Hi there (again),

I tried going about a proof to the following proposition,

Any symmetric integral polynomial in $\displaystyle d\beta_1, \ldots, d\beta_n$ is a symmetric polynomial in $\displaystyle d\beta_1, \ldots, d\beta_n, 0, 0 \dots, 0$,

and it was this

Let us denote our symmetric integral polynomial by $\displaystyle f(d\beta_1, \ldots, d\beta_n).$ Then, we know that there exists some polynomial g that satisfies the relation $\displaystyle f(d\beta_1, \ldots, d\beta_n) = g(d\beta_1, \ldots, d\beta_n 0,0 , \dots, 0).$ We conclude that g is also a symmetric integral polynomial, since the zeros are not labelled, and so are unaffected by any permutation $\displaystyle \sigma$ on the set $\displaystyle \{1,2,\ldots,n\}$

Now, this is wrong. I have to admit, it was a total stab at a proof. Does anyone know why this is wrong, and how would you go about doing this?

EDIT: Of course, all of $\displaystyle d, \beta_1, \ldots, \beta_n $ are integers