The Plucker image is a closed projective algebraic set

For $r=0$ or $r=n$, the Grassmannian and $\mathbf{P}({\mathrm{\Lambda }}^{r}V)$ are both one point, so the assertion is immediate. Suppose henceforth that $0<r<n$. For ordered tuples $(i_1,\dots ,i_{r-1})$ and $(j_1,\dots ,j_{r+1})$, alternating multilinearity gives the signed Plucker relation ${\sum }_{a=1}^{r+1}(-1{)}^a{p}_{i_1\dots i_{r-1}{j}_a}{p}_{j_1\dots \widehat{j_a}\dots j_{r+1}}=0.$ The ordered-coordinate convention supplies every insertion sign and makes repeated indices contribute zero.

On a chart with $p_{1\cdots r}\mathrm{\ne }0$, rescale so that $p_{1\cdots r}=1$ and define the entries of $A$ by the coordinates with exactly one index outside $\{1,\dots ,r\}$, using their alternating signs. The maximal minors of $(I_r\mid A)$ agree with these coordinates when zero or one index is outside the base set. Induct on the number of outside indices in an ordered $r$-tuple $I$. For an outside index $j\in I$, apply the relation of step 1.1 to the $(r-1)$-tuple $I\setminus \{j\}$ and the $(r+1)$-tuple $(1,\dots ,r,j)$. The term omitting $j$ is, up to sign, $p_I{p}_{1\cdots r}=p_I$; every other term is a product of a coordinate with fewer outside indices and a one-outside coordinate. The minors of $(I_r\mid A)$ satisfy the same relation, so the induction identifies $p_I$ with its corresponding minor. Thus the row span has precisely the given projective coordinates.

The analogous charts cover every nonzero coordinate point satisfying the relations. Thus every such point is decomposable, while step 1.1 proved the reverse inclusion; homogeneous quadratic equations make this locus closed, and the injective Plucker map realizes it as the required projective algebraic set.