Characterizing kernel of ring morphism

Question

Let $K$ be a field and define a ring morphism $$\psi: K[x_1,x_2, \dots , x_n, y_1, y_2, \dots , y_n] \rightarrow K(x_1,x_2, \dots , x_n)$$ by $\psi(x_i) =x_i$ and $\psi(y_i) =\frac{1}{x_i}$.

I think that $\ker(\psi) = \langle x_iy_i - 1\rangle _ {1 \leq i \leq n}$ but I've not been able to prove it.

The inclusion $\ker(\psi) \supseteq \langle x_iy_i - 1\rangle _ {1 \leq i \leq n}$ is trivial.

For the other inclusion all I could do was to prove that if $p$ is in $\ker(\psi)$ then there exist $p_1, p_2, \dots , p_n \in K(x_1,x_2, \dots , x_n)$ such that $p = (x_1y_1 - 1)p_1 +(x_2y_2 - 1)p_2 +\dots +(x_ny_n - 1)p_n$.

Any idea?

Answer 1

Try arguing by induction on $n$.

Write $\psi_{n}$ rather than $\psi$ (to indicate the dependence on the choice of $n$).

Also, write $\phi$ to denote morphism $$K(x_1,\ldots,x_{n-1}) [x_n,y_n] \to K(x_1,\ldots,x_n)$$ given by $y_n \mapsto x_n^{-1}$; so $\phi$ is an analogue of $\psi_1$ but with the coefficient field $K$ replaced by the larger field $K(x_1,\ldots,x_{n-1})$, and with the variables $x_n,y_n$ being used rather than $x_1,y_1$.

Notice that $\psi_n$ factors as $\psi_n = \phi \circ \psi_{n-1}$.

Now suppose you have proved the base case $n = 1$. Then (taking into the account the preceding remark that $\phi$ is just $\psi_1$, but with a different coefficient field and differently labelled variables) you know the kernel of $\phi$. By induction, you can also assume that you know the kernel of $\psi_{n-1}$. Now by a fairly easy argument using the factorization of $\psi_n$, you can compute the kernel of $\psi_n$.

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This reduces you to the case $n = 1$, which is basically covered by user26857's comment above (and is easy to check in any case).

Answer 2

The problem can be solved using the theory of Gröbner Bases. The basic idea is to linearly order the monomials (ie $x_1 < x_2 < \cdots < x_n < y_1 < \cdots<y_n$) in such a way that multiplication respects the order, ie $x_1x_2 < x_1y_3$. It follows that every polynomial has a largest monomial term with respect to the given order (called the leading term). Like in the 1-variable case, there is a multivariable polynomial division algorithm so that $$p=(x_1y_1−1)p_1+(x_2y_2−1)p_2+⋯+(x_ny_n−1)p_n + r$$ ($p_i \in K[x_1, \dots,y_n]$) where $r$ is a polynomial such that no monomial of $r$ is divisible by any of the leading terms of $x_1y_1-1, \dots , x_ny_n−1$, ie not divisible by $x_1y_1, \dots , x_ny_n$. Thus distinct monomials of $r$ are mapped under $\psi$ to distinct rational monomials in $K(x_1, \dots,x_n)$. Thus there is no cancellation of rational monomial terms of $\psi (r)$. Hence $\psi (r) = 0$ iff $r=0$. So $p \in \ker(\psi)$ iff $r=0$. Thus $$\ker(\psi) = \langle x_1y_1−1, \dots,x_ny_n−1\rangle.$$

Answer 3

The image of $\psi$ is the localization $K[x_1^{\pm 1},\dotsc,x_n^{\pm 1}]$ of $K[x_1,\dotsc,x_n]$ at the elements $x_1,\dotsc,x_n$. Remember that in general the image is isomorphic to the quotient ring by the kernel. Thus, it suffices to prove that

$R[y_1,\dotsc,y_n]/(x_i y_i - 1)_i \to R[x_1^{-1},\dotsc,x_n^{-1}], ~y_i \mapsto x_i^{-1}$

is an isomorphism for every commutative ring $R$ and every sequence of elements $x_1,\dotsc,x_n \in R$. But this is true simply because both rings satisfy the same universal property: They are the "smallest" ring extensions of $R$ in which $x_i$ become invertible. More formally, if $S$ is a commutative ring, then we have natural bijections (by the universal properties of quotient rings and polynomial rings and localizations): $$\hom(R[y_1,\dotsc,y_n]/(x_i y_i - 1)_i,S)$$ $$ \cong \{f \in \hom(R[y_1,\dotsc,y_n],S) : f(x_i y_i-1)=0 \text{ i.e. } f(x_i) f(y_i)=1\}$$ $$ \cong \{(g,s_1,\dotsc,s_n) : g \in \hom(R,S), s_i \in S, g(x_i) s_i = 1 \}$$ $$ \cong \{g \in \hom(R,S) : g(x_i) \in S^*\} \cong \hom(R[x_1^{-1},\dotsc,x_n^{-1}],S).$$ By the Yoneda Lemma, this tells us $R[y_1,\dotsc,y_n]/(x_i y_i - 1)_i \cong R[x_1^{-1},\dotsc,x_n^{-1}]$. There is no need to fiddle around with elements!