rightarrow f(\alpha^n)=0\left(f(x)\mid f(x^n)\right)\\\\ &\Rightarrow f\left(\alpha^\{n^2\}\right)=f\left((\alpha^n)^n\right)=0\Rightarrow\cdots\\\\ &\Rightarrow f(\alpha^\{n^m\})=0\left(\forall\ m\geq 0\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这表明 $\displaystyle \partial(f)$ 次多项式 $\displaystyle f$ 有根 $\displaystyle \alpha,\alpha^n, \cdots,\alpha^\{n^m\},\cdots$. 它们不会全都相同 (否则 $\displaystyle f$ 有无穷多个根, 与代数基本定理矛盾),
\begin\{aligned\} \exists\ i\neq j,\mathrm\{ s.t.\} \alpha^\{n^i\}=\alpha^\{n^j\} &\Rightarrow \alpha^\{n^j\}\left(\alpha^\{n^i-n^j\}-1\right)=0\left(\mbox\{不妨设 $\displaystyle i > j$\}\right)\\\\ &\Rightarrow \mbox\{ $\displaystyle \alpha=0$ 或为单位根\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
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---
931、 1、 填空题 (每题 5 分, 共 50 分). (1)、 多项式 $\displaystyle x^4+4x^3-2x^2-12x+9$ 的有理根是 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$. (中国矿业大学(徐州)2023年高等代数考研试题) [多项式 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设题中多项式为 $\displaystyle f(x)$, 则计算 $\displaystyle f(\pm 1),f(\pm 3), f(\pm 9)$ 后发现只有 $\displaystyle f(1)=0, f(-3)=0$. 故 $\displaystyle f$ 的有理根是 $\displaystyle 1,-3$.跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
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932、 2、 (10 分) 设复系数多项式 $\displaystyle f(x)$ 没有重因式, 证明: $\displaystyle 2f(x)+f'(x)$ 与 $\displaystyle f(x)$ 是互素的. (中国矿业大学(徐州)2023年高等代数考研试题) [多项式 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由 $\displaystyle f$ 没有重因式知
\begin\{aligned\} (f,f')=1\Rightarrow (2f+f',f)=(f',f)=1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
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---
933、 1、 判断题. (1)、 设 $\displaystyle f(x)=g(x)h(x)$, 其中 $\displaystyle f(x),g(x)$ 都是整系数多项式, $\displaystyle h(x)$ 是有理系数多项式, 则 $\displaystyle h(x)$ 也是整系数多项式. (中国人民大学2023年高等代数考研试题) [多项式 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / $\displaystyle \times$. 比如 $\displaystyle f(x)=x^2, g(x)=2x, h(x)=\frac\{1\}\{2\}x$.跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
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---
934、 (2)、 多项式 $\displaystyle x^6+5x-54$ 在任意数域上可约. (中国人民大学2023年高等代数考研试题) [多项式 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / $\displaystyle \surd$. 设 $\displaystyle f(x)=x^6+5x-54$, 则 $\displaystyle f(-2)=64-10-54=0$. 而 $\displaystyle (x+2)\mid f(x)$. 进一步,
\begin\{aligned\} f(x)=(x+2)(x^5-2x^4+4x^3-8x^2+16x-27). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle f$ 在 $\displaystyle \mathbb\{Q\}$ 中可约, 而在任意数域中可约.跟锦数学微信公众号. [在线资料](
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935、 2、 填空题. (1)、 设 $\displaystyle n$ 为正整数, 则 $\displaystyle x^\{2n+1\}-1$ 在实数域 $\displaystyle \mathbb\{R\}$ 上的标准分解式为 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$. (中国人民大学2023年高等代数考研试题) [多项式 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由 $\displaystyle x^\{2n+1\}-1$ 的全体复根为 $\displaystyle \mathrm\{e\}^\{\mathrm\{ i\} \frac\{2k\pi\}\{2n+1\}\}, -n\leq k\leq n$ 知
\begin\{aligned\} x^\{2n+1\}-1=&(x-1)\prod\_\{k=1\}^n \left\[\left(x-\mathrm\{e\}^\{\mathrm\{ i\} \frac\{2k\pi\}\{2n+1\}\}\right) \left(x-\mathrm\{e\}^\{-\mathrm\{ i\}\frac\{2k\pi\}\{2n+1\}\}\right)\right\]\\\\ =&(x-1)\prod\_\{k=1\}^n \left(x^2-2x\cos\frac\{2k\pi\}\{2n+1\}+1\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
936、 (2)、 设 $\displaystyle \alpha\_1,\alpha\_2,\alpha\_3$ 是多项式 $\displaystyle x^3+px^2+qx+r$ 的三个根, 其中 $\displaystyle r\neq 0$, 则
\begin\{aligned\} \frac\{1\}\{\alpha\_1^2\}+\frac\{1\}\{\alpha\_2^2\}+\frac\{1\}\{\alpha\_3^2\}=\underline\{\ \ \ \ \ \ \ \ \ \ \}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(中国人民大学2023年高等代数考研试题) [多项式 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由
\begin\{aligned\} &x^3+px^2+qx+r=(x-\alpha\_1)(x-\alpha\_2)(x-\alpha\_3)\\\\ \Rightarrow&\frac\{1\}\{x^3\}+p\frac\{1\}\{x^2\}+q\frac\{1\}\{x\}+r=\left(\frac\{1\}\{x\}-\alpha\_1\right)\left(\frac\{1\}\{x\}-\alpha\_2\right)\left(\frac\{1\}\{x\}-\alpha\_3\right)\\\\ \Rightarrow&rx^3+qx^2+px+1=(1-\alpha\_1 x)(1-\alpha\_2x)(1-\alpha\_3x) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle \frac\{1\}\{\alpha\_1\}, \frac\{1\}\{\alpha\_2\}, \frac\{1\}\{\alpha\_3\}$ 是 $\displaystyle rx^3+qx^2+px+1=0$ 的根, 而
\begin\{aligned\} &\frac\{1\}\{\alpha\_1^2\}+\frac\{1\}\{\alpha\_2^2\}+\frac\{1\}\{\alpha\_3^2\}\\\\ =&\left(\frac\{1\}\{\alpha\_1\}+\frac\{1\}\{\alpha\_2\}+\frac\{1\}\{\alpha\_3\}\right)^2 -2\left(\frac\{1\}\{\alpha\_1\alpha\_2\}+\frac\{1\}\{\alpha\_2\alpha\_3\}+\frac\{1\}\{\alpha\_3\alpha\_1\}\right)\\\\ =& \left(-\frac\{q\}\{r\}\right)^2-2\frac\{p\}\{r\}=\frac\{q^2-2pr\}\{r^2\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
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---
937、 3、 设 $\displaystyle a,b$ 是互异常数. (1)、 求 $\displaystyle (x-a)(x-b)$ 除多项式 $\displaystyle f(x)$ 的余式. (2)、 求 $\displaystyle x^2-1$ 除 $\displaystyle f(x)=x^4+x^3+x+1$ 的商和余式; (3)、 求 $\displaystyle 99999999$ 除 $\displaystyle 10001000000010001$ 的商和余数. (中国人民大学2023年高等代数考研试题) [多项式 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 余式为 $\displaystyle \frac\{[f(b)-f(a)]x+[bf(a)-af(b)]\}\{b-a\}$. 事实上,
\begin\{aligned\} &f(x)=q(x)(x-a)(x-b)+Ax+B\left(\mbox\{辗转相除法\}\right)\\\\ \Rightarrow&\left\\{\begin\{array\}\{llllllllllll\} f(a)=Aa+B\\\\ f(b)=Ab+B \end\{array\}\right. \Rightarrow \left\\{\begin\{array\}\{llllllllllll\} A=\frac\{f(b)-f(a)\}\{b-a\}\\\\ B=\frac\{bf(a)-af(b)\}\{b-a\} \end\{array\}\right.. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 按第 1 步的公式代入即知余式为 $\displaystyle 2x+2$. (3)、 硬算得
\begin\{aligned\} 10001000000010001 =100020001\times 99999999+20002. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故商为 $\displaystyle 100020001$, 余数为 $\displaystyle 20002$.跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
938、 5、 设 $\displaystyle f(x)=4x^3-21x-2019$. (1)、 证明: $\displaystyle f(x)$ 在有理数域 $\displaystyle \mathbb\{Q\}$ 上不可约. (2)、 设 $\displaystyle \alpha$ 是 $\displaystyle f(x)$ 在复数域 $\displaystyle \mathbb\{C\}$ 的一个根, 记
\begin\{aligned\} \mathbb\{Q\}[\alpha]=\left\\{a\_0+a\_1\alpha+a\_2\alpha^2; a\_0,a\_1,a\_2\in\mathbb\{Q\}\right\\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
证明: 对任意 $\displaystyle g(x)\in \mathbb\{Q\}[x]$, 有 $\displaystyle g(\alpha)\in\mathbb\{Q\}[\alpha]$. (3)、 证明: 若 $\displaystyle 0\neq \beta\in \mathbb\{Q\}[\alpha]$, 存在 $\displaystyle \gamma\in\mathbb\{Q\}[\alpha]$, 使得 $\displaystyle \beta\gamma=1$. (中国人民大学2023年高等代数考研试题) [多项式 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 用反证法. 若 $\displaystyle f$ 在 $\displaystyle \mathbb\{Q\}$ 中可约, 则
\begin\{aligned\} \exists\ g,h\in \mathbb\{Z\}[x],\mathrm\{ s.t.\} &f(x)=g(x)h(x), 1\leq \deg g\leq \deg h\leq 2\\\\ \Rightarrow& \deg g=1, \deg h=2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle f$ 有有理根. 但 $\displaystyle f$ 的有理根必定是 $\displaystyle \frac\{s\}\{r\}, s\in \left\\{\pm 1, \pm 29\right\\}, r\in \left\\{1,2,4\right\\}$. 验算后发现上述 $\displaystyle \frac\{s\}\{r\}$ 满足 $\displaystyle f\left(\frac\{s\}\{r\}\right)\neq 0$. 这是一个矛盾. 故有结论. (2)、 对 $\displaystyle \forall\ g\in\mathbb\{Q\}[x]$, 由带余除法知
\begin\{aligned\} &g(x)=q(x)f(x)+r(x), r(x)=0\mbox\{或\} \deg r\leq 2\\\\ \Rightarrow&g(\alpha)=q(\alpha)f(\alpha)+r(\alpha)=r(\alpha)\in \mathbb\{Q\}[\alpha]. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(3)、 对 $\displaystyle \forall\ 0\neq \beta\in \mathbb\{Q\}[\alpha]$, 则存在不全为零的 $\displaystyle a\_0,a\_1,a\_2\in\mathbb\{Q\}$ 使得 $\displaystyle \beta=a\_0+a\_1\alpha+a\_2\alpha^2$. 设 $\displaystyle g(x)=a\_0+a\_1x+a\_2x^2$, 则由 $\displaystyle f$ 在 $\displaystyle \mathbb\{Q\}$ 中不可约知 $\displaystyle f\mid g$ 或 $\displaystyle (f,g)=1$. 但 $\displaystyle \deg g\leq 2\Rightarrow f\nmid g$, 而
\begin\{aligned\} (f,g)=1\Rightarrow& \exists\ u,v\in\mathbb\{Q\}[x],\mathrm\{ s.t.\} u(x)f(x)+v(x)g(x)=1\\\\ \Rightarrow&1=u(\alpha)f(\alpha)+v(\alpha)g(\alpha)=v(\alpha)g(\alpha)=v(\alpha)\beta. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由第 2 步知 $\displaystyle \gamma\equiv v(\alpha)\in\mathbb\{Q\}[\alpha]$, 且 $\displaystyle 1=\gamma\beta$.跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
939、 1、 (16 分) 设 $\displaystyle f\_1,\cdots,f\_n\ (n\geq 2)$ 均为实多项式. 若
\begin\{aligned\} (1+x+\cdots+x^n)\mid \left\[\begin\{array\}\{c\}x^\{n-1\}f\_1(x^\{n+1\}) +x^\{n-2\}f\_2(x^\{n+1\})+\cdots\\\\+xf\_\{n-1\}(x^\{n+1\})+f\_n(x^\{n+1\})\end\{array\}\right\], \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
证明: $\displaystyle f\_i(1)=0, 1\leq i\leq n$. (中南大学2023年高等代数考研试题) [多项式 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 令 $\displaystyle x=\omega\_k\equiv \mathrm\{e\}^\{\mathrm\{ i\} \frac\{2k\pi\}\{n+1\}\}, 1\leq k\leq n+1$, 则
\begin\{aligned\} &\omega\_k^\{n-1\}f\_1(1)+\omega\_k^\{n-2\}f\_2(1)+\cdots+\omega\_k f\_\{n-1\}(1)+f\_n(1)=0, 1\leq k\leq n\\\\ \Leftrightarrow&\left(\begin\{array\}\{cccccccccccccccccccc\}\omega\_1^\{n-1\}&\omega\_1^\{n-1\}&\cdots&\omega\_1&1\\\\ \omega\_2^\{n-1\}&\omega\_2^\{n-1\}&\cdots&\omega\_2&1\\\\ \vdots&\vdots&&\vdots&\vdots\\\\ \omega\_n^\{n-1\}&\omega\_n^\{n-1\}&\cdots&\omega\_n&1\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}f\_1(1)\\\\ f\_2(1)\\\\ \vdots\\\\ f\_n(1)\end\{array\}\right)=\left(\begin\{array\}\{cccccccccccccccccccc\}0\\\\0\\\\\vdots\\\\0\end\{array\}\right)\\\\ \Rightarrow&f\_k(1)=0, 1\leq k\leq n. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
940、 3、 (15 分) 已知
\begin\{aligned\} f(x)=x^4+2x^3-4x^2-7x-2, g(x)=x^3-2x-4. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(1)、 求 $\displaystyle f(x)$ 与 $\displaystyle g(x)$ 的首一最大公因式 $\displaystyle \left(f(x),g(x)\right)$; (2)、 求多项式 $\displaystyle u(x),v(x)$, 使得
\begin\{aligned\} u(x)f(x)+v(x)g(x)=\left(f(x),g(x)\right), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
且 $\displaystyle \deg u(x) < \deg g(x), \deg v(x) < \deg f(x)$. (中山大学2023年高等代数考研试题) [多项式 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由
\begin\{aligned\} f(x)=&(x+2)g(x)+r\_1(x), r\_1(x)=-2x^2+x+6,\\\\ g(x)=&\left(-\frac\{x\}\{2\}-\frac\{1\}\{4\}\right)r\_1(x)+r\_2(x), r\_2(x)=\frac\{5\}\{4\}x-\frac\{5\}\{2\},\\\\ r\_1(x)=&\left(-\frac\{8\}\{5\}x-\frac\{12\}\{5\}\right)r\_2(x) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知
\begin\{aligned\} x-2=&(f,g)=\frac\{4\}\{5\}r\_2(x)=\frac\{4\}\{5\}\left\[g(x)+\left(\frac\{x\}\{2\}+\frac\{1\}\{4\}\right)r\_1(x)\right\]\\\\ =&\frac\{4\}\{5\}g(x)+\frac\{1\}\{5\}(2x+1)[f(x)-(x+2)g(x)]\\\\ =&\frac\{2x+1\}\{5\}f(x)+\frac\{-2x^2-5x+2\}\{5\}g(x). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
取 $\displaystyle u(x)=\frac\{2x+1\}\{5\}, v(x)=\frac\{-2x^2-5x+2\}\{5\}$ 就满足题设要求.跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
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---
941、 1、 (12 分) 设由多项式
\begin\{aligned\} f(x)=x^2-x-2, g(x)=x^3+x^2-x-1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
求 $\displaystyle u(x),v(x)$ 使得
\begin\{aligned\} u(x)f(x)+v(x)g(x)=\left(f(x),g(x)\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(重庆大学2023年高等代数考研试题) [多项式 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由
\begin\{aligned\} g(x)=&(x+2)f(x)+r\_1(x), r\_1(x)=3(x+1),\\\\ f(x)=&\frac\{x-2\}\{3\}r\_1(x) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知
\begin\{aligned\} x+1=&\left(f(x),g(x)\right)=\frac\{1\}\{3\}[g(x)-(x+2)f(x)]\\\\ =&-\frac\{x+2\}\{3\}f(x)+\frac\{1\}\{3\}g(x). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
942、 7、 (14 分) 设 $\displaystyle a\_1,\cdots,a\_n$ 是互不相同的整数, 且满足
\begin\{aligned\} f(x)=(x-a\_1)\cdots(x-a\_n)-1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
证明: $\displaystyle f(x)$ 在有理数域上不可约. (重庆大学2023年高等代数考研试题) [多项式 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 我们证明
\begin\{aligned\} f(x)=(x-a\_1)(x-a\_2)\cdots(x-a\_n)-1 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
在有理数域上不可约. 用反证法. 若 (任一非零有理系数多项式均可写成一个有理数与一个本原多项式的乘积)
\begin\{aligned\} f(x)=g(x)h(x),\quad g(x)\in\mathbb\{Z\}[x],\ h(x)\in \mathbb\{Z\}[x],\quad 1\leq \partial(g),p(h) < n. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则
\begin\{aligned\} g(a\_i)h(a\_i)=f(a\_i)=-1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
注意到 $\displaystyle g(a\_i),h(a\_i)\in\mathbb\{Z\}$, 而
\begin\{aligned\} g(a\_i)=\pm 1,\quad h(a\_i)=\mp 1\Rightarrow g(a\_i)+h(a\_i)=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(1)、 若 $\displaystyle g+h=0$, 则 $\displaystyle f=gh=-g^2$. 注意到 $\displaystyle -g^2$ 的首项系数为负数, 与 $\displaystyle f$ 的首项系数为 $\displaystyle 1$ 矛盾. (2)、 若 $\displaystyle g+h\neq 0$, 则次数 $\displaystyle < n$ 的多项式 $\displaystyle g(x)+h(x)$ 有 $\displaystyle n$ 个互不相同的根. 这与代数学基本定理矛盾. 不论何种情形, 都得到矛盾. 故有结论.跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
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---
943、 1、 (12 分) 设 $\displaystyle f(x)$ 是一个三次多项式, 问 $\displaystyle f(x)$ 重根如何? [题目不全, 跟锦数学微信公众号没法做哦.] (重庆师范大学2023年高等代数考研试题) [多项式 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / [题目不全, 跟锦数学微信公众号没法做哦.]跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
发表于 2023-3-5 09:16:05
