Degeneration of transcendence degree in polynomial rings

292 Views Asked by At

Before ask the general question, let us check the motivating example.

Consider two transcendental elements (and they are algebraically independent) over $\mathbb C$, say $x$ and $y$. Then $\mathbb C[x,y]$ has transcendence degree 2 over $\mathbb C$. Now take subrings. One thing is

$$\mathbb C[x,y^2-x]$$

and another one is

$$\mathbb C[x+y,(x+y)^2].$$

The first one has transcendence degree 2 over $\mathbb C$, but the second one has degree 1 obviously.

So in general setting, consider a subring $A=k[y_1,\ldots,y_d]$ in $k[x_1,\ldots,k_d]$. Thus $y_i$ for $1 \leq i \leq d$ is a polynomial over $k$ with variables $x_1,\ldots,x_d$. How we determine that $A$ has transcendence degree $d$, or $<d$.

1

There are 1 best solutions below

4
On BEST ANSWER

Let $B = k[x_1,\ldots,x_d]$ a polynomial ring and $y_i \in B$ for $i=1,\ldots,d$ and $A=k[y_1,\ldots,y_d]$. Consider the exact sequence

$$0 \to I \to k[W_1,\ldots,W_d] = C \xrightarrow{\phi} B$$

with $\phi(W_\nu) = y_\nu$. The ideal $I$ can be easily computed in a concrete example e.g. with Macaulay2. This is nothing more but a suitable gröbner base calculation. Iff $I=0$ then $\mathrm{trdeg}_k A = d$.

More generally $\mathrm{trdeg}_k A = \dim A = \dim C/I$ as $A = \phi(C) \cong C/I$ and $\dim C/I$ can be computed with Macaulay2 too - in general it can be read off from the leading terms of a gröbner base of $I$.