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Let $k$ be a field and consider the following two $k$-algebras:

  1. $R_1 = k[a,b,c,d] / (ab - cd). $
  2. $R_2$ is the unital $k$-subalgebra of $k(t)[x,y]$ generated (as an algebra) by $x,y,tx,t^{-1}y$.

Are $R_1$ and $R_2$ isomorphic?

Clearly we have a surjective map $\varphi \colon k[a,b,c,d] \rightarrow R_2$ which maps $$a \mapsto x, b \mapsto y, c \mapsto tx, d \mapsto t^{-1}y$$ such that $(ab - cd) \subseteq \ker \varphi$ but I'm not sure if the kernel is strictly larger or not.

The algebra $R_1$ is generated by four elements $a,b,c,d$ where the "only" relation we impose is $ab = cd$. The algebra $R_2$ is also generated by four elements which satisfy the relation but a priori might satisfy "more" relations (which come from the elements being specific elements in $k(t)[x,y]$). For example, we have the relation $\left( tx + t^{-1}y \right)^2 = t^2 x^2 + 2xy + t^{-2}y^{-2}$. However, this relation written in terms of $a,b,c,d$ says that $(c+d)^2 = c^2 + 2ab + d^2$ which is already a consequence of the basic relation $ab = cd$ so this isn't a "new" relation.

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  • $\begingroup$ If you accept that $R_2$ is a subalgebra of $k[x, y, t, u] / (tu-1)$, then in principle you could do a Groebner basis calculation on $(a-x, b-y, c-tx, d-uy, tu-1)$ as an ideal of $k[a, b, c, d, x, y, t, u]$, with respect to an elimination monomial order where $x, y, t, u > a, b, c, d$ and extract the generators with only $a,b,c,d$ to get generators of the kernel. $\endgroup$ May 26 at 23:02

3 Answers 3

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The answers by John and Stefan are very slick, but may involve things the OP hasn't learned yet. Here is a proof that doesn't involve Krull dimension or algebraic geometry. Let $f : R_1 \to R_2$ be the homomorphism induced by your ring homomorphism from $k[a, b, c, d]$ to $R_2$, so that: $$ f([a]) =x; \quad f([b]) =y; \quad g([c]) = tx; \quad g([d]) = t^{-1}y. \quad $$ (where I am writing $[x]$ for the class $x + (ab - cd)$ of $x$ in the quotient ring).

Let us construct an inverse to $f$. $R_2$ has a $k$-vector space basis comprising monomials of the form:

$$ x^py^q(tx)^r(t^{-1}y)^s = x^{p+r}y^{q+s}t^{r-s} $$ where $p, q, r, s \ge 0$. (Note that different quadruples $(p, q, r, s)$ may give the same monomial.)

I claim that the following defines a well-defined mapping from our monomial basis for $R_2$ to $R_1$:

$$ x^{p+r}y^{q+s}t^{r-s} \mapsto [a^pb^qc^rd^s] $$

For, if $x^{p+r}y^{q+s}t^{r-s} = x^{p'+r'}y^{q'+s'}t^{r'-s'}$, then we have $$ p' = p - k; \quad q' = q - k; \quad r' = r + k; \quad s' = s + k. $$ for some $k$, but then: \begin{align*} [a^{p'}b^{q'}c^{r'}d^{s'}] &= [a^{p - k}b^{q-k}c^{r+k}d^{s+k}] \\ &= [a^{p-k}b^{q-k}][cd]^k[c^rd^s]\\ &= [a^{p-k}b^{q-k}][ab]^k[c^rd^s]\\ &= [a^{p-k}b^{q-k}a^kb^kc^rd^s]\\ &= [a^pb^qc^rd^s] \end{align*} since $[ab] = [cd]$. So we have a well-defined mapping from our monomial basis for $R_2$ to $R_1$, which extends by linearity to a linear map $g : R_2 \to R_1$. But it is easy to check that $g$ respects multiplication of monomials, and hence is a ring homomorphism. It is also easy to check that $g$ and $f$ are mutual inverses, and hence that $R_1$ is isomorphic to $R_2$.

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Prove first that $\dim R_2=3$. We have $$\dim R_2=\mathrm{trdeg}_kk(x,y,tx,t^{-1}y)=\mathrm{trdeg}_kk(x,y,t)=3.$$

Then the height of $\ker\varphi$ equals $1$ since $k[a,b,c,d]/\ker\varphi\simeq R_2$. Moreover $\ker\varphi$ is a prime ideal and contains $(ab-cd)$ which is also a prime ideal, so they are equal.

We got that $R_1\simeq R_2$.

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Here is an answer in terms of geometry: the ring $R_1$ is the coordinate ring of the variety of $2\times 2$ degenerate matrices. On the other hand, any such matrix is of the form $$ \begin{pmatrix} v_1& tv_1\\ v_2 & tv_2 \end{pmatrix} $$ for some $t\in k^\times$. Now putting $v_1=x$, $v_2=t^{-1}y$, we see that $R_2$ is also the coordinate ring of the same variety. Hence $R_1\simeq R_2$.

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