Let $K$ be a division ring and $\text{char}(K)\not = 2$. Show that $x^2+y^2-1$ is irreducible in $K[X,Y].$

Question

Let $K$ be a division ring and $\text{char}(K)\not = 2$. Show that $x^2+y^2-1$ is irreducible Let $K$ be a division ring and $\text{char}(K)\not = 2$. Show that $x^2+y^2-1$ is irreducible in $K[X,Y].$

My Attempt: Since $K $is a division ring we have that every non zero element has a multiplicative inverse. However, I am not sure what exactly I have to prove in order to show that the given polynomial is irreducible over $K[X, Y].$ I read the Wikipedia page but I am still not sure what needs to be proved here. Do I have to show that there are no roots of this polynomial in this ring $K[X, Y]?$ Or is there something else?

Answer 1

Suppose one can factor $x^2+y^2-1$ in $K[x,y].$ Then that factorization has to be of the form $$(a_1 x+b_1 y+c_1)(a_2 x+b_2 y+c_2).$$ Now expand this product and comparing its coefficients with $x^2+y^2-1$ to find possible constants $a_1, a_2, b_1, b_2, c_1, c_2\in K$ and derive a contradiction.

Note:
Since $K[x,y]$ is a vector space (more precisely a module) over the division ring $K$ (so, there are no zero divisors!) spanned by the basis $\{1,x, y, xy, x^2, y^2, x^2y, xy^2, \cdots\},$ we are allowed to equate corresponding coefficients of two equal polynomials (finite linear combinations of basis elements).

More:
If your ground division ring is of characteristic $2,$ we have a (unique) non-trivial factorization $$x^2+y^2-1=(x+y)^2-1=(x+y-1)(x+y+1).$$