Is it true that the number of Plücker relations for a Grassmannian $Gr(k,n)$ is equal to the dimension $k(nk)$ of said Grassmannian? So far, for $Gr(2,5)$, I get exactly five Plücker relations: $$p_{12}p_{34}+ p_{23}p_{14} p_{13}p_{24}=0,$$ $$p_{12}p_{35}+p_{23}p_{15} p_{13}p_{25}=0,$$ $$p_{12}p_{45}+p_{24}p_{15} p_{14}p_{25}=0,$$ $$p_{13}p_{45}+p_{34}p_{15} p_{14}p_{35}=0,$$ $$p_{23}p_{45}+p_{34}p_{25} p_{24}p_{35}=0.$$ I'm not sure, but I think that three of the above relations are algebraically independent, no? Are there supposed to be six Plücker relations for $Gr(2,5)$?

3$\begingroup$ See Chapter 14.2 of Miller and Sturmfels' Combinatorial Commutative Algebra for a detailed discussion of the minimal Plücker relations. $\endgroup$ – Sam Hopkins Feb 9 '16 at 23:10

$\begingroup$ The part of Chapter 14.2 concerning Gröbner bases is well beyond me. $\endgroup$ – Libertron Feb 9 '16 at 23:25
"The number of Plucker relations" is a little ambiguous, but there is no sense in which it is $k(nk)$.
The number of Plucker coordinates is $\binom{n}{k}$, so the number of degree $2$ monomials in Plucker coordinates is $\tfrac{1}{2} \left( \binom{n}{k}^2 + \binom{n}{k} \right)$. The vector space they span inside the homogenous coordinate ring of the Grassmannian has dimension $$\frac{1}{k+1} \binom{n}{k} \binom{n+1}{k}.$$ (Derivation available on request.) So a minimal list of relations between them would be of size $\tfrac{1}{2} \left( \binom{n}{k}^2 + \binom{n}{k} \right)  \tfrac{1}{k+1} \binom{n}{k} \binom{n+1}{k}$. Note that, if we fix $k$ and let $n$ grow, then $\tfrac{1}{2} \left( \binom{n}{k}^2 + \binom{n}{k} \right) \approx \tfrac{n^{2k}}{2 (k!)^2}$ and $\tfrac{1}{k+1} \binom{n}{k} \binom{n+1}{k} \approx \tfrac{n^{2k}}{(k+1) (k!)^2}$. So the number of relations is growing like $\tfrac{k1}{k+1} \tfrac{n^{2k}}{(k!)^2}$, not $kn$.
Now, many people mean specific lists of relations when they say "the Plucker relations". They don't always agree on which relations they mean, and they don't always mean an irredundant list. But, adding redundant relations would just make the list longer.
In your particular case of $G(2,5)$, the $5$ relations you gave form a basis for the space of relations. In general, there is an $\binom{n}{4}$dimensional space of relations for $G(2,n)$, and I think everyone would agree that the best basis is the relations of the form $p_{ab} p_{cd}  p_{ac} p_{bd} + p_{ad} p_{bc}$, for $1 \leq a < b < c < d \leq n$.

$\begingroup$ Of the five relations I gave for $Gr(2,5)$, how do I determine which ones are algebraically independent? $\endgroup$ – Libertron Feb 10 '16 at 17:16

$\begingroup$ I would use the Jacobian criterion math.stackexchange.com/questions/1258530/… $\endgroup$ – David E Speyer Feb 10 '16 at 17:35

$\begingroup$ When considering index sets, are there particular criteria that lead to trivial Plücker relations? Evidently, for $k=2$ if the index sets have at most $2$ indices in common, then that should result in trivial Plücker relations I think. $\endgroup$ – Libertron Feb 11 '16 at 2:12

$\begingroup$ @Libertron Given we know that $G(2,5) \subset \mathbb{P}^9$ has dimension $6$ shouldn't we be able to determine there are three independent Plücker relations? I think a complete intersection of three hypersurfaces in general position $ \mathbb{P}^9$ should have dimension $6.$ $\endgroup$ – user100272 Aug 28 '17 at 20:29

2$\begingroup$ @JacobA.Gross But $G(2,5)$ is not a complete intersection! It has degree $5$, not $8$. If you intersect three of the Plucker relations for $G(2,5)$, there will be a second, degree $3$, component. $\endgroup$ – David E Speyer Aug 28 '17 at 22:11