Let $R$ be a unital commutative ring and $A$ a finitely generated $R$-algebra. I found out that if $R$ is a field, then any surjective $R$-endomorphism over $A$ must be injective, too. Does that hold for general $R$? I think it suffices to consider the case when $R$ is a local ring, but I'm not sure if this tip is useful. Any hint will be appreciated.

  • $\begingroup$ Possibly related: mathoverflow.net/questions/144567/… $\endgroup$ – darij grinberg Apr 2 '15 at 16:52
  • $\begingroup$ noetherianness is not assumed! $\endgroup$ – darij grinberg Apr 2 '15 at 17:53
  • $\begingroup$ @darijgrinberg Thanks a lot! I think your link have perfectly solved the case when there is a $R$-subalgebra $B$ of a such that $B$ is a polynomial $R$-algebra and $A$ is a integral $B$-algebra. Unfortunately, it does't work for all cases, for example, taking $R=\mathbb Z[X]$ and $A=\mathbb Z[X,Y]/(3Y-X)$. $\endgroup$ – Censi LI Apr 2 '15 at 18:18
  • $\begingroup$ Ah! Well, it might actually be useful as a step for a nonconstructive proof of the OP's conjecture, provided that one can convincingly argue that an $R$-algebra endomorphism of $A$ must have an "$S$-form" over some finitely generated $\mathbb{Z}$-subalgebra of $R$. (This sounds possible because if we fix a finite system of $R$-algebra generators of $A$, then ... $\endgroup$ – darij grinberg Apr 2 '15 at 19:20
  • $\begingroup$ ... the surjectivity of the map means that every of these generators is an image of a polynomial in those generators; now, one could have $S$ be generated by all coefficients of these polynomials, along with whatever coefficients are necessary to build up our element of $A$ that gets sent to $0$.) I don't want to dig into the details of this because I wouldn't like such a proof anyway, but if you are looking for a quick way to verify the result classically, this might do it. $\endgroup$ – darij grinberg Apr 2 '15 at 19:22

Let $A$ be a commutative ring, and $R$ a finitely generated $A$-algebra. Then every surjective $A$-endomorphism of $R$ is injective.

Let $f:R\to R$ be a surjective $A$-endomorphism, and $0 \neq x_0 \in R$. It suffices to prove $f(x_0) \neq 0$.

Let $x_1, \dots, x_n$ be generators for $R$ as an $A$-algebra. Then $x_0=\sum a_{i_1,\dots,i_n}x_1^{i_1}\cdots x_n^{i_n}$.

Let $x'_i\in R$ such that $f(x'_i) = x_i$ and write $x'_i=\sum a'_{i;i_1,\dots,i_n}x_1^{i_1}\cdots x_n^{i_n}$. Also write $f(x_i)=\sum a_{i;i_1,\dots,i_n}x_1^{i_1}\cdots x_n^{i_n}$.

Let $A' = \mathbb{Z}[a_{i;i_1,\dots,i_n}, a'_{i;i_1,\dots,i_n}, a_{i_1,\dots,i_n}]$. $A'$ is a Noetherian subring of $A$.

Let $R'=A'[x_1,\dots,x_n]$ and $f':R'\to R'$ the restriction of $f$ to $R'$. Since $R'$ is noetherian, and $x_0\ne 0$ we have $f'(x_0)\ne 0$, so $f(x_0)\ne 0$.

  • $\begingroup$ If some necessary elements are missing from $R'$ please feel free to add them :) $\endgroup$ – user26857 Apr 5 '15 at 16:58
  • $\begingroup$ This time I get it. Thank you veeeeerrrrry much... $\endgroup$ – Censi LI Apr 5 '15 at 17:37
  • $\begingroup$ For reference: This proof reduces the general situation to the case when $R$ is Noetherian. In this case, the result is proven in Section 4 of Kazimierz Szymiczek, NAK and injectivity of surjections. $\endgroup$ – darij grinberg Feb 5 '18 at 17:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.