Recognizing algebraic independence among Schur polynomials Given a set of integer partitions $\{\lambda_1, \lambda_2,\dots \lambda_n\}$. Are there combinatorial criteria for deciding whether the associated Schur polynomials $s_{\lambda_1}, s_{\lambda_2},\dots s_{\lambda_n}$ are algebraically independent?
If this is too hard, are there any surprising necessary or sufficient criteria?
 A: I'll add further details to my comment and make it an answer. This will give algebraic relations between certain collections of six Schur polynomials. Hence, for algebraic independence it will give a necessary condition of avoiding these six Schur polynomials together. This relation comes from Bijective proofs for Schur function identities which imply Dodgson's condensation formula and Plücker relations by Fulmek and Kleber (see Theorem 2 of this paper). I will change notation from the linked paper to agree with the notation in the question. Take $a_1 \geq a_2 \geq \cdots \geq a_{r+1}$ if we let
$$\lambda_1 = (a_1, a_2, \dots, a_r)$$
$$\lambda_2 = (a_2, a_2, \dots, a_{r+1})$$
$$\lambda_3 = (a_2, a_3, \dots, a_r)$$
$$\lambda_4 = (a_1, a_2, \dots, a_{r+1})$$
$$\lambda_3 = (a_2-1, a_3-1, \dots, a_{r+1}-1)$$
$$\lambda_4 = (a_1+1, a_2+1, \dots, a_r+1)$$
then
$$s_{\lambda_1}s_{\lambda_2} = s_{\lambda_3}s_{\lambda_4} + s_{\lambda_5}s_{\lambda_6}.$$
Letting $\lambda_1 = \lambda_2 = \cdots = \lambda_{r+1} = c$ we find that $\lambda_1 = \lambda_2 = (c)^r$, $\lambda_3 = (c)^{r-1}$, $\lambda_4 = (c)^{r+1}$, $\lambda_5 = (c-1)^r$, and $\lambda_6 = (c+1)^r$ which are all rectangle shapes recovering a result of Kirillov (which is cited in the linked paper).
