Irreducibility of $Y^2-2X^2$ over $\mathbb Q$

Question

Irreducibility of $Y^2-2X^2$ over $\mathbb Q$

''Were it reducible, it would have a root in $\mathbb Q^2$'' Is this also valid here or in general with more indeterminates ?

Answer 1

Over the reals this factors completely as $(y-\sqrt{2}x)(y+\sqrt{2}x)$. If it factored over the rationals, that would give a second factorization over the reals, violating unique factorization.

Alternatively: if it factored, then because it's quadratic it could only factor into two linear terms, and lines defined over the rationals clearly have more than one rational point.

Of course neither of these arguments generalizes very well.

Answer 2

By Eisenstein criterion, $2|2X^2$ but $2^2\nmid 2X^2$. So $Y^2-2X^2$ is irreducible over $\Bbb{Z}[X]$ and so over $\Bbb{Z}$. Since it is monic, $Y^2-2X^2$ is irreducible over $\Bbb{Q}$.

Answer 3

$Y^2 - 2X^2$ is a univariate polynomial over the field $\mathbb{Q}(X)$. Since it doesn't have a root in $\mathbb{Q}(X)$, it is irreducible over $\mathbb{Q}(X)$.

Consequently, if $Y^2 - 2X^2$ is reducible over $\mathbb{Q}$, one of its factors must be a unit in $\mathbb{Q}(X)$ — that is, a polynomial in $X$ alone. By looking at the coefficient on $Y^2$, it's clear that $Y^2 - 2X^2$ has no such factors.

Therefore, $Y^2 - 2X^2$ is irreducible over $\mathbb{Q}$.

Furthermore, it is irreducible over $\mathbb{Z}$ for a similar reason; if it were reducible, you could factor out an integer, but we see that is impossible by looking at the coefficient on $Y^2$.

(more generally, we can use the content of the polynomial to make these deductions)