# An algebra given by generators and relations and an algebra generated as a subalgebra by some elements. Are they isomorphic?

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.

• 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. May 26 at 23:02

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 $$R_2$$.

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 $$R_1\simeq R_2$$.

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 $$R_1\simeq R_2$$.