Let $Gr(k,n)$ denote the complex Grassmannian embedded into $\mathbb{P} \Lambda^k \mathbb{C}^n $ via the Plücker embedding.
Using the fact that $Gr(k,n)$ is the union of open subsets isomorph to $ \mathbb{A}^{k(n-k)}$ (non zero loci of the plückercoordinates) one can see that the Grassmannian is normal.
As the title suggests, i am wondering if the same is true for the affine cone $X := X(k,n)$ of the Grassmannian. If i am not mistaken $X \setminus \{0\}$ is the union of open subsets isomorph to $\mathbb{C}^* \times \mathbb{A}^{k(n-k)} $, the latter beeing smooth. Again, if i am not mistaken this suffices to conclude that the local rings $\mathcal{O}_{X,p}$ ,where $p \in X \setminus \{0\}$, are normal.
This would leave $\mathcal{O}_{X,0}$ to be inspected. Here are my thoughts so far:
$\mathcal{O}_{X,0} \cong K[X]_{m_0} =: R$ viewed as a subring of $K(X)$ is the set $\{ \frac{p}{q} \mid p,q \in K[X],q(0) \neq 0 \} $. Now let $a = \frac{p}{q} \in K(X)$ be the root of a monic polynomial over $R$, lets say $$ a^n + \frac{p_{n-1}}{q_{n-1}}a^{n-1} + \cdots + \frac{p_1}{q_1}a + \frac{p_0}{q_0} = 0 $$ then we need to show $q(0) \neq 0. $ Defining $p_n = q_n = 1$ the above can be rewritten as $$ \left( \sum_{i=0}^n p_i p^i q^{n-i} \prod_{j \neq i} q_j \right) \big/ \left( q^n \prod_{i=0}^nq_i \right)= 0 $$ hence $ \sum_{i=0}^n p_i p^i q^{n-i} \prod_{j \neq i} q_j = 0$. If we now assume $q(0) = 0$ then $p(0)= 0$ follows, since $q_i(0) \neq 0$ for all $i$.
This is where i am stuck. Maybe there is a better approach ? Or is the statement in question just wrong ?