By analyzing the locus x2+y2=1 prove that the polynomial x2+y2−1 is irreducible in C[x, y]

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By analyzing the locus x2+y2=1 prove that the polynomial x2+y2−1 is irreducible in C[x, y]

Durst37 · Accepted Answer

Step 1  Let p(x, y)=x2+y2−1.  Suppose that it is not irreducible. Then there exists f(x, y)∈C[x, y] which is not a unit, nor is it associated with p(x, y), and f(x, y) divides p(x, y). That is, there exists some g(x, y)∈C[x, y] such that  p(x, y)=f(x, y)g(x, y)  If f(x, y) is constant, then it a unit.  If g(x, y) is constant, then it is a unit, so p(x, y) and f(x, y) are associated.  Thus, f(x, y) and g(x, y) are not constant. Since p(x, y) is of degree 2, f(x, y) and g(x, y) must be of degree 1. Thus,  f(x, y)=a1x+b1y+c1  g(x, y)=a2x+b2y+c2  From factorization p(x, y)=f(x, y)g(x, y), we now get that  {p=0}={f=0}∪{g=0}  However, {p=0} is a circle, while {f=0} and {g=0} are lines.  Thus, we get that a circle is a union of two lines, which is absqrt.  Therefore, p(x, y)=f(x, y)g(x, y) cannot hold.  To conclude, we now get that p(x, y) is irreducible in C[x, y]

By analyzing the locus x^{2}+y^{2}=1 prove that the polynomial x^{2}+y^{2}-1

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