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

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.

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

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 $$K[X]_X$$.

• $K[X]_X$ is localization wrt the multiplicative closed set $\{1,X,X^2,\ldots\}$ – user371231 Mar 6 '19 at 12:27
• Ah, I read it as the localization w.r.t. to (the complement of) the prime ideal $(X)$. See my new answer. – Servaes Mar 6 '19 at 12:29
• Anyone care to explain the downvote? – Servaes May 5 '19 at 20:54

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.