Proofs of The Fundamental Theorem of Symmetric Polynomials I have been considering a few proofs of this theorem, and I noticed that a few of them (for example Proof 1, and Proof 2) prove the theorem first for homogeneous symmetric polynomials and then generalise it. However I have seen other proofs which don't seem to require this step (e.g. Proof 3) and it seems to me superfluous.
Could anyone explain why so many proofs include this step and/or if it is necessary?
 A: As you observe regarding "Proof 3", the lexicographic order proof (which goes back to Gauss and may be the earliest clean, clear proof of this theorem), there is no need to reduce to the homogeneous case to prove the theorem. The lex-order algorithm operates happily on any symmetric polynomial, homogeneous or not. There are many other proofs that also avoid this reduction. In fact, as you correctly sense, both your other linked proofs ("Proof 1" at PlanetMath, which is really the same lex-order proof, and "Proof 2" at Wikipedia, which I first encountered in the algebra textbooks by Serge Lang and Michael Artin) can be easily rewritten without this assumption. The theorem can also be proven non-constructively from Galois theory (as is done e.g. in the textbook by Hungerford), and in such a proof it would be kind of ridiculous to bother with the assumption of homogeneity.
However, when one wants in practice to represent a symmetric polynomial in terms of the elementary ones, one does in fact always end up working separately on each homogeneous component, for the simple reason that the elementary symmetric polynomials are themselves homogeneous. So while the reduction to the homogeneous case can be eliminated to make the exposition of the proof more economical, it cannot be eliminated from the actual calculation of a representation (of a symmetric polynomial in terms of the elementary ones).
Thus, to the constructively-minded, there is no real loss of efficiency in a proof that reduces to the homogeneous case first. If the proof is at all constructive (as are all the proofs you link), rigging it to avoid this step could be seen as a kind of deception, since when you unpack the calculations behind the proof, you must work separately on each homogeneous component. I'm guessing that's why those proofs are written that way.
A: The lex order proof starts with a symmetric polynomial $f_0$. It then subtracts off something to make a new symmetric polynomial $f_1$ whose leading term is less than that of $f_0$. Then we make another symmetric polynomial $f_2$ whose leading term is less than $f_1$, and so on. You need to know that this process stops. In other words, you need to show that lex order has no infinite descending chains.
This isn't so hard to prove for lex order. But it is immediate if you restrict to the homogenous case first, since there are only finitely many monomials of a given degree.
