Let $Y_\ell = \mathbb C[y_1,\cdots, y_\ell]$ be an unital associative commutative algebra in $\ell$ variables $y_1,\cdots, y_\ell$. From the theory of symmetric polynomials, we know that the elementary symmetric polynomials $e_j(X_1,\cdots, X_n)$ and the complete homogeneous polynomials $h_k(X_1,\cdots, X_n)$ in $n$ variables $X_i$ are related by the equality $$ \sum_{j=0}^m (-1)^je_j(X_1,\cdots,X_n)h_{m-j}(X_1,\cdots, X_n) = 0, m\geq 1.$$ I was wondering what is the dimension of $Y$ as a $\mathbb C$-vector space if I force similar relations to hold in a quotient of $Y_\ell$:
Given scalars $a_1,\cdots, a_\ell \in \mathbb C^\times$, set $F_m^{(\ell)} = \sum_{j=0}^m(-1)^j h_{m-j}(y_1,\cdots,y_\ell)e_j(a_1,\cdots,a_\ell), 1\leq m \leq \ell,$ and let $I_\ell$ be the ideal generated by $F_1^{(\ell)},\cdots, F_\ell^{(\ell)}$. What is the dimension of $Y_\ell/I_\ell?$
I suspect that the dimension of $Y_\ell/I_\ell$ will be $\ell!$, as I have verified for the cases $\ell=1,2,3$ below. However, as $\ell$ increases, counting basis elements one by one is quickly getting out of hand, so I must find a new method. I'm looking forward to see some new ideas on how to prove or disprove this conjecture, as I'm not familiar with techniques of counting dimension of quotients of polynomial rings.
Relevant edit: Setting $a_1=\cdots=a_\ell=1$ (I'm not sure yet if the dimension of $Y_\ell/I_\ell$ is independent of the choice of the $a_i$'s), and using the GAP software, I could find a good candidate for a linearly independent set of cardinality $\ell!$ in $Y_\ell/I_\ell$ by iteration of powers of $y_\ell$ with the basis of $Y_{\ell-1}/I_{\ell-1}$. For example, for $\ell=1$, the basis consists of $\{1\}$ alone, whilst the basis for the case $\ell=2$ consists of $\{1, y_2\}$, which rewrites as $\{1.y_2^0,1.y_2^1\}$. Now the basis for $\ell=3$ consists of $\{1,y_3,y_3^2,y_2,y_2y_3,y_2y_3^2 \}$, which rewrites as $\{1.y_3^0,1.y_3,1.y_3^2,y_2y_3^0,y_2y_3,y_2y_3^2\}$. This suggests that the following pattern must hold:
Suppose that $\{f_1,\cdots,f_{(\ell-1)!}\}$ is the basis of $Y_{\ell-1}/I_{\ell-1}$. Then, the basis of $Y_\ell/I_\ell$ is the set $\{f_jy_\ell^k: 1\leq j \leq (\ell-1)!\, \,,0 \leq k \leq \ell-1\}$ of cardinality $\ell(\ell-1)!=\ell!$ . In more explicit terms, this basis rewrites as the set $\{y_2^{s_1}y_3^{s_2}\cdots y_\ell^{s_{\ell-1}}: 0 \leq s_1 \leq 1, 0 \leq s_2 \leq 2, \cdots, 0 \leq s_{\ell-1}\leq \ell-1 \}.$
This conjecture still holds for $\ell=4$, as the software gave me the basis of 24 elements $\{1,y_4,y_4^2,y_4^3,y_3,y_3y_4,y_3y_4^2,y_3y_4^3,y_3^2,y_3^2y_4,y_3^2y_4^2,y_3^2y_4^3,y_2,y_2y_4,y_2y_4^2,y_2y_4^3,y_2y_3,y_2y_3y_4,y_2y_3y_4^2,y_2y_3y_4^3,y_2y_3^2,y_2y_3^2y_4,y_2y_3^2y_4^2,y_2y_3^2y_4^3\}.$
Even though Im not sure if the dimension depends or not on the $a_i$'s, the set $\{f_jy_\ell^k: 1\leq j \leq (\ell-1)!\, \,,0 \leq k \leq \ell-1\}$ is probably the basis we are looking for (or at least a LI set) in the general case. Since this basis take the basis of the case $\ell-1$ and iterates with each element of it by powers of $y_\ell$, a proof of this conjecture will probably follow using induction. The difficulty I'm facing now is to show that $f_1,\cdots,f_{(\ell-1)!}$ is still a linearly indepedent set in $Y_{\ell}/I_\ell$. I tried to use the canonical inclusion $Y_{\ell-1} \rightarrow Y_\ell$ to obtain a map $Y_{\ell-1} \rightarrow Y_\ell/I_\ell$, however, the elements $F_1^{(\ell-1)},\cdots, F_{\ell-1}^{(\ell-1)}$ do not seem to lie in the ideal $I_\ell$, so a map $Y_{\ell-1}/I_{\ell-1}\rightarrow Y_\ell/I_\ell$ cannot be guaranteed. Also, I have no idea why iterating with linearly indepedent elements by powers of $y_\ell$ will still give linearly independent elements.
In the following I show some calculations I did with my own hands and independently of the $a_i$'s, confirming the conjecture for the cases $\ell=1,2,3$.
$1. \,\, \ell=1$: From $h_1(y_1) - e_1(a_1) =0$ in $Y/I$ we clearly see that it is $1$-dimensional.
$2.\,\,\ell=2:$ In $Y/I$, we have $F_1=0$ and $0 = F_2 = y_2(y_2+y_1-(a_1+a_2)) +(y_1-a_1)(y_1-a_2) = y_2F_1 + (y_1-a_1)(y_1-a_2)$, so $(y_1-a_1)(y_1-a_2)=0$. Therefore, $Y/I$ is 2-dimesional, as every element of the form $y_1^ry_2^k$ is written in terms of $y_1$ by the relation $F_1=0,$ and $(y_1-a)(y_1-b) =0$ shows that every power of $y_1$ reduces to a sum of elements of degree $\leq 1$
$3. \,\, \ell=3:$ We have $F_1=0$ and $0=F_2 = y_3F_1+y_1^2+y_2^2+y_1y_2-e_1(a_1,a_2,a_3)(y_1+y_2)+e_2(a_1,a_2,a_3),$ so $G_{1,2} = y_1^2+y_2^2+y_1y_2-e_1(a_1,a_2,a_3)(y_1+y_2)+e_2(a_1,a_2,a_3) = 0.$ Similary, we can deduce that $G_{1,3} = y_1^2+y_3^2+y_1y_3-e_1(a_1,a_2,a_3)(y_1+y_3)+e_2(a_1,a_2,a_3)=0$ as well, so $0 = F_3 = (y_1-a_1)(y_1-a_2)(y_1-a_3) + y_2F_2 +y_3G_{1,3}$ shows that $G_1 = (y_1-a_1)(y_1-a_2)(y_1-a_3)=0$ in $Y/I$. We can proceed similarly to also deduce that $G_2 = (y_2-a_1)(y_2-a_2)(y_2-a_3)=0$. From the relations $G_1=G_2=0$, we see that $\alpha = \{1,y_1,y_1^2,y_2,y_2^2\}$ is a LI set in $Y/I$. Since $y_3 = -y_1-y_2 +a_1+a_2+a_3,$ we concern ourselves only with monomials of the form $y_1^ky_2^s$. By $G_{1,2} =0$, it is clear that $y_1y_2$ lies in the span of $\alpha$. Also, multiplying $G_{1,2}$ by $y_1$, we see that $y_1^2y_2$ lies in the span of $\alpha'= \alpha \cup \{y_1y_2^2\}$. The remaining monomials of the form $y_1^ky_2^s$ will all lie in the span of $\alpha'$, by relations $G_{1,2}= G_1= G_2=0$. Thus, $Y/I$ is $6$-dimensional.
We can prove this using some more sophisticated notions from commutative algebra. In particular, lets consider the rings $R=k[x_1,x_2,..,x_n]$, $S=k[x_1,..,x_n,y_1,..,y_n]$, with the natural inclusion $R\rightarrow S$. Then lets consider the ideal $I=(f_m)_{1\leq m \leq n}$ in $S$, where $$f_m:=\sum_{j=0}^m (-1)^je_j(x_1,..x_n)h_{m-j}(y_1,..y_n)$$
We can check that $R\rightarrow S\rightarrow S/I$ is an inclusion, and by checking over the fraction field of $R$, we can check that this is finite, in that $S/I$ is a finitely generated as an $R$ module.
Since our ideal was actually expressible in terms of symmetric polynomials, our ring $S/I$ is tensor product of $S$ and $\Sigma/(g_1,..,g_n)$, where $\Sigma=k[e_1,..,e_n,h_1,..h_n]$, and $g_m:=\sum_{0=j}^m (-1)^je_j h_{m-j}$, along the natural map $\Sigma\rightarrow S$ induced by $$e_i\mapsto e_i(x_1,..x_n)$$ $$h_j\mapsto h_j(y_1,..y_n)$$
Now we claim that $S/I$ is a Cohen-Macauley $k$ algebra. In fact we will prove something stronger, that the $f_m$ are a regular sequence, so $S/I$ is a complete intersection. So we need to check that $f_m$ is not a zerodivisor on $S/(f_1,..,f_{m-1})$.
So we want to show that $S/(f_1,..,f_{m-1})\xrightarrow{f_m\cdot}S/(f_1,..,f_{m-1})$ is injective for all $1\leq m\leq n$.
This map is isomorphic to $$S\otimes_{\Sigma}\Sigma/(g_1,..,g_{m-1})\xrightarrow{S\otimes_{\Sigma}g_m\cdot} S\otimes_{\Sigma}\Sigma/(g_1,..,g_{m-1})$$
So now notice that $S/\Sigma$ is flat, since this inclusion is the tensor product of the (flat) invariant inclusion maps $$k[e_1,..,e_n]\rightarrow k[x_1,..,x_n]$$ $$k[h_1,..,h_n]\rightarrow k[y_1,..,y_n]$$
So we just need to check that $g_m$ form a regular sequence in this polynomial ring $\Sigma$. Changing our (free) variables to $\alpha$ and $\beta$, this is equivalent to the regularity of the sequence $$g_m:=\sum_{j=0}^m (-1)^j\alpha_j\beta_{m-j}$$ in $k[\alpha_i,\beta_j]_{1\leq i,j\leq n}$, where $\alpha_0=1$ and $\beta_0=1$.
But now we can just compute the quotients, since each $g_m=\alpha_m+\beta_m+\text{stuff in }k[\alpha_i,\beta_j]_{1\leq i,j\leq m-1}$, we see the inclusion of $k[\alpha_i]_{1\leq i\leq m}\rightarrow k[\alpha_i,\beta_j]_{1\leq i,j\leq m}/(g_1,..,g_m)$ is an isomorphism, so our sequence is regular, so we've proved our Cohen Macauley assertion about our original algebra.
So now to prove your claim, Cohen-Macauleyness implies that $S/I$ is a free $R$ module of some rank $n$ (this is known as miracle flatness), so when we specialise $R$ down to $k$, the algebra we get is obtained by tensoring over $R/\mathfrak{m}$, so this will be a vector space of dimension $n$ (since tensoring free modules gives free modules and preserves their rank). So finally since we know for any specialisation $a_1,a_2,..,a_n$ of our variables $x_1,x_2,..,x_n$ we will get an algebra of the same dimension over $k$, it remains to show this dimension is $n!$, for some choice of specialisation. For this, lets pick all $x_i=0$, then our quotient algebra is $k[y_1,y_2,..,y_n]/(h_1,h_2,..,h_n)$ where $h_i$ are the symmetric homogeneous polynomials in the $y_i$.
Since these symmetric polynomials generate the positive degree invariant elements (Newtons formulas show that $e_i$ generate imply $h_i$ generate), we finally arrive at the claim that the co-invariant algebra of the symmetric group acting on its standard representation has dimension $n!$. This is a classical fact, there are lots of proofs, but a proof in this way of thinking is that $k[x_1,..x_n]$ as free as a $k[x_1,..,x_n]^{S_n}=k[h_1,h_2,..,h_n]$ module, since $k[x_1,..,x_n]$ is Cohen-Macauley, so the coinvariant algebra's dimension is the rank of this free module, and looking at the associated fraction fields, by Galois theory, the associated field extension is degree $n!$.