Can't understand proof of fundamental theorem of symmetric polynomials

Question

I'm reading through Introduction to Abstract Algebra 4th edition by Nicholson, and I'm having trouble understanding the proof for the fundamental theorem of symmetric polynomials in section 4.5 (Never before in reading this book have I had such a hard time understanding a proof, and this is only paragraph 1 of 7!! Normally after a minute things would click when something's not clear when I'm reading a proof, but this isn't happening here.). In particular I've been stuck on the first paragraph for like 30 minutes yet the final few sentences are not very clear.

Let $g = g(x_1, \dots, x_n) \neq 0$ be symmetric. If $k_1, \dots, k_m$ are the (distinct) integers that occur as degrees of monomials in $f$, then $g = g_1 + \cdots + g_m$, where $g_i$ is homogeneous of degree $k_i$ for each $i$. Given $\sigma \in S$, and a monomial $h(x_1, \dots, x_n)$, the fact that $h(x_1, \dots, x_n)$ and $h(x_{\sigma 1}, \dots, x_{\sigma n})$ have the same degree shows that each $g_i$ is itself symmetric. Hence, we may assume that $g$ is homogeneous.

Ok, I'm mainly confused with the conclusions the author is drawing here, and here are my two questions:

1. How did the author conclude that each $g_i$ is symmetric?
2. How does the fact that $g$ is homogeneous follow from this fact?

Thanks!

And please don't explain at the graduate level. I'm not a graduate. There's a reason I'm reading an introductory book. This has happened numerous times so I'm just putting this here.

Answer 1

Let's take a look at a concrete example, say $g(x, y) = x + y + xy$. This is a symmetric polynomial, in fact it's the sum of a homogeneous symmetric polynomial $x + y$ of degree $1$ and another homogeneous symmetric polynomial $xy$ of degree $2$. You might wonder: is it always true that a symmetric polynomial is a sum of homogeneous symmetric polynomials?

The answer is yes. The reason is that permuting the variables preserves degree: permuting the variables in a degree $k$ monomial produces another degree $k$ monomial. Above permuting $x$ produces $y$ (degree $1$) and permuting $xy$ produces itself (degree $2$). So the action of permuting variables operates in each degree "independently" (there's a more formal way to say this in terms of direct sums but I'll avoid it); the permutation actions in different degrees don't interact. This means that a polynomial is symmetric iff its degree $k$ part is symmetric for each $k$, which means that a symmetric polynomial must be a sum of homogeneous symmetric polynomials. You could write this out in a lot more detail but it's pretty tedious.

The conclusion is not that $g$ is homogeneous; it is that to prove the fundamental theorem of symmetric polynomials it suffices to prove it for homogeneous symmetric polynomials, because every symmetric polynomial is a sum of homogeneous symmetric polynomials. The last step is a WLOG assumption, reducing from the general case to the homogeneous case.