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Perhaps there is a more standard terminology, but let us say a subset $S$ of a commutative ring $R$ is a $\mathbb Z$-basis for $R$ if

  • for every $a\in R$, there exist polynomial $p(\bar X)\in\mathbb Z[\bar X]$ and $\bar x\in S$ such that $a=p(\bar x)$; and
  • for all polynomials $p(\bar X)\in\mathbb Z(\bar X)$ and all $\bar x\in S$, if $p(\bar x)=0$, then $p(\bar X)$ is the zero polynomial.

When does an ordered commutative ring have a $\mathbb Z$-basis? More generally, when can one find a $\mathbb Z$-basis which contains a given non-zero element for an ordered commutative ring?

The motivation is to have a cheap way of constructing ring automorphisms. What I can find do not seem to help. So references would be helpful too.

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    $\begingroup$ Does $\mathbb{Z}[\overline{X}]$ mean polynomials in several variables? $\endgroup$ Commented Feb 15, 2016 at 10:30
  • $\begingroup$ @Pierre-GuyPlamondon: Yes, and "several" implies finitely many, in case it is not clear. $\endgroup$ Commented Feb 15, 2016 at 10:47
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    $\begingroup$ And is this number of variables given with $S$? Or should the definition read "$S$ is a basis if there exists $N$ (number of variables) such that [the two conditions hold]? $\endgroup$ Commented Feb 15, 2016 at 11:22
  • $\begingroup$ @Pierre-GuyPlamondon: The number of variables is not meant to be fixed. $\endgroup$ Commented Feb 15, 2016 at 12:21

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Let $A=\mathbb{Z}[X_s]$ be the polynomial ring with one variable for each element of $S$, and let $f:A\to R$ be the ring homomorphism sending $X_s$ to $s$.

Your first condition says that $f$ is surjective, and your second condition says that $f$ is injective. Therefore $f$ is an isomorphism, so $R$ must be a polynomial ring.

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  • $\begingroup$ I have an issue with injectivity of $f$, which comes from an uncertainty in the definition of basis in the question (namely, is the number of variables given or not). To illustrate: take $R=\mathbb{Z}[x^r \ | \ r\in\mathbb{Q}_{>0}]$. Now take $S= \{x^r \ | \ r\in\mathbb{Q}_{>0}\}$, and let $1$ be the number of variables in $\mathbb{Z}[\overline{X}]$. Then the two conditions in the question are satisfied, but the map $f$ in your construction is not injective. $\endgroup$ Commented Feb 15, 2016 at 11:42
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    $\begingroup$ @Pierre-GuyPlamondon I agree, a lot of clarification is in order. I don't think that it's likely that the number of variables in $\overline{X}$ is supposed to be fixed, unless it's fixed to $|S|$. This produces some strange things, for example any subset of $\mathbb{Z}$ would be a $\mathbb{Z}$-basis, by taking the number of variables to be zero. $\endgroup$ Commented Feb 15, 2016 at 11:53

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