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The Ring of Symmetric Functions over a commutative ring $R$, $\Lambda$, is the subring of the ring of formal power series $R[[x_1, x_2, \dots]]$ such that $f \in \Lambda$ if $f$ is invariant under every permutation of of the indeterminants and the degrees of the monomials occurring in $f$ are bounded.

Why do we need this second condition? I know the second condition gives that $\Lambda \cong R[\sigma_1,\sigma_2, \dots]$, the polynomial ring in the infinite indeterminants $\sigma_i$, but is there another reason we require this? Thanks!

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Because we don't want to include power series like $$\frac{1}{1-\sigma_1}=1+\sigma_1+\sigma_1^2+\sigma_1^3+\cdots.$$ We barely even want power series at all; we're just using them as a notational convenience to allow ourselves to talk about $\sigma_1=x_1+x_2+\cdots$ as a stand-in for "$\sigma_1=x_1+\cdots+x_N$ where $N$ is an undetermined quantity that we arbitrarily increase whenever necessary".

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  • $\begingroup$ This is to say, $\Lambda$ is a direct limit of the $\Lambda_N = R[x_1,\dots,x_N]^{S_N}$. As this limit is characterized by a universal mapping property, there is no choice in the definition of $\Lambda$. $\endgroup$ – andrewBee Nov 14 '17 at 18:46
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    $\begingroup$ $\Lambda$ is only a direct limit in the category of graded rings. So the question remains "why graded?". $\endgroup$ – Friedrich Knop Nov 14 '17 at 20:31
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    $\begingroup$ Of course what you're saying is absolutely true; that said, I often want to think of such power series, and I kind of think you do too. In particular, the "symmetric function" associated to a tensorial species/FB-module does not have bounded degree. And the formalism of symmetric functions is very useful when thinking about tensorial species. $\endgroup$ – Dan Petersen Nov 15 '17 at 5:47
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  1. One reason for considering $\Lambda$ is that many systems of symmetric polynomials form an honest $R$-basis of $\Lambda$, for example elementary symmetric, monomial symmetric, power symmetric, and Schur functions. The base change matrices are very important in combinatorics.
  2. On the other hand, it is not true that $R[[x_i]]^{S_\infty}$ is irrelevant. For example a formula like $$ \log(1+e_1+e_2+\ldots)=\log\prod_i(1+x_i)=\sum_i\log(1+x_i)=\sum_{n\ge1}\frac{(-1)^{n-1}}np_n $$ is an identity in the latter ring. By comparing homogeneous components, it can be transformed into a sequence of identities in $\Lambda$, though.
  3. So the answer to your question might be that $\Lambda$ has strictly more structure than $R[[x_i]]^{S_\infty}$, namely a grading, and that most objects occuring in nature respect this structure, i.e., have a degree.
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