# Characterizing kernel of ring morphism

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?

• Isn't $p_1,\ldots,p_n\in K[x_1,\ldots,x_n,y_1,\ldots,y_n]$? – Hamou Aug 29 '14 at 7:16
• Your question is basically solved here and in some other places on M.SE. – user26857 Aug 29 '14 at 7:20
• The link has nothing to do with the question. – Martin Brandenburg Sep 16 '14 at 10:21

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 $\psi_n$.

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).

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 $$\ker(\psi) = \langle x_1y_1−1, \dots,x_ny_n−1\rangle.$$

• @user172412: You were not logged in when making the last Edit, so the clarification you added is going through a Review process. If you were logged in, you'd have permission to make the edit. – hardmath Sep 20 '14 at 14:47

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!