Hi there (again),

I tried going about a proof to the following proposition,

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

and it was this

Let us denote our symmetric integral polynomial by f(d\beta_1, \ldots, d\beta_n). Then, we know that there exists some polynomial g that satisfies the relation 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 \sigma on the set \{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 d, \beta_1, \ldots, \beta_n are integers