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.

Let $A$ be a subring of a field $K$, and suppose that $A$ is a local ring with maximal ideal $\mathfrak{m}$. Let $x \in K, \, x \neq 0$. Let $\phi: A \rightarrow L$ be a homomorphism of $A$ into the algebraically closed field $L$ with kernel $\mathfrak{m}$. Suppose that $\mathfrak{m} A[x] \neq A[x]$. Then $\mathfrak{m} A[x]$ is contained into a maximal ideal $\mathfrak{P}$ of $A[x]$ and $\mathfrak{P} \cap A = \mathfrak{m}$. Now, there exists an embedding $\psi : A/ \mathfrak{m} \rightarrow L$ such that $A \rightarrow A/ \mathfrak{m} \stackrel{\psi}{\rightarrow} L$ is equal to $\phi$. Suppose that the . According to Lang's Algebra p. 348, we can extend $\psi$ to $A[x]/ \mathfrak{P}$ whether the image of $x$ in $A[x]/ \mathfrak{P}$ is transcendental over $A/ \mathfrak{m}$ or not, but i can not see why this is possible. Any insights?

Thanks

Edited:

Related to the question is the observation that, according to the proof of Theorem 5.21 in Atiyah's and McDonald's "Introduction to Commutative Algebra", where they follow a similar construction, it is mentioned that the image of $x$ in $A[x]/ \mathfrak{P}$ is algebraic over $A/ \mathfrak{m}$ (notation is different). But i can't see why that's the case.

share|improve this question
    
This question is also related to Theorem 5.21 of Atiyah's and MacDonald's "Introduction to Commutative Algebra", p. 66. In the proof it is mentioned that $\bar{x}$ is algebraic over $k$, but i can't see why. –  Manos Aug 12 '12 at 15:32
    
Ok, the part relating to Atiyah's proof, is answered nicely here: math.stackexchange.com/questions/157099/… –  Manos Aug 13 '12 at 15:20
    
Also, as precisely Atiyah claims, $\bar{x}$ will be algebraic and not transcendental, which could be a possibility according to Lang's phrasing. –  Manos Aug 13 '12 at 15:22

1 Answer 1

up vote 1 down vote accepted

Let $t$ be the image of $x$ in $A[x]/\mathfrak{P}$. Let $F = A/\mathfrak{m}$. Then $A[x]/\mathfrak{P}$ is identified with $F[t]$. If $t$ is transcendental, $F[t]$ is not a field. This is a contradiction. Therefore $t$ is algebraic over $F$. Let $f(X)$ be the minimal polynomial of $t$ over $F$. Let $\alpha$ be a root of $f'(X)$ in $L$, where $f'(X)$ is a polynomial in $\psi(F)[X]$ corresponding $f(X)$ by $\psi$. Then there exists an embedding $\Psi:F[t] \rightarrow L$ extending $\psi$ such that $\Psi(t) = \alpha$.

share|improve this answer
    
In the case where $t$ is transcendental over $F$, how can you guarantee that the kernel of the map $\Psi$ is zero? –  Manos Aug 14 '12 at 13:49
    
@Manos Your are right. I think it was Lang's mistake. I edited my answer. –  Makoto Kato Aug 14 '12 at 16:42
    
So we found a mistake in Lang. That's no small achievement :-) –  Manos Aug 14 '12 at 17:00

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.