# Extension of Homomorphisms (Lang, Atiyah and McDonald)

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.

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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

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 $\Psi(t) = \alpha$.

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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