Let A be a finite dimensional separable commutattive algebra over a field k . The book A

tr2os8x 2022-06-26 Answered
Let A be a finite dimensional separable commutattive algebra over a field k. The book A Course in Computational Algebraic Number Theory by Cohen says that A has a single generator as an algebra over k. I tried to prove this but failed. A is a finite product of separable extension fields K i over k. It is well known that each K i has a single generator over k, but I don't know how to use this fact to prove it.
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Answers (1)

pheniankang
Answered 2022-06-27 Author has 22 answers
This can be false when k has characteristic p. For example, F 4 × F 4 cannot be generated by a single element over F 2 , because we would have F 4 × F 4 F 2 [ X ] / p for some polynomial p. This can only happen if p is the product of two distinct degree 2 irreducibles, but there is only one degree 2 irreducible over F 2 .
On the other hand, if k has characteristic 0, we can get around this problem. If A is a finite product i K i , we can write K i k [ X ] / p i for irreducible polynomials p i . If the p i are distinct, then A k [ X ] / i p i , by the Chinese Remainder Theorem. If not, we can replace p i ( X ) by p i ( X + n ) for some n Z. In characteristic 0, these are all distinct polynomials, so we can ensure that i p i has no repeated factors.
Here is a proof in the more general case when k is infinite: as above, we are done if we can show that there are infinitely many distinct polynomials among { p i ( X + a ) a k }. Suppose that there are only finitely many. By the pigeonhole principle, there must be an infinite set S k such that p i ( X + a ) = p i ( X + b ) for all a , b S.
But then p i ( a ) = p i ( b ) for all a , b S, so p i is constant, as a function, on an infinite subset of k, which can only happen if p i itself is constant.

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