Showing that $x^2+y^2-1$ is irreducible in $\mathbb{Q}[x,y]$ [Direct approach].

Question

I want to check if my reasoning is correct and makes sense. I know other approaches, but I wonder if this one works also:

1. To show that $x^2+y^2-1$ is irreducible in $\mathbb{Q}[x,y]$, it's enough to check that $x^2+(y^2-1)$ is irreducible in $\mathbb{Q}[y][x]$ as there is a ring isomorphism between $\mathbb{Q}[x,y]$ and $\mathbb{Q}[y][x]$.

2. As $x^2+y^2-1$ is a quadratic polynomial, then it's enough to show that $\pm\sqrt{y^2-1}$ does not belong to $\mathbb{Q}[y]$.

3. If $\sqrt{y^2-1}$ belongs to $\mathbb{Q}[y]$, then there exists $f(y)\in \mathbb{Q}[y]$ s.t. $$\sqrt{y^2-1}=f(y)\text{ i.e. }y^2-1=f^2(y)\text{ }[*].$$

4. But, then $(y-f(y))(y+f(y))=1$. So, $y-f(y)=c$ for some $c\in\mathbb{Q}$ with $c\neq0$ i.e. $f(y)=y-c$.

5. From $[*]$, follows that $y^2-1=(y-c)^2$ i.e. $y=\frac{c^2-1}{2c}\in\mathbb{Q}$ which is a contradiction.

Thank you!

Answer 1

You actually need to show that $1 - y^2$ is not a square in $\mathbb{Q}[y]$ (you need to pay attention to the sign here because you're not working over $\mathbb{C}$), which is even easier since its quadratic coefficient is negative. Of course it is still not a square over $\mathbb{C}$ and there are many ways to see this, e.g.

• it has two distinct roots, or
• by comparing degrees we see that a square root must be linear, and by comparing the quadratic and constant terms it must have the form $\pm iy \pm 1$, whose square has the wrong linear term.