This was proved by me during the VSRP with a hint from Prof Rajan.

Theorem: Prove that the set of algebraic integers in an algebraic number field forms a ring.

A different Proof:

Denote by K, the given algebraic number field and the set of algebraic numbers in K, by \mathcal {O}_K.

Suppose \alpha \in \mathcal{O}_K satisfies
f(x)=x^n+a_{n-1} x^{n-1} + \cdots + a_0 \in \mathbb{Z}[x] .
Then, ( -\alpha ) satisfies
f'(x)= (-1)^n a_n x^n + \cdots + (-1)a_1 x + a_0. Clearly, f' \in \mathbb Z so no problem there.

Now, given \alpha and \beta in \mathcal{O}_K, we need to show that \alpha\beta and \alpha+\beta are in \mathcal{O}_K.

So assume that \alpha and \beta satisfy the monic polynomials f(x) and g(x) in \mathbb{Z}[x].

Since we can factor f and g in \mathbb{C} using the Fundamental Theorem of Algebra, we can factor f and g as
f(x) = x^m+s_{m-1} x^{m-1} + \cdots + s_0 = (x-\alpha_1) (x-\alpha_2) \cdots (x-\alpha_m)

g(x) = x^n+t_{n-1} x^{n-1} + \cdots + t_0 = (x-\beta_1) (x-\beta_2) \cdots (x-\beta_n)

where \alpha=\alpha_1 and \beta=\beta_1 for notational convenience.

Note: The s_i‘s and t_j‘s are elementary symmetric functions of \alpha_i‘s and \beta_j‘s respectively.

Now, consider the polynomials
h(x)= \displaystyle\prod_{i \in \{1,2,\cdots ,m\}} \prod_{j\in \{1,2,\cdots ,n\}} \left(x-\alpha_i \beta_j\right)


h'(x)=\displaystyle\prod_{i \in \{1,2,\cdots ,m\}} \prod_{j \in \{1,2,\cdots , n\}} (x-(\alpha_i+\beta_j) )

The polynomials h and h' are invariant under any permutation \sigma \in S_m. So, h and h' can be viewed as symmetric polynomials in the variables \alpha_1, \alpha_2, \cdots, \alpha_m. Hence, by the Fundamental Theorem on Symmetric Functions, (not proved!) we have,

h(x) = h(s_1, s_2, \cdots , s_m) (x)

Now, since the equation on the right is invariant under any permutation on n characters applied to \beta_1, \beta_2, \cdots, \beta_n, it is symmetric in \beta_j‘s. So, invoking the theorem on symmetric functions again, we have,

h(x) =  h(s_1, s_2, \cdots , s_m) (x) =  h(s_1, s_2, \cdots , s_m) (t_1, t_2, \cdots , t_n) (x)

Of course, a similar argument holds for h'.

Since the coefficients s_i‘s and t_j‘s are integers, hence h (x)\in \mathbb{Z}[x] and h' (x) \in \mathbb{Z}[x].

Clearly, \alpha\beta and \alpha+\beta are roots of h and h' respectively and hence, \alpha, \beta \in \mathcal{O}_K. \blacksquare

Remark: With a little addition, the above proof can be extended to show that the set of all algebraic numbers forms a field.