## 这是Math 4120，abstract agebra的一次作业分析。

Problem 1.

Let $u$ be a root of the polynomial $x^{3}+3 x+3 .$ In $\mathbf{Q}(u)$, express $\left(7-2 u+u^{2}\right)^{-1}$ in the form $a+b u+c u^{2}$.

Proof .

Solution: Dividing $x^{3}+3 x+3$ by $x^{2}-2 x+7$ gives the quotient $x+2$ and remainder $-11$. Thus $u^{3}+3 u+3=(u+2)\left(u^{2}-2 u+7\right)-11$, and so $\left(7-2 u+u^{2}\right)^{-1}=(2+u) / 11=(2 / 11)+(1 / 11) u .$

Problem 2.

(a) Show that $\mathbf{Q}(\sqrt{2}+i)=\mathbf{Q}(\sqrt{2}, i)$.
(b) Find the minimal polynomial of $\sqrt{2}+i$ over $\mathbf{Q}$.

Proof .

(a) Show that $\mathbf{Q}(\sqrt{2}+i)=\mathbf{Q}(\sqrt{2}, i)$.
Solution: Let $u=\sqrt{2}+i$. Since $(\sqrt{2}+i)(\sqrt{2}-i)=2-i^{2}=3$, we have $\sqrt{2}-i=3(\sqrt{2}+i)^{-1} \in \mathbf{Q}(u)$, and it follows easily that $\sqrt{2} \in \mathbf{Q}(u)$ and
$i \in \mathbf{Q}(u)$, so $\mathbf{Q}(\sqrt{2}, i) \subseteq \mathbf{Q}(u) .$ The reverse inclusion is obvious.
(b) Find the minimal polynomial of $\sqrt{2}+i$ over $\mathbf{Q}$. Solution: We have $\mathbf{Q} \subseteq \mathbf{Q}(\sqrt{2}) \subseteq \mathbf{Q}(\sqrt{2}, i)$. Thus $[\mathbf{Q}(\sqrt{2}): \mathbf{Q}]=2$ since $\sqrt{2}$
is a root of a polynomial of degree 2 but is not in $\mathbf{Q} .$ We have $[\mathbf{Q}(\sqrt{2}, i)$ :
$\mathbf{Q}(\sqrt{2})]=2$ since $i$ is a root of a polynomial of degree 2 over $\mathbf{Q}(\sqrt{2})$ but is not in $\mathbf{Q}(\sqrt{2})$. Thus $[\mathbf{Q}(\sqrt{2}+i): \mathbf{Q}]=4$, and so the minimal polynomial for $\sqrt{2}+i$ must have degree $4 .$

Since $u=\sqrt{2}+i$, we have $u-i=\sqrt{2}, u^{2}-2 i u+i^{2}=2$, and $u^{2}-3=2 i u$. Squaring again and combining terms gives $u^{4}-2 u^{2}+9=0$. Thus the minimal polynomial for $\sqrt{2}+i$ is $x^{4}-2 x^{2}+9$

Problem 3.

Find the minimal polynomial of $1+\sqrt[3]{2}$ over $\mathbf{Q}$.

Proof .

Solution: Let $x=1+\sqrt[3]{2} .$ Then $x-1=\sqrt[3]{2}$, and so $(x-1)^{3}=2$, which yields $x^{3}-3 x^{2}+3 x-1=2$, and therefore $x^{3}-3 x^{2}+3 x-3=0$. Eisenstein’s criterion (with $p=3$ ) shows that $x^{3}-3 x^{2}+3 x-3$ is irreducible over $\mathbf{Q}$, so this is the required minimal polynomial.

Problem 4.

Find a basis for $\mathbf{Q}(\sqrt{5}, \sqrt[3]{5})$ over $\mathbf{Q}$.

Solution: The set $\{1, \sqrt[3]{5}, \sqrt[3]{25}\}$ is a basis for $\mathbf{Q}(\sqrt[3]{5})$ over $\mathbf{Q}$, and since this extension has degree 3 , the minimal polynomial $x^{2}-5$ of $\sqrt{5}$ remains irreducible in the extension $\mathbf{Q}(\sqrt[3]{5})$. Therefore $\{1, \sqrt{5}$ is a basis for $\mathbf{Q}(\sqrt{5}, \sqrt[3]{5})$ over $\mathbf{Q}(\sqrt[3]{5})$, and so the proof of Theorem $6.2 .4$ shows that the required basis is $\{1, \sqrt{5}, \sqrt[3]{5}, \sqrt{5} \sqrt[3]{5}, \sqrt[3]{25}, \sqrt{5} \sqrt[3]{25}\}$.

Problem 5.

Show that $x^{3}+6 x^{2}-12 x+2$ is irreducible over $\mathbf{Q}$, and remains irreducible over $\mathbf{Q}(\sqrt[5]{2})$.

Solution: Eisenstein’s criterion works with $p=2 .$ Since $x^{5}-2$ is also irreducible by Eisenstein’s criterion, $[\mathbf{Q}(\sqrt[5]{2}): \mathbf{Q}]=5 .$ If $x^{3}+6 x^{2}-12 x+2$ could be factored over $\mathbf{Q}(\sqrt[5]{2})$, then it would have a linear factor, and so it would have a root in $\mathbf{Q}(\sqrt[5]{2})$. This root would have degree 3 over $\mathbf{Q}$, and that is impossible since 3 is not a divisor of 5 .

##### Math 4120 abstract agebra

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# MATH 4120 – Algebra I

3 Credits (3 Contact Hours)

Provides a first introduction to algebra with topics including modular arithmetic, ring theory and group theory. Preq: MATH 3110 and MATH 3190, each with a C or better.This 4000-level course has a 6000-level counterpart. Students should refer to the Graduate Announcements for the 6000-level description and requirements.