This question is related to this other question.
When understanding how it is defined the ring of symmetric functions, I can not see why is so much important to take the inverse limit in the category of graded rings.
MY WORK
Consider $\Lambda$ to be the ring of symmetric functions.
$\Lambda_n$ to be the symmetric polynomials in $n$ independent variables.
Moreover, I know that in the category of rings, the objects are rings and the arrows are ring homomorphisms.
I also know that in the category of graded rings, the objects are rings and the arrows are graded rings homomorphisms. I.e. if $f:R\to S$ is a ring homomorphisms. A graded ring homomorphisms is $f$ such that $f(R)\subseteq S$.
Then, in the category of graded rings,
$$\Lambda = \varprojlim\Lambda_n = \left\{a \in \prod_{i\in I}\Lambda_i \mathrel{\Bigg|} \forall i \leq j: f_{i,j}(a_j)=a_i \right\}$$
In the category of rings,
$$\Lambda ^* = \varprojlim\Lambda_n = \left\{a \in \prod_{i\in I}\Lambda_i \mathrel{\Bigg|} \forall i \leq j: f_{i,j}(a_j)=a_i \right\}$$
And $\Lambda \subset \Lambda^*$ (my teacher told me).
But I can not see what makes the difference in considering the inverse limit in these two different categories. I can not see how it affects the arrows in the categories to these sets.
Any help?
Best Answer
Taking an inverse limit in the category of graded rings corresponds to taking the inverse limit at each degree and taking the direct sum.
So if $\Lambda_n^k$ is the group of degree $k$ symmetric polynomials in $n$ variables then we let
$$ \Lambda^k = \lim_{\gets} \Lambda_n^k \quad \text{and} \quad \Lambda = \bigoplus_k \Lambda^k.$$
And the effect of this is that we still have a direct sum at the end of the day. I.e. elements of $\Lambda$ consist of symmetric functions with bounded top-degree.
On the other hand, consider the sequence
\begin{align} &1 + x_1, \\ &1+(x_1+x_2)+x_1x_2, \\ &1 + (x_1 + x_2 + x_3) + (x_1x_2 + x_1x_3 + x_2x_3) + (x_1x_2x_3), \\ &\dots \end{align}
In $\Lambda^*$ this is an element, which you can see because if you set $x_n = 0$ in the $n$-th term of the sequence, you get the previous term.
This sequence converges to
$$ 1 + \sum_i x_i + \sum_{i < j} x_ix_j + \sum_{i < j < k} x_ix_jx_k + \cdots $$
but this is not an element of $\Lambda$.