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 , the given algebraic number field and the set of algebraic numbers in , by .

Suppose satisfies

.

Then, satisfies

. Clearly, so no problem there.

Now, given and in , we need to show that and are in .

So assume that and satisfy the **monic **polynomials and in .

Since we can factor and in using the Fundamental Theorem of Algebra, we can factor and as

where and for notational convenience.

Note: The ‘s and ‘s are elementary symmetric functions of ‘s and ‘s respectively.

Now, consider the polynomials

and

The polynomials and are invariant under any permutation . So, and can be viewed as symmetric polynomials in the variables Hence, by the Fundamental Theorem on Symmetric Functions, (not proved!) we have,

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

Of course, a similar argument holds for .

Since the coefficients ‘s and ‘s are integers, hence and .

Clearly, and are roots of and respectively and hence,

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

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