$a+b\sqrt{2}$ not a root of monic polynomial over $\mathbb{Z}$

Question

Consider $a+b\sqrt{2}$ for $a,b \in \mathbb{Q}-\mathbb{Z}$ . I need to show that it cannot be a root of any monic polynomial with coefficients in $\mathbb{Z}$

Answer 1

The (possibly complex) roots of $X^2+pX+q$ are $\frac{-p\pm\sqrt{p^2-4q}}{2}$. Conclude (not as obvious as at first glance, perhaps) that $a=-\frac p2$ and $|4b|=\sqrt{2p^2-8q}$. The square root of an integer is either an integer or irrational, hence $2p^2-8q$ is the square of an integer and is even, hence the square of an even integer, hence a multiple of $4$, hence $p$ is even and $a\in\mathbb Z$, contrary to assumption.

Answer 2

Hints:

1. Assume $p\in\mathbb Z[t]$ is a monic polynomial, such that $p(a+b\sqrt{2})=0$. Check, that $p(a-b\sqrt{2})=0$.
2. If $x$ and $y$ are roots of monic polynomials in $\mathbb Z$, then so are $xy$ and $x+y$.
3. If $x\in\mathbb Q$ is a root of a monic polynomial over $\mathbb Z$, then $x\in\mathbb Z$.

(By the way, those elements are called integral elements)

EDIT: How to conclude the solution from this hints:

Let $a+b\sqrt{2}$ be a root of a monic polynomial, by (1) so is $a-b\sqrt{2}$ and by (2), so are $$\left(a+b\sqrt{2}\right)+\left(a-b\sqrt{2}\right)=2a$$ and $$\left(a+b\sqrt{2}\right)\left(a-b\sqrt{2}\right)=a^2-2b^2$$ Since $2a,a^2-2b^2\in\mathbb Q$, it follows, that $2a,a^2-2b^2\in\mathbb Z$ using (3). So $a=\frac{a'}2$ for some $a\in\mathbb Z$ and, since $$8b^2=(2a)^2-4\cdot(a^2-2b^2)\in\mathbb Z$$ also $b=\frac{b'}{2}$ for some $b'\in\mathbb Z$. Inserting this into $a^2-2b^2\in\mathbb Z$ yields $4\mid a'^2-2b'^2$. Calculating modulo $4$ shows that, this fails unless both $a'$ and $b'$ are even, so $a,b\in\mathbb Z$.