$K[X]_X$ is not integral over $K[X]$

Question

The localization $K[X]_{X}$ is a ring extension of $K[X].$ I want to show that $K[X]_X$ is not integral over $K[X]$ using lying above.

I tried to find a maximal ideal in $K[X]_X$ whose contraction in $K[X]$ is not maximal ideal, or maybe we can produce two prime ideals above one containing another whose contraction is the same prime ideal. I need some help to construct such prime ideals. Thanks.

Answer 1

Your second approach cannot work because this does not imply that the lying over property fails. For example, the extension $\Bbb{Z}\subset\Bbb{Z}[i]$ is integral and hence has the lying over property. But the two prime ideals $(2+i),(2-i)\subset\Bbb{Z}[i]$ satisfy $$(2+i)\cap\Bbb{Z}=(2-i)\cap\Bbb{Z}=5\Bbb{Z},$$ so their contraction is the same prime ideal.

Instead, consider the prime ideal $(X)\subset K[X]$. If there is a prime ideal $\mathfrak{q}\subset K[X]_X$ such that $\mathfrak{q}\cap K[X]=(X)$ then in particular $X\in\mathfrak{q}$. But $X$ is a unit in $K[X]_X$ and so $\mathfrak{q}=K[X]_X$, a contradiction. So there is no prime ideal lying over $(X)$.

For some more perspective; in general when localizing a (commutative unital) ring $R$ with respect to a multiplicative subset $S$, the set of prime ideals of $R_S$ corresponds bijectively to the set of prime ideals of $R$ that are disjoint from $S$. The bijection is given by taking contractions/extensions w.r.t. the localization map $R\ \longrightarrow\ R_S$.

In this particular case, the set of prime ideals of $K[X]_X$ corresponds bijectively to the set of prime ideals of $K[X]$ that do not contain any power of $X$. These are all prime ideals except $(X)$, and every prime ideal of $K[X]$ except $(X)$ is the contraction of a prime ideal of $K[X]_X$.

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[Original answer, where I mistook $K[X]_X$ for $K[X]_{(X)}$.]

As you say, the ring $K[X]_X$ is local, so there is only one maximal ideal, which is the ideal generated by $X$. Its contraction in $K[X]$ is the ideal generated by $X$ there, which is also maximal, so this approach won't work.

The only other prime ideal of $K[X]_X$ is the zero ideal, which contracts to the zero ideal in $K[X]$, which is also prime. So this approach won't work either.

However, this does show that the extension does not have the lying over property; the ring $K[X]_X$ has only two prime ideals whereas $K[X]$ has infinitely many. So there must be some prime ideal of $K[X]$ that is not the contraction of a prime ideal of $K[X]_X$.

Answer 2

Here is a conceptual view: a crucial ($\rm\color{#c00}{characteristic})$ property of integral ring extensions is that they don't alter unit properties in the base ring, i.e. a nonunit $\,r\in R\,$ remains a nonunit in any integral extension ring $S.\,$ [For example, this may allow us to deduce that Diophantine equations in $\Bbb Z$ are unsolvable by deducing a parity contradiction $\,2\,|\,1\,$ in any number ring with parity, or a proof that $(9+4\sqrt{5})^n + (9-4\sqrt{5})^n$ is even]. Thus the nonunit $X\in R[X]$ must remain a nonunit in any integral extension (so the OP extension is not integral since it inverts $X$)

A proof is easy: below specialize $\,u \in R\,$ to get $\,\bbox[3px,border:1px solid #c00]{u^{-1}\text{ is integral over }R\! \iff\! u^{-1}\in\ R\,}$

Lemma $ $ Suppose $\,R\subset S\,$ is an integral extensions of commutative rings and $\,u\,$ is a unit in $S$. Then $u^{-1}$ is integral over $R\iff u^{-1}\in R[u]$

$\begin{align}{\bf Proof}\ \ \ u^{-1}\ \text{is integral over } R&\iff u^{-n} = r_{1} u^{-(n-1)}+\cdots +\, r_n,\:\! \ \ \ \ {\rm for\ some}\ \ r_i\in R\\ &\!\!\overset{\times\ u^{\large n-1}\!\!}\iff u^{-1} =\, r_{1}\ +\ \cdots\,\ +\,\ r_n\, u^{n-1},\ \ {\rm for\ some}\ \ r_i\in R \end{align}$

Generalization $ $ More generally it is easy to prove that $R[u]\cap R[u^{-1}]\,$ is integral over $R$.

Viewed in terms of ideals the key property is: $ $ principal ideals $\:\!(r)\:\!$ survive in integral extensions, i.e. $\, rR\neq 1\,\Rightarrow\, rS\neq 1.\,$ Integral extensions can be nicely $\rm\color{#c00}{characterized}$ by various universal forms of such ideal survivability, e.g. see the paper of Coykendall and Dobbs cited here.

Answer 3

Haven't seen a proof like this yet so let me write down:

Suppose otherwise $K[x]_x$ is integral over $K[x]$. Since $1/x\in K[x]_x$, it is the solution of a monic polynomial over $K[x]$, that is: $$\left(\frac{1}{x}\right)^n+f_{n-1} \left(\frac{1}{x}\right)^{n-1}+\cdots+f_0\equiv 0$$ where $f_i\in K[x], i=0, \cdots, n-1$ are the coefficients. Clearing denominators (multiplying both sides with $x^n$) to obtain a polynomial relation in $K[x]$: $$1+f_{n-1} x+\cdots+f_0 x^n\equiv 0.$$ However this is impossible since the left side has a non-zero constant term: letting $x=0$ gets an equation $1=0$.