Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm having difficulty understanding the relationship between two corollaries in Lang's Algebraic Number Theory, on page 16 for those with the book. They can also be found in his Algebra.

The first is:

Let $A$ be a ring integrally closed in its quotient field $K.$ Let $L$ be a finite Galois extension of $K$ and let $B$ be the integral closure of $A$ in $L.$ Let $\mathfrak{p}$ be a maximal ideal of $A.$ Let $\varphi:A\rightarrow A/\mathfrak{p}$ be the canonical map, and let $\psi_1,\psi_2$ be two homomorphisms of $B$ extending $\varphi$ in an algebraic closure of $A/\mathfrak{p}.$ Then, there exists and automorphism $\sigma$ of $L$ over $K$ such that $\psi_1=\psi_2\circ\sigma.$

The second, under the same assumptions:

Let $\mathfrak{B}$ be the only prime of $B$ lying above $\mathfrak{p}.$ Let $f(X)$ be a monic polynomial in $A[X].$ Assume that $f$ is irreducible in $K[X]$ and that it has a root $\alpha$ in $B.$ Then the reduced polynomial $\overline{f}$ is a power of an irreducible polynomial in $\overline{A}[X].$

Lang goes on to state that it follows from the first corollary that any two roots of $\overline{f}$ are conjugate under an isomorphism of $\overline{B}$ over $\overline{A},$ the reductions mod the the respective primes. I'm not sure how this follows immediately. Also, it seems to me that this could be deduced from the fact that the Galois group of $L$ over $K$ acts transitively on the roots of $f,$ and under these assumptions all such mappings descend to mappings on $\overline{B}$ over $\overline{A}.$ Could someone help me see if my reasoning is off, and also explain the relationship between these two corollaries? If it's helpful, these are corollaries to the proposition that the decomposition group surjects onto the Galois group of $\overline{B}$ over $\overline{A}.$

share|improve this question
add comment

1 Answer 1

up vote 3 down vote accepted

Proposition Let $A$ be an integral domain. Let $K$ be the field of fractions of $A$. Let $L$ be an algebraic(not necessarily finite) extension of $K$. Let $B$ be the integral closure of $A$ in $L$. Let $\mathfrak{p}$ be a maximal ideal of $A$. Let $\bar A = A/\mathfrak{p}$. Assume that $\mathfrak{P}$ is the only prime ideal of $B$ lying over $\mathfrak{p}$. Let $f(X) \in A[X]$ be a monic polynomial. Let $\bar f(X) \in \bar A[X]$ be the reduction of $f(X)$ mod $\mathfrak{p}$. Assume that $f(X)$ is irreducible in $K[X]$ and has a root $\alpha$ in $B$. Then $\bar f(X)$ is a power of an irreducible polynomial in $\bar A[X]$.

Proof: Let $\varphi$ be the canonical homomorophism $A \rightarrow \bar A$. Let $\Omega$ be the algebraic closure of $\bar A$. Let $\omega_1$ be a root of $\bar f(X)$ in $\Omega$. Assume $g(X) \in A[X]$ and $g(\alpha) = 0$. Since $f(X)$ is irreducible in $K[X]$, there exists $h(X) \in K[X]$ such that $g(X) = f(X)h(X)$. Since $f(X)$ is monic, $h(X) \in A[X]$. Hence $\bar g(X) = \bar f(X) \bar h(X)$. Hence $\bar g(\omega_1) = 0$. Therefore there exists a homomorphism $\psi_1:A[\alpha] \rightarrow \Omega$ extending $\varphi$ such that $\psi_1(\alpha) = \omega_1$. Let $P_1$ be the kernel of $\psi_1$. By the lying-over theorem, there exists a prime ideal $Q_1$ of $B$ lying over $P_1$. By the assumption, $Q_1 = \mathfrak{P}$. Since $B$ is integral over $A$, $\mathfrak{P}$ is a maximal ideal and $B/\mathfrak{P}$ is algebraic over $\bar A$. Hence $B/\mathfrak{P}$ is isomorphic to a subfield of $\Omega$. Hence $\psi_1$ can be extended to a homomorphis $\Psi_1:B \rightarrow \Omega$.

Let $\omega_2$ be another root of $\bar f(X)$ in $\Omega$. By the similar argument as above, there exists a homomorphis $\Psi_2:B \rightarrow \Omega$ extending $\varphi$ such that $\Psi_2(\alpha) = \omega_2$. Since Ker($\Psi_1$) = Ker($\Psi_2$), there exists an isomorphism $\sigma:\Psi_1(B) \rightarrow \Psi_2(B)$ over $\bar A$ such that $\Psi_2 = \sigma\Psi_1$. Hence $\omega_1$ and $\omega_2$ are conjugate over $\bar A$. Hence $\bar f(X)$ is a power of an irreducible polynomial in $\bar A[X]$. QED

share|improve this answer
    
This is similar to the arguments used to prove the corollary and theorem I mentioned. Perhaps Lang meant it's a corollary of the proof rather than the result. In any case, thank you for another nice answer. –  Mike B Aug 9 '12 at 4:26
    
@MakotoKato: Why it is that we can extend $\psi_1$ to $\Psi_1$? –  Manos Aug 9 '12 at 15:21
1  
@Manos We can identify $\Psi_1$ as the canonical homomorphism $B \rightarrow B/\mathfrak{P}$. Similarly we can identify $\psi_1$ as the canonical homomorphism $A[\alpha] \rightarrow A[\alpha]/P_1$. Since $\mathfrak{P}$ lies over $P_1$, $\Psi_1$ extends $\psi_1$. Regards, –  Makoto Kato Aug 9 '12 at 15:45
1  
@Manos The argument is necessary to prove that there exists a homomorphism $\psi_1:A[\alpha] \rightarrow \Omega$ extending $\varphi$ such that $\psi_1(\alpha) = \omega_1$. Regards, –  Makoto Kato Aug 9 '12 at 17:35
1  
@Manos That's about it. Suppose $g(\alpha) = h(\alpha)$, Since $(g - h)(\alpha) = 0$, ($\bar g - \bar h)(\omega_1) = 0$. Hence $\bar g(\omega_1) = \bar h(\omega_1)$. Therefore we get a well-defined map sending $g(\alpha)$ to $\bar g(\omega_1)$. It's easy to see that this map is a homomorphism extending $\varphi$. Regards, –  Makoto Kato Aug 9 '12 at 18:03
show 4 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.