 # 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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This can be false when $k$ has characteristic $p$. For example, ${\mathbb{F}}_{4}×{\mathbb{F}}_{4}$ cannot be generated by a single element over ${\mathbb{F}}_{2}$, because we would have ${\mathbb{F}}_{4}×{\mathbb{F}}_{4}\cong {\mathbb{F}}_{2}\left[X\right]/p$ 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 ${\mathbb{F}}_{2}$.
On the other hand, if $k$ has characteristic $0$, we can get around this problem. If $A$ is a finite product $\prod _{i}{K}_{i}$, we can write ${K}_{i}\cong k\left[X\right]/{p}_{i}$ for irreducible polynomials ${p}_{i}$. If the ${p}_{i}$ are distinct, then $A\cong k\left[X\right]/\prod _{i}{p}_{i}$, by the Chinese Remainder Theorem. If not, we can replace ${p}_{i}\left(X\right)$ by ${p}_{i}\left(X+n\right)$ for some $n\in \mathbb{Z}$. In characteristic $0$, these are all distinct polynomials, so we can ensure that $\prod _{i}{p}_{i}$ 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 $\left\{{p}_{i}\left(X+a\right)\mid a\in k\right\}$. Suppose that there are only finitely many. By the pigeonhole principle, there must be an infinite set $S\subset k$ such that ${p}_{i}\left(X+a\right)={p}_{i}\left(X+b\right)$ for all $a,b\in S$.
But then ${p}_{i}\left(a\right)={p}_{i}\left(b\right)$ for all $a,b\in S$, 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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