Let $f(x,y)=x+y$ and $g(x,y)=xy$, prove that the closure of $\mathbb{Q} \cup\{\sqrt{2}\}$ under $\{f, g\}$ is $\mathbb{Q}(\sqrt{2})$

Question

$Proof$.

Let $H$ be the set of all supersets of $\mathbb{Q} \cup\{\sqrt{2}\}$ closed under $\lbrace f, g\rbrace $ and
$Z=\mathbb{Q}(\sqrt{2})= \lbrace{a+b\sqrt{2}\mid a,b\in\mathbb{Q}\\}\rbrace$
Then $\cap H$ is the closure of $\mathbb{Q} \cup\{\sqrt{2}\}$ under $\lbrace f,g\rbrace$. We prove that $\cap H=Z$.

1. $\cap H \subseteq Z$.
   We show $Z \in H$. For any $a \in \mathbb{Q}, a=a+0 \sqrt{2}$, so $a \in Z$. Also $\sqrt{2}=0+1 \sqrt{2} \in Z$. Hence $\mathbb{Q} \cup\lbrace\sqrt{2}\rbrace \subseteq Z$. Now take any $x, y \in Z$ and write them as $x=a+b \sqrt{2}$ and $y=c+d \sqrt{2}$. Then $f(x, y)=a+b \sqrt{2}+c+d \sqrt{2}=(a+c)+\sqrt{2}(b+d) \in Z$
   and
   $g(x, y)=(a+b \sqrt{2})(c+d \sqrt{2})=a c+c b \sqrt{2}+a d \sqrt{2}+2 d b=(a c+2 d b)+\sqrt{2}(c b+a d) \in Z$.
   Hence $Z$ is closed under $\lbrace f, g\rbrace$, and so $Z \in H$, which implies $\cap H \subseteq Z$.

2. $Z\subseteq\cap H$.

   Let $x \in Z$ and $C \in H$. If $x \in \mathbb{Q} \cup\{\sqrt{2}\}$, then $x \in \cap H$ since $\mathbb{Q} \cup\lbrace\sqrt{2}\rbrace \subseteq \cap H$. Otherwise $x$ is irrational. We can write $x=a+b \sqrt{2}$ for some $a, b \in \mathbb{Q}$. Since $x$ is irrational, $b \neq 0$, and therefore $x=a+b \sqrt{2}=b\left(\frac{a}{b}+\sqrt{2}\right)$. Division of two rationals yields a rational, so $\frac{a}{b} \in \mathbb{Q}$. Then $\frac{a}{b} \in \mathbb{Q} \cup\lbrace\sqrt{2}\rbrace \subseteq C$ and $\sqrt{2} \in \mathbb{Q} \cup\lbrace\sqrt{2}\rbrace \subseteq C$, so since $C$ is closed under $f, f\left(\frac{a}{b}, \sqrt{2}\right)=\frac{a}{b}+\sqrt{2} \in C$. Then $b \in C$, so $f\left(b, \frac{a}{b}+\sqrt{2}\right)=a+b \sqrt{2}=x \in C$. Therefore $x \in \cap H$, and thus $Z\subseteq \cap H$.

Is the proof valid?