Exercise 1.3

Question

Let $\mathbb Q[\alpha,\beta]$ denote the smallest subring of $\mathbb C$ containing the rational numbers $\mathbb Q$ and the elements $\alpha = \sqrt{2}$ and
$\beta = \sqrt{3}$. Let $\gamma = \alpha + \beta$. Is $\mathbb Q[\alpha,\beta] = \mathbb Q[\gamma]$? Is $\mathbb Z[\alpha,\beta] = \mathbb Z[\gamma]$

Abdullah Zubair · Accepted Answer

First, we show that $\mathbb Q[\alpha,\beta]$ is contained in $\mathbb Q[\gamma]$. Note that 
\begin{align*}
	\alpha = \dfrac{1}{2}(\alpha + \beta)^3 - 9(\alpha + \beta)
.\end{align*}
So $\alpha \in \mathbb Q[\gamma]$. Similarly, 
\begin{align*}
	\beta = (\alpha + \beta)^{3} - 3\alpha - 8(\alpha + \beta)
.\end{align*}
So $\beta \in \mathbb Q[\gamma]$. Conversely, note that $\alpha + \beta \in \mathbb Q[\alpha,\beta]$, and so we conclude that $\mathbb Q[\alpha,\beta] = \mathbb Q[\gamma]$.

We claim that $\mathbb Z[\alpha,\beta] 
eq \mathbb Z[\gamma]$. To see this, note that $\mathbb Z[\alpha,\beta] = \mathbb Z(\alpha,\beta)$ and $\mathbb Z[\gamma] =
\mathbb Z(\gamma)$, and so both $\mathbb Z(\gamma)$ and $\mathbb Z(\alpha, \beta)$ are field extensions of $\mathbb Z$. The basis for the vector space $\mathbb Z(\gamma)$
over $\mathbb Z$ is $\{1,\gamma,\gamma^2,\gamma^{3}\}$ and the basis for the vector space $\mathbb Z(\alpha,\beta)$ over $\mathbb Z$ is $\{1,\alpha,\beta,\alpha\beta\}$. Assuming towards a
contradiction, if $\mathbb Z(\gamma) = \mathbb Z(\alpha, \beta)$ then there exists scalars $t_3,t_2,t_1,t_0 \in \mathbb Z$
such that
\begin{align*}
	t_3\gamma^{3} + t_2\gamma^2 + t_1\gamma + t_0 = \alpha
	\implies t_3\gamma^{3} + t_2\gamma^2 + t_1\gamma + t_0 - \alpha = 0
.\end{align*}
This would imply that the set $\{1,\alpha,\beta,\alpha\beta\}$ is linearly dependent over $\mathbb Z$, which contradicts the fact that $\{1,\alpha,\beta,\alpha\beta\}$ is a
basis for $\mathbb Z(\alpha,\beta)$ over $\mathbb Z$.

Richard Ganaye · Answer

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{\U}{\mathbb{U}}

\begin{proof} 
Since $\gamma = \alpha + \beta \in \Q[\alpha,\beta]$, $\Q[\gamma]$ being the smallest subring of $\C$ containing $\Q$ and $\gamma$, we obtain first
$$\Q[\gamma] \subseteq \Q[\alpha,\beta].$$
Starting from 
$$\gamma^3 = (\sqrt{2} + \sqrt{3})^3 = 2 \sqrt{2} + 3 \sqrt{3} + 6 \sqrt{3} + 9 \sqrt{2} = 11 \sqrt{2} + 9 \sqrt{3},$$
 we obtain the system
\begin{align*}
\gamma &= \sqrt{2} + \sqrt{3},\
\gamma^3 &= 11\sqrt{2} + 9 \sqrt{3}.
\end{align*}
This gives
\begin{align}
\alpha &= \sqrt{2} = \frac{1}{2} (\gamma^3 - 9 \gamma),\
\beta &= \sqrt{3} = \frac{1}{2}(11\,  \gamma - \gamma^3).
\end{align}
This shows that $\alpha \in \Q[\gamma], \beta \in \Q[\gamma]$. Since $\Q[\alpha,\beta]$ is the smallest subring of $\C$ containing $\Q$ and $\alpha,\beta$, this gives
$$\Q[\alpha,\beta] \subseteq \Q[\gamma],$$
so $$\Q[\alpha,\beta] = \Q[\gamma].$$

\medskip
Reasoning by contradiction, assume that $\Z[\alpha,\beta] = \Z[\gamma]$.  Then $\alpha \in \Z[\gamma]$.

But $\Z[\gamma]$ is the subring of $\C$ whose elements are of the form $P(\gamma)$, where $P \in \Z[x]$. Therefore 
$$\alpha = P(\gamma), \qquad P \in \Z[x].$$

\medskip
To obtain the minimal polynomial of $\gamma$ over $\Q$, note that
$$\gamma - \sqrt{2} = \sqrt{3},\quad \gamma^2 - 2 \sqrt{2} \gamma + 2 = 3,\quad  \gamma^2 - 1 = 2 \sqrt{2} \gamma,\quad  (\gamma^2 - 1)^2 = 8 \gamma^2,$$ so
$$\gamma^4 - 10 \gamma^2 + 1 = 0.$$
Moreover $Q(x) = x^4 - 10x^2 + 1$ is irreducible over $\Q$:
\begin{verbatim}
sage: L.<x> = QQ['x']
sage: P = x^4 -10*x^2 + 1
sage: P.is_irreducible()
True
\end{verbatim}
(We can give a more theoretical justification by proving $[\Q(\sqrt{2},\sqrt{3}):\Q] = 4$.)

Since the leading coefficient of $Q$ is $1$, the Euclidean division of $P$ by $Q$ gives
$$P(x) = Q(x) S(x) + R(x), \qquad R  \in \Z[x],\quad \deg(R)<4.$$
\begin{align}
\alpha =P(\gamma) = R (\gamma) = a \gamma^3 + b \gamma^2 + c \gamma + d\qquad (a,b,c,d \in \Z)
\end{align}
This writing of $\alpha$ is unique, because $Q(x)$ is the minimal polynomial of $\gamma$ over $\Q$: 
if $a \gamma^3 + b \gamma^2 + c \gamma + d = a' \gamma^3 + b' \gamma^2 + c' \gamma + d'$, then $\gamma$ is a root of $S(x) = (a-a')x^3 + (b-b')x^2 + (c-c')x + (d-d')$, whose degree is less than the degree of the minimal polynomial of $\gamma$, thus $S(x) = 0$ and $a=a', b= b', c= c', d= d'$.

(This shows that $(1,\gamma, \gamma^2, \gamma^3)$ is a $\Z$-basis of the $\Z$-module $\Z[\gamma]$.)

The comparison of (2) and (3) gives
$$\alpha = \frac{1}{2} \gamma^3 - \frac{9}{2} \gamma = a \gamma^3 + b \gamma^2 + c \gamma + d.$$

The unicity of the decomposition of $\alpha$ gives $\frac{1}{2} = a \in \Z$: this is a contradiction. Therefore
$$\Z[\alpha,\beta] 
e \Z[\alpha + \beta].$$
\end{proof}

Exercise 1.3

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Exercise 1.3

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