Is $k[x,y]/\langle y-x^2\rangle = k[x,x^2]/\langle x^2-x^2\rangle = k[x,x^2] = k[x]$ a valid argument?

Question

Apologies for my limited algebra knowledge.

I previously asked why polynomial rings $k[x,y]/\langle y-x^2\rangle$ and $k[x]$ with $k$ a field are isomorphic. I ran into a related answer where the accepted answer used a suitable substitution to make the ideal zero and hence obtain a description of the quotient ring.

In the present case the suitable substitution would be $y=x^2$ as the ideal $\langle y-x^2\rangle$ we are taking the quotient against vanishes under that substitution. Does this sort of thing work in general? Consider a quotient $$ k[x_1, \dots, x_n] / \langle f_1, \dots, f_m \rangle, $$ with $f_1, \dots, f_m \in k[x_1, \dots, x_n]$. Assume that we are able to find substitutions $x_i = g_i(y_1, \dots, y_r)$ in terms of $1 \leq r < n$ free parameters $y_i$ such that each polynomial $f_i$ vanishes. Does that imply the isomorphism $$ k[x_1, \dots, x_n] / \langle f_1, \dots, f_m \rangle \simeq k[x_1, \dots, x_r]? $$

Answer 1

Did you mean $x_i=g_i(x_1,x_2,\ldots,x_r)$? Even so, the answer is definitely no. Even if you have such substitutions, you cannot conclude that $ k[x_1, \dots, x_n] / \langle f_1, \dots, f_m \rangle \cong k[x_1, \dots, x_r] $. For example, $k[x,y]/\langle x^2-y^2\rangle$ is not isomorphic as a ring to $k[x]$ even if you have a substitution $y=x$ that kills $x^2-y^2$.

However, if you have polynomials $f_i$ and $g_i$ for $i=1,2,\ldots,n-r$ such that $$f_i(x_1,x_2,\ldots,x_n)=x_{r+i}-g_i(x_1,x_2,\ldots,x_r)$$ or $$f_i(x_1,x_2,\ldots,x_n)=x_{r+i}-g_i(x_1,x_2,\ldots,x_{r+i-1})\,,$$ then the answer is yes, namely, $$k[x_1,x_2,\ldots,x_n]/\langle f_1,f_2,\ldots,f_{n-r}\rangle \cong k[x_1,x_2,\ldots,x_r]\,.$$ This is the case with your original problem.

Answer 2

In general, if you have a substitution $x_i = g_i(y_1,...,y_r)$, what it really means is that there is a polynomial map $G:k^r\to k^n$ such that $f_1\circ G=\cdots =f_m \circ G =0$

Then consider the induced ring homomorphism $k[x_1,...,x_n] \to k[y_1,...,y_r]$ by $f\mapsto f\circ G$. The kernel of this map includes $\langle f_1,...,f_m\rangle$, but is not necessarily equal to it. If the kernel were equal to the ideal, then you had your isomorphism. As @Batominovski pointed out, even though the substitution $x=a, y=a$ eliminatea $x^2 - y^2$, $\langle x^2 - y^2\rangle$ is not the kernel of the map $k[x, y] \to k[a]$ by $x\mapsto a$ and $y\mapsto a$. So we don't have an isomorphism here.

A sufficient condition (as seen in the examples you provided) is that, if the ideal is of the form $\langle x_2 - f_2(x_1),..., x_n - f_n(x_1) \rangle$. But I'm not sure what would be the necessary condition for the kernel to be exactly equal to the ideal.

Answer 3

Here is a very useful fact:

Given a commutative ring $R$, and let $r\in R$, we have $$R[x]/(x-r) \cong R.$$ Note that in the ideal $(x-r)$, the coefficient of $x$ must be 1.

It is useful because we only require $R$ to be a commutative ring. In your first example, we can take $R=k[y]$, then we have $$k[y][x]/(x-y_2) \cong k[y].$$ And you can easily see when you could apply this to the more general case $k[x_1, \cdots, x_n]$ quotient some polynomial: coefficient of one of the variables being 1.

Proof: we define a map

$$\phi:R[x] \rightarrow R\\ \;x\mapsto r.$$

It is clearly onto, we just need to show the $\ker \phi = (x-r)$. Let $p(x)\in \ker\phi$, because in $x-r$ the coefficient of $x$ is $1$, there is nothing stopping us to use the division algorithm to rewrite $p(x) = (x-r)g(x) + q$, to obtain this formula, we only need to perform the long division method we learned in high school. Now it is clear $q$ must be zero for $p(x)$ to be in the kernel, thus by Isomorphism Theorem, we have $$R = Img(\phi) \cong R[x]/\ker\phi = R[x]/(x-r).$$