How does corollary symmetry functions follow? 

This is from Artin Algebra. How does the corollary follow from the theorem? Does that mean all polynomials in $n$ variables are symmetric? (I know they are not). What’s the difference between $R[z]$ and $R[u]$
 A: Let $R[u_1, \ldots, u_n]^{S_n}$ be the ring of symmetric polynomials, i.e., those invariant under the action of $S_n$. Then $R[s_1, \ldots, s_n] \subseteq R[u_1, \ldots, u_n]^{S_n}$ and existence part of the theorem gives the reverse inclusion, so $R[s_1, \ldots, s_n] = R[u_1, \ldots, u_n]^{S_n}$.
It's the uniqueness part of the theorem that gives the corollary. Let $R[z_1, \ldots, z_n]$ be another polynomial ring in $n$ variables and consider the homomorphism
\begin{align*}
\psi: R[z_1, \ldots, z_n] &\to R[s_1, \ldots, s_n]\\
z_i &\mapsto s_i
\end{align*}
so $\psi(z_1) = s_1(u_1, \ldots, u_n) = u_1 + \cdots + u_n$, etc. A polynomial relation among the $s_i$, i.e., an element $f \in R[z_1, \ldots, z_n]$ with $f(s_1, \ldots, s_n) = 0$, is exactly an element of $\ker(\psi)$. But the uniqueness statement in the theorem precludes a nontrivial polynomial relation: the zero polynomial evaluated at $s_1, \ldots, s_n$ also yields $0$, so by uniqueness we must have $f = 0$. Thus $\ker(\psi) = \{0\}$ and since $\psi$ is also surjective, then it is an isomorphism.
As you rightly point out, there is no real difference between $R[u_1, \ldots, u_n]$ and $R[z_1, \ldots, z_n]$ aside from how we've named the variables. But that does not mean $R[u_1, \ldots, u_n] = R[s_1, \ldots, s_n]$! We've simply shown that $R[u_1, \ldots, u_n]$ is isomorphic to a proper subring of itself. This may seem strange at first, but you probably have no problem with the analogous fact that $\mathbb{Z}$ and $2\mathbb{Z}$ are isomorphic as groups even though $2 \mathbb{Z} \subsetneq \mathbb{Z}$.
The fact that the ring of symmetric polynomials is isomorphic to a polynomial ring is quite special, though. For example, consider the polynomial ring $R[x,y]$ in $2$ variables and define the action of the cyclic group $C_2 = \{\pm 1\}$ on $R[x,y]$ by $(-1) \star f(x,y) = f(-x,-y)$. Then the ring of invariants $R[x,y]^{C_2}$ is generated by $x^2, xy,$ and $y^2$. As I mentioned in my comment, this situation is different since these invariants do have nontrivial polynomial relations, namely $x^2 \cdot y^2 - (xy)^2 = 0$. In this case the map
\begin{align*}
\psi: R[z_1, z_2, z_3] &\mapsto R[x^2, xy, y^2]\\
z_1, z_2, z_3 &\mapsto x^2, xy, y^2
\end{align*}
has kernel $z_1 z_3 - z_2^2$, so
$$
R[x,y]^{C_2} = R[x^2, xy, y^2] \cong \frac{R[z_1, z_2, z_3]}{(z_1 z_3 - z_2^2)} \, .
$$
This is a glimpse into invariant theory. If you'd like to see more, chapter 7 of Cox, Little, and O'Shea's Ideals, Varieties, and Algorithms gives a nice introduction to the invariant theory of finite groups.
