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Let $k$ be a field and $R=k[x_1,\cdots,x_i]$. Suppose we have two unital $R$-algebras $A$ and $B$ which are isomorphic to $R^{\oplus j}$ as modules, $Z(A)\cong Z(B)\cong R$, and there exists an injective map $f:A\to B$ which maps $1_A$ to $1_B$ (in particular, it maps $Z(A)=R\cdot 1_A$ isomorphically onto $Z(B)=R\cdot 1_B$). Is it always true that such a map $f$ is an isomorphism? If not, what extra conditions are necessary to guarantee that it will be an isomorphism?

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After posting my answer it occurred to me that you didn't specify, whether you want $f$ to be an isomorphism of $R$-modules or $R$-algebras? – Jyrki Lahtonen Apr 22 '12 at 9:59
$f$ ought to be an $R$-algebra isomorphism. – KReiser Apr 22 '12 at 10:00
up vote 3 down vote accepted

No, such a mapping $f$ need not respect multiplication. For example, we could have $A=B={\mathcal M}_{2\times2}(R)$, when $Z(A)=Z(B)$ consists of the scalar matrices, and define $f$ by $$ f:\pmatrix{a&b\cr c&d\cr}\mapsto\pmatrix {a&-b\cr 3c&d\cr}.$$

This is an isomorphism of $R$-modules, its restriction to the scalar matrices is the identity mapping. But it does not respect multiplication. For example, the nilpotent matrix $$ A=\pmatrix{1&-1\cr1&-1\cr}$$ gets mapped to an invertible matrix.

Adding an example, where $f$ is a homomorphism of $R$-algebras, but is not onto.

Let us introduce an element $y$ with the property $y^2=x_1$. Let us introduce another element $u$ subject to relations $u^2=1$, $ur=ru$ for all $r\in R$ and $uy=-yu$. Let $B$ be the free $R$-module of rank four with a basis $\{1,y,u,uy\}$. We can faithfully represent $B$ in the ring of $2\times2$ matrices with entries in the ring $R[y]$ as follows: $$ u\mapsto \pmatrix{0&1\cr1&0\cr},\qquad y\mapsto \pmatrix{y&0\cr0&-y\cr}, $$ and extending this $R$-linearly, so $$a=r_1+r_2y+r_3u+r_4uy\mapsto \pmatrix{r_1+r_2y&r_3-r_4y\cr r_3+r_4y&r_1-r_2y\cr}. $$ We easily check that $B$ is closed under multiplication, $R\subseteq Z(B)$, and hence $B$ is an $R$-algebra.

Claim. $Z(B)=R$.

Proof. Assume that $a=r_1+r_2y+r_3u+r_4uy\in Z(B)$. Then $au=r_3-r_4y+r_1u-r_2uy$ and $ua=r_3+r_4y+r_1u+r_2uy$, so we must have $r_2=r_4=0$. But also $ay=r_1y+r_3uy$ and $ya=r_1y-r_3uy$, so we must also have $r_3=0$. Q.E.D.

The submodule $A$ with basis $\{1,y,x_1u,x_1uy\}$ is also isomorphic to $R^4$ as an $R$-module. Furthermore, it is closed under multiplication, so it is a subalgebra of $B$. The above argument shows that $Z(A)=R$. So as a counterexample of an $R$-algebra homomorphism that meets all the criteria without being an isomorphism of algebras I offer the inclusion mapping $f=i:A\rightarrow B$.

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IOW, in the latter example $B$ is an $R$-order in a cyclic algebra, and $A$ is a smaller $R$-order. – Jyrki Lahtonen Apr 22 '12 at 11:08

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