# Integral element in $\mathbb{Q}(\sqrt{2})$ over $\mathbb{Z}$ has form $m+n\sqrt{2}$ [duplicate]

I have studied about integral element. And I am not clear about an example.

Could you please help me explain about it?

Let R is a subring of S. $s \in S$ is integral if $s$ is a zero of some monic polynomial in $R[x]$.

Example:

Every elements $s=a+b\sqrt{2} \in \mathbb{Q}(\sqrt{2})$ that is integral over $\mathbb{Z}$ is in $\mathbb{Z}[\sqrt{2}]$.

## 1 Answer

Note that the minimal polynomial for any element of $\mathbb{Q}(\sqrt{2})$ can be written as $$m(x) = x^2 - Tx + N = \big(x - (a + b\sqrt{2})\big)\big(x - (a - b\sqrt{2})\big)$$ where $T = (a + b \sqrt{2}) + (a - b\sqrt{2}) = 2a$ and $N = (a + b \sqrt{2}) \cdot (a - b\sqrt{2}) = a^2 - 2b^2$, and with $a, b \in \mathbb{Q}$.

If $a + b\sqrt{2}$ is integral, then we must have $T, N \in \mathbb{Z}$. So what conclusions can you draw from this?

• Thank you! I think I got it. I calculate in case $a=\frac{1}{2} +$ integer and $a \in \mathbb{Z}$. Then both $a,b \in \mathbb{Z}$. – user69833 Feb 3 '16 at 6:18