Proof that ideal of Plücker relations is a prime ideal I am reading section 8.4 of Fulton's Young tableaux where he defines a certain ring as follows. Fix a complex vector space $E$ of dimension $m$ and integers $d_1,\ldots d_s$ such that $m \geq d_1 > \ldots > d_s > 0$. Define
$$S^{\bullet}(m;d_1,\ldots,d_s) := \Bbb{C}[X_{i_1,\ldots,i_p}, p \in \{d_1,\ldots,d_s\}]/Q$$
where $Q$ is the ideal generated by all quadratic relations $$X_{i_1,\ldots,i_p}X_{j_1,\ldots,j_q} - \sum X_{i_1',\ldots,i_p'}X_{j_1',\ldots,j_q'} \hspace{1in} (\ast)$$
with the sum over all exchanges of $j_1,\ldots,j_k$ with $k$ of the indices $i_1,\ldots,i_p$, with $p\geq q \geq k \geq 1$ and $p,q\in \{d_1,\ldots,d_s\}$. Furthermore we need to regard the variables $X_{i_1,\ldots,i_p}$ as alternating functions of the subscripts.


My question is: On the last paragraph of page 125 he says: "Let $\lambda$ be a partition whose columns have length among $\{d_1,\ldots,d_s\}$. That is the conjugate $\tilde{\lambda}$ has the form $(d_1^{a_1},\ldots,d_s^{a_s})$ for some nonnegative integers $a_1,\ldots,a_s$." We have seen that the representation $E^\lambda$ is the quotient
    $$\text{Sym}^{a_1}(\wedge^d_1 E) \otimes \ldots \otimes \text{Sym}^{a_s}(\wedge^{d_s} E)/Q.$$
    Now I don't understand the sentence highlighted in bold. I've looked through all of chapter 8 to find this but can't find a single result that shows this isomorphism. What am I missing here? This is somewhat important for me to know because $E^\lambda$ can be realized as the subring of some polynomial ring, thus showing that $Q$ is actually a prime ideal. Furthermore this $Q$ is actually important in showing that the Grassmannian is a projective variety. Can someone help me;  what am I missing here?


 A: Just as you do in your comment, I will denote by $\mu_1,\mu_2,...,\mu_l$ the lengths of the columns of $\lambda$.
In Lemma 1 of §8.1, you can identify $F$ with $E^{\otimes \mu_1} \otimes E^{\otimes \mu_2} \otimes ... \otimes E^{\otimes \mu_l}$ (by sending every $e_S \in F$ to
$\left(e_{S_{1,1}} \otimes e_{S_{2,1}} \otimes ... \otimes e_{S_{\mu_1,1}}\right) \otimes \left(e_{S_{1,2}} \otimes e_{S_{2,2}} \otimes ... \otimes e_{S_{\mu_2,2}}\right) \otimes ... \otimes \left(e_{S_{1,l}} \otimes e_{S_{2,l}} \otimes ... \otimes e_{S_{\mu_l,l}}\right)$
$\in E^{\otimes \mu_1} \otimes E^{\otimes \mu_2} \otimes ... \otimes E^{\otimes \mu_l}$,
where $S_{i,j}$ denotes the entry of $S$ in cell $\left(i,j\right)$). Lemma 1 says $E^{\lambda} \cong F/Q$.
Now, for any $R$-submodule $Q'$ of $Q$, you have
(1) $F/Q \cong \left(F/Q'\right) / \left(Q/Q'\right)$.
Let $Q'_1$ be the submodule of $Q$ generated by the elements $e_T + e_{T'}$ for all fillings $T'$ and $T$ such that $T'$ is obtained from $T$ by interchanging two entries in a column. Then, it is easy to see that
(2) $F/Q'_1 \cong \wedge^{\mu_1}E \otimes \wedge^{\mu_2}E \otimes ... \otimes \wedge^{\mu_l}E$.
(In order to prove this strictly, you need to know that if $A_1$, $A_2$, ..., $A_l$ are any $R$-modules, and $B_1$, $B_2$, ..., $B_l$ are submodules of $A_1$, $A_2$, ..., $A_l$, then $\bigotimes\limits_{i=1}^l \left(A_i / B_i\right) \cong \left(\bigotimes\limits_{i=1}^l A_i\right) / \left(\sum\limits_{j=1}^l A_1 \otimes A_2 \otimes ... \otimes A_{j-1} \otimes B_j \otimes A_{j+1} \otimes A_{j+2} \otimes ... \otimes A_l\right)$.)
But you could also let $Q'_2$ be the submodule of $Q$ generated by the elements $e_T - e_S$ for all fillings $T$ and $S$ such that $S$ is obtained from $T$ by interchanging two adjacent columns of equal length. (That all these elements belong to $Q$ is clear because they are particular cases of (iii) of Lemma 1.) It is then easy to see that
(3) $F/Q'_2 \cong \mathrm{Sym}^{a_1}\left(E^{\otimes d_1}\right) \otimes \mathrm{Sym}^{a_2}\left(E^{\otimes d_2}\right) \otimes ... \otimes \mathrm{Sym}^{a_s}\left(E^{\otimes d_s}\right)$.
Finally, since $Q'_1$ and $Q'_2$ are $R$-submodules of $Q$, so is $Q'_1+Q'_2$. And combining the arguments used for proving (2) and (3), you can see that
(4) $F/\left(Q'_1+Q'_2\right) \cong \mathrm{Sym}^{a_1}\left(\wedge^{d_1} E\right) \otimes \mathrm{Sym}^{a_2}\left(\wedge^{d_2} E\right) \otimes ... \otimes \mathrm{Sym}^{a_s}\left(\wedge^{d_s} E\right)$.
But
$E^{\lambda} \cong F/Q \cong \left(F/\left(Q'_1+Q'_2\right) \right) / \left(Q/\left(Q'_1+Q'_2\right)\right)$
(by (1), applied to $Q=Q'_1+Q'_2$). If we substitute (4) into this, and abuse notation by writing $Q$ for $Q/\left(Q'_1+Q'_2\right)$, we get the formula Fulton is claiming.
