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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.

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  • $\begingroup$ Rings of fractions are almost never integral over the original ring. $\endgroup$ – user26857 Mar 6 '19 at 15:01
  • $\begingroup$ Do you mean if $S$ contains a non unit then $S^{-1}A$ is not integral over $A$ $\endgroup$ – user371231 Mar 6 '19 at 15:23
  • $\begingroup$ When $A$ is an integral domain, yes. $\endgroup$ – user26857 Mar 6 '19 at 16:15
  • $\begingroup$ See my answer for details on the fundamental property mentioned by @user26857 $\endgroup$ – Bill Dubuque Mar 6 '19 at 17:13
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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 $K[X]_X$.


[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 $K[X]_X$.

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    $\begingroup$ $K[X]_X$ is localization wrt the multiplicative closed set $\{1,X,X^2,\ldots\}$ $\endgroup$ – user371231 Mar 6 '19 at 12:27
  • $\begingroup$ Ah, I read it as the localization w.r.t. to (the complement of) the prime ideal $(X)$. See my new answer. $\endgroup$ – Servaes Mar 6 '19 at 12:29
  • $\begingroup$ Anyone care to explain the downvote? $\endgroup$ – Servaes May 5 '19 at 20:54
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A crucial 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 $\,2\mid 1\,$ in some convenient extension ring of algebraic integers. Thus in the OP the $ $ nonunit $X\in R[X]$ remains a nonunit in any integral extension.

The proof is quite simple: specialize $\,u \in R\,$ below to conclude $\,u^{-1}\in R[u] = R$

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

$\begin{align}{\bf Proof}\ \ \ u^{-1}\ \text{is integral over } R&\iff u^{-n}-\, r_{1} u^{-(n-1)}-\cdots - r_n = 0,\ \ {\rm for\ some}\ \ r_i\in R\\ &\!\!\overset{\ \ \times\ u^{\large n-1}}\iff u^{-1} =\, r_{1}\, u\,\ +\,\ \cdots\,\ +\,\ r_n\, u^{n-1},\ \ {\rm for\ some}\ \ r_i\in R \end{align}$

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

We can view this property ideal theoretically as: $ $ principal ideals $(r)$ survive in integral extensions, i.e. $\, (r)\neq 1\,\Rightarrow\, (r)S\neq 1.\,$ In fact integral extensions can be characterized by various universal forms of such ideal survivability, e.g. see the paper of Coykendall and Dobbs cited here.

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