How do I prove that $x^2 + y^2 - 1$ irreducible is $\mathbb{R}[x,y]$

Question

I have the polynomial $x^2 + y^2 - 1$ in the ring $R = \mathbb{R}[x,y]$.
I have to prove that it's irreducible in $\mathbb{R}[x,y]$ and I want to do it without using Eisenstein's Criterion otherwise i'd have to prove said criterion on my assignment.

Answer 1

Suppose $x^2+y^2-1$ has a nontrivial factorization in $\mathbb R[x,y]=\mathbb R[y][x]$, say $x^2+y^2-1=(a(y)x-f(y))(b(y)x-g(y))$ for some $a,b,f,g\in\mathbb R[y]$. Comparing the coefficients we obtain $$a(y)b(y)=1,$$ $$a(y)g(y)+b(y)f(y)=0,$$ $$f(y)g(y)=y^2-1.$$ It follows that $a=a(y)$ and $b=b(y)$ are constants, and $f(y)=-\frac abg(y)$, so $$-\frac ab g(y)^2=y^2-1=(y+1)(y-1).$$ Note that $\mathbb R[y]$ is a UFD, so it is a contradiction.

Answer 2

$$x^2 + y^2 -1 = x^2 + (y^2-1) = x^2 + (y+1)(y-1) \in \mathbb{R}[y][x]$$ Defini $f\colon \mathbb{R}[y] \to \mathbb{R}$ as $f(p(y)) = p(1)$. This is a ring homomorphism with $\ker f = (y-1)$, i.e. $\mathbb{R}[y]/(y-1) \cong \mathbb{R}$. (Isomorphism theorem.) Since $\mathbb{R}$ is a field, $(y-1) \subset \mathbb{R}[y]$ is maximal ideal, so it is prime ideal.

Thus, by Eisenstein, $x^2 + y^2 -1 $ is irreducible in $\mathbb{R}[y][x]$.

Answer 3

Suppose $f = x^2+y^2-1$ were the product of two nonconstant polynomials $P, Q$. Then clearly, $P$ and $Q$ must have both total degree $1$, since otherwise $PQ$ has total degree at least $3$.

Consider the set of points $(x, y)$ with $f(x, y) = 0$, call it $V(f)$. This is a circle. Because $f = PQ$, we have $V(f) = V(P) \cup V(Q)$. But $V(P)$ and $V(Q)$ are straight lines! Clearly, this is impossible: a circle is not the union of two straight lines. (A line and a circle intersect in at most $2$ points; not infinitely many.)

Answer 4

Expanding on Angina Seng's comment: We have $\mathbb{R}[x,y]=\mathbb{R}[y][x]$. Considering $\deg_x$, we see that a non trivial factor of $f$ must have degree $1$ in $x$. In particular, if $f$ is not irreducible, it has a root in $\mathbb{R}(y)$, meaning that $1-y^2$ is a square in the field $\mathbb{R}(y)$.

This is false: since $y-1$ is irreducible in $\mathbb{R}[y]$, one may consider the $y-1$-adic valuation. If $1-y^2$ is a square, its $1-y$-adic valuation must be even. But it is equal to one (since $y-1$ and $y+1$ are not associate), hence a contradiction.