ideal projective closure saturation

Dehomogenization on $D_{+}(x_0)$ identifies $V_{+}(I^h)$ there with $A$.

Let G be homogeneous of degree d. It vanishes on the embedded A exactly when $g=G(1,x)\in I(A)=I$. If g is nonzero of degree e, homogenization gives $G=x_0^{d-e}{g}^h\in I^h$. If g=0, the same homogenization identity gives G=0. Conversely, if $x_0^{r}G\in I^h$, dehomogenizing an ideal expression gives $g\in I$, because every generator $f^h$ dehomogenizes to f in I. Thus the homogeneous elements vanishing on A are exactly the homogeneous elements of $I^h:x_0^{\mathrm{\infty }}$.

A homogeneous equation vanishes on A exactly when its projective closed zero locus contains the projective closure of A. Both the homogeneous vanishing ideal of that closure and $I^h:x_0^{\mathrm{\infty }}$ are homogeneous ideals; agreement in every homogeneous degree therefore proves equality for all polynomials. If A is empty, I is the unit ideal, as are both ideals in the formula. No Nullstellensatz or choice principle is needed because $I=I(A)$ is given.