Give an example of an irreducible polynomial \(f \in \RR[x,y]\) whose zero set \(Z(f)\) in \(\AA^2_\RR\) is not irreducible.
(Real points and the affine scheme)
The ideal of this real point set is \((x,y^2-1)\). Indeed, reducing a polynomial modulo \(x\) leaves a polynomial in \(y\), which vanishes at both \(1\) and \(-1\) exactly when it is divisible by \(y^2-1\). It strictly contains \((f)\): the element \(x\) belongs to the former and not the latter, as its \(x\)-degree is smaller than that of \(f\). By contrast, \((f)\) is a prime ideal because \(\RR[x,y]\) is a UFD and \(f\) is irreducible, so \(\Spec(\RR[x,y]/(f))\) is integral and irreducible. The example concerns the real point set, not a failure of irreducibility for that scheme; the algebraically closed-field hypothesis in the Nullstellensatz is what distinguishes the two ideal computations [@Har10a, Chapter I, §1].
Solution
Here \(Z(f)\) is the set of real solutions in \(\RR^2\), with the Zariski topology whose closed sets are common zero sets of real polynomials.
1. An irreducible polynomial with the required property is\[ \boxed{f(x,y)=x^2+(y^2-1)^2.} \]
Regard \(f\) as a monic polynomial of degree two in \(x\) over the domain \(\RR[y]\). Suppose \(f=gh\) with neither factor a unit in \(\RR[y][x]\). The \(x\)-degrees add, and the leading coefficients in \(x\) multiply to \(1\). A factor of \(x\)-degree zero would therefore be a unit of \(\RR[y]\), which is excluded. Both factors consequently have \(x\)-degree one, and their leading coefficients are nonzero real constants. Rescaling the two factors by reciprocal constants gives \[ f=(x+a(y))(x+b(y))\qquad(a,b\in\RR[y]). \] Comparison of the coefficient of \(x\) gives \(b=-a\). Comparison of the constant coefficient then gives \(-a(y)^2=(y^2-1)^2\). Evaluating this polynomial identity at \(y=0\) gives \(-a(0)^2=1\), impossible over \(\RR\). Thus no nontrivial factorization exists, proving irreducibility.
2. The real zero set is \(Z(f)=\{(0,1),(0,-1)\}\) and is reducible.
At a real point, both summands \(x^2\) and \((y^2-1)^2\) are nonnegative. Their sum is zero exactly when \(x=0\) and \(y^2-1=0\). This gives the two stated points. The singleton subsets are Zariski closed, since they are \(Z(x,y-1)\) and \(Z(x,y+1)\) respectively. Each is a proper nonempty closed subset of \(Z(f)\), and their union is all of \(Z(f)\). This is a decomposition showing that \(Z(f)\) is not irreducible.
3. Q.E.D.
Step 1 proves that the displayed polynomial is irreducible in \(\RR[x,y]\), and step 2 proves that its real zero set is reducible.
Source Collections
Dependencies
T-JRTS2Hilbert's Nullstellensatzuses