$A^B =\{\,b \in B : \text{$b$ is integral over $A$}\,\}$
Let $A=\Bbb{Z}$ and $B=\Bbb{Q}[\sqrt{2}]$ . Prove that $A^B=\Bbb{Z}[\sqrt{2}]$ Thank you for your helping. i am sorry for my poor English
i have tried to show $\Bbb{Z}[\sqrt{2}]$ is integrally closed in $\Bbb{Q}[\sqrt{2}]$ but i dont know how to start and what kind kind of element i should take
You can show that an element $a+b \sqrt{2}$ with $a, b \in \mathbb{Z}$ is integral over $\mathbb{Z}$.
Conversely suppose $r+s\sqrt{2}$ is integral over $\mathbb{Z}$, where $r, s \in \mathbb{Q}$. We'll show that $r, s \in \mathbb{Z}$. Let $K = \mathbb{Q}(\sqrt{2})$. Now any $\mathbb{Q}$-embedding of $K$ into $\mathbb{C}$ sends integral elements to integral elements. Also a sum or product of integral elements is integral. Hence the trace and norm maps $K \rightarrow \mathbb{Q}$ send $r+s \sqrt{2}$ to $\mathbb{Z}$.
The only $\mathbb{Q}$-embeddings of $K$ into $\mathbb{C}$ are the identity map and the map which sends $\sqrt{2}$ to $-\sqrt{2}$ (so $K$ is in fact a Galois extension of $\mathbb{Q}$). Thus $$Tr_{K/\mathbb{Q}}(r+ s \sqrt{2}) = (r+s \sqrt{2}) + (r - s \sqrt{2}) = 2r$$ is in $\mathbb{Z}$, as well as $$N_{K/ \mathbb{Q}}(r+s \sqrt{2}) = (r+ s \sqrt{2})(r - s \sqrt{2}) = r^2 - 2s^2$$ Alternatively this can be seen by finding the minimal polynomial of $r+ s \sqrt{2}$ as suggested in the comments.
Now $4r^2 \in \mathbb{Z}$, so $8s^2 \in \mathbb{Z}$ as well. Now $8s^2 = 2(2s)^2$. If we factor $2s$ as a product of primes with integral exponents, then if $2s$ isn't an integer, there will be primes which occur in the denominator, and hence primes with multiplicity at least two which occur in the denominator of $(2s)^2$. The number $2$ in the numerator cannot possibly cancel them out, so we must conclude $2s \in \mathbb{Z}$.
So $(2r)^2 - 2(2s)^2 = 4r^2 - 8s^2 = 4(r^2 - 2s^2)$ is an integer, divisible by $4$, with $2r, 2s$ both integers. The squares modulo $4$ are $0$ and $1$, so you can see that $(2r)^2$ and $2(2s)^2$ both have to be congruent to $0$ modulo $4$. This means that $2r$ has to be even, so $r$ is an integer. Also $2(2s)^2 \equiv 0$ (mod $4$) implies $(2s)^2$ is even, which means $s$ has to also be an integer.