^2a2660 ----- > [!proposition] Proposition. ([[integral extensions, units, and fields]]) > Let $A \subset B$ be an [[integral algebra|integral extension]] of ([[commutative ring|commutative]]) [[ring|rings]]. Then: >1. $A \cap B^{*}=A^{*}$; >2. If $A,B$ are [[integral domain|integral domains]], then $B$ is a [[field]] $\iff$ $A$ is a [[field]]. ^proposition > [!proposition] Corollary. > For an [[integral algebra|integral extension]] $A \subset B$ and a [[prime ideal]] $\mathfrak{q}$ of $B$, the [[contraction of an ideal|contraction]] $\mathfrak{q} \cap A$ is a [[maximal ideal]] of $A$ if and only if $\mathfrak{q}$ is a [[maximal ideal]] of $B$. > > > > [!proof]- Proof of Corollary. > > The [[kernel of a ring homomorphism|kernel]] of the composition $A \hookrightarrow B \to B / \mathfrak{q}$ is $\mathfrak{q} \cap A$ and [[first isomorphism theorem for rings|hence]] induces an embedding $\frac{A}{\mathfrak{q} \cap A} \hookrightarrow \frac{B}{\mathfrak{q}}$. Now: > > > > - $\mathfrak{q}$ and $\mathfrak{q}^{c}=\mathfrak{q} \cap A$ are [[prime ideal|prime ideals]] of $B$ and $A$ respectively; thus $\frac{A}{\mathfrak{q} \cap A}$ and $\frac{B}{\mathfrak{q}}$ are [[integral domain|integral domains]]. > > - By the properties in [[integral algebra|integral extension]], $\frac{B}{\mathfrak{q}}$ is integral over $\frac{A}{\mathfrak{q} \cap A}$. > > > > $(2)$ in the original proposition now says that $\frac{A}{\mathfrak{q} \cap A}$ is a [[field]] $\iff$ $\frac{B}{\mathfrak{q}}$ is a [[field]]. Equivalently, $\mathfrak{q}\cap A$ is maximal $\iff$ $\mathfrak{q}$ is maximal. ^corollary Summary: For **1.**: for the nonobvious inclusion, take $a \in A \cap B^{*}$; know $ab=1$ for some $b \in B$, show that in fact $b \in A$ by multiplying integrality equation through by $a^{n-1}$ For **2.**: one direction can be quickly gotten from (1). For the other, let $b \in B$ nonzero; WTS $b \in B^{*}$. Write out integrality equation *with $n \geq 1$ minimal*, manipulate to get $b \cdot (\text{nonzero thing})=-a_{n}$. Justify why the LHS is nonzero, hence the RHS is nonzero, hence why done. For the **corollary.** Show that there's an integral extension of integral domains $\frac{A}{\mathfrak{q} \cap A}\hookrightarrow \frac{B}{\mathfrak{q}}$, then apply **2**. > [!proof]+ Proof. ([[integral extensions, units, and fields]]) > **1.** > $\supset$. [[unit|Units]] of $A$ certainly live in $A$, and certainly are also [[unit|units]] of $B$. > > $\subset.$ Take $a \in A \cap B^{*}$. We know there exists nonzero $b \in B$ such that $ab=1$. Want to show that, in fact, $b \in A$. > > By hypothesis, $b$ is $A$-[[integral element of an algebra|integral]], so consider an integrality equation $b^{n}+a_{1}b^{n-1}+\dots+a_{n}b^{0}=0, a_{i} \in A.$ > Multiply both sides by $a^{n-1}$ to get $b + \overbrace{ a_{1} + a_{2}a \dots+ a_{n}a^{n-1}}^{\in A}=0$ > and rearranging witness $b$ to be a combination of elements in $A$, thus $b\in A$. > > **2.** > $\implies.$ Assume $B$ is a [[field]]. Then $B^{*}=B - \{ 0 \}$, hence $A \cap B^{*}=A-\{ 0 \}$. But by **(1)**, $A \cap B^{*}=A^{*}$, hence $A-\{ 0 \}=A^{*}$, which is precisely the definition of $A$ being a [[field]]. > > $\impliedby$. Assume $A$ is a [[field]]. Let $b \in B$, $b \neq 0$; want to show $b$ has an [[unit|inverse]] in $B$. Using that $b$ is $A$-integral, write an integrality equation *with $n \geq 1$ minimal* $b^{n}+a_{1}b^{n-1}+\dots+a_{n}b^{0}=0.$ > That is, $b(\underbrace{b^{n-1}+ a_{1}b^{n-2} + \dots + a_{n-1}}_{:= \Delta})=-a_{n}.$ > Note that $\Delta \neq 0$ because we chose $n$ minimal[^1], and $b \neq 0$ by assumption; since $B$ is an [[integral domain]], $-a_{n} \neq 0$. Since $A$ is a [[field]], this means $a_{n}$ has an inverse $a_{n} ^{-1} \in A$ and we have $b(-a_{n}^{-1}\Delta)=1$. > ----- #### [^1]: If $\Delta=0$ then $a_{n}=0$, and so $b^{n}+a_{1}b^{n-1}+\dots+a_{n-1}b=0$ is an integrality equation for $b$. But then $b(a_{1}b^{n-1}+\dots+a_{n-1}b^{0})=0$, and since $B$ is an [[integral domain]] and $b \neq 0$ this means $a_{1}b^{n-1}+\dots+a_{n-1}b^{0}=0$, producing an integrality equation for $b$ of degree less than $n$. This cannot happen because we chose $n$ to be minimal. ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```