# Is a surjective $R$-endomorphism over a finitely generated $R$ algebra always bijective?

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.

• Possibly related: mathoverflow.net/questions/144567/… – darij grinberg Apr 2 '15 at 16:52
• noetherianness is not assumed! – darij grinberg Apr 2 '15 at 17:53
• @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 $A=\mathbb Z[X,Y]/(3Y-X)$. – Censi LI Apr 2 '15 at 18:18
• 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 ... – darij grinberg Apr 2 '15 at 19:20
• ... 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. – 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 $f(x_0)\ne 0$.

• If some necessary elements are missing from $R'$ please feel free to add them :) – user26857 Apr 5 '15 at 16:58
• This time I get it. Thank you veeeeerrrrry much... – Censi LI Apr 5 '15 at 17:37
• 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. – darij grinberg Feb 5 '18 at 17:48