How are these two monomials $u_1u_2^2$ and $u_2u_3^2$ in the same orbit?

Question

From Artin's Algebra:

A polynomial is symmetric if two monomials that are in the same orbit, such as $u_1u_2^2$ and $u_2u_3^2$, have the same coefficient in $g$.

How are the two monomials $u_1u_2^2$ and $u_2u_3^2$ in the same orbit? I don't seem to understand.

If we have $g(u_1,u_2, u_3)=u_1u_2^2+u_2u_3^2$, then $u_1u_2^2$ and $u_2u_3^2$ have the same coefficient, but $g$ is not symmetric.

Answer 1

The cyclic permutation $(1\,2\,3)$ transforms $u_1u_2^2$ into $u_2u_3^2$ so these monomials are in the same orbit of the action of the symmetric group on the monomials. But a symmetric polynomial needs to be invariant under every permutation of the variables, not just one: the quote from Artin should be read as "a polynomial is symmetric if every pair of monomials in the same orbit has the same coefficient".

Answer 2

You have $g(u_1,u_2, u_3)=u_1u_2^2+u_2u_3^2$ but they are only two of the $6$ monomials in the same orbit. The quote should have stated

A polynomial is symmetric iff any two monomials that are in the same orbit, such as $u_1u_2^2$ and $u_2u_2^3$ have the same coefficient $g$, including $0$.

Answer 3

In relation to the question about the $u_1 u_2^2$ and $u_2 u_3^2$ being in the same orbit:

Consider the group action of the symmetric group $S_3$ acting on the polynomials of the ring $R[x_1, x_2, x_3]$ $$ \star : S_3 \times R[x_1, x_2, x_3] \rightarrow R[x_1, x_2, x_3] $$

Where this action is defined for some $\sigma \in S_3$ and $f(x_1, x_2, x_3) \in R[x_1, x_2, x_3] $ $$ \sigma \star f(x_1, x_2, x_3) = f(x_{\sigma(1)}, x_{\sigma(2)}, x_{\sigma(3)}) $$

Then the orbit of $x_1 x_2^2$ is the set $\{\sigma \star x_1 x_2^2 : \sigma \in S_3 \} = \{x_{\sigma(1)} x_{\sigma(2)}^2 : \sigma \in S_3 \} $ then expanding this to individual terms the orbit is $$ \{x_1 x_2^2,\, x_2 x_3^2,\, x_1 x_3^2,\, x_3 x_2^2,\, x_2 x_1^2,\, x_3 x_1^2\} $$

As this is every permutation of $S_3$ acting on $x_1 x_2^2$.