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Let $A \subset B$ be a finite ring extension. Fix a ring homomorphism $\nu : A \to \Omega$, where $\Omega$ is an algebraically closed field.

  1. I want to show there exists a non-zero homomorphism $v : B \to \Omega$ of $A$-modules.

  2. I want show that the space $V$ of all the homomorphisms $v$ in (i) is a non-zero finite dimensional vector space over $\Omega$. Show that the algebra $B$ acts on this space. Show that eigenvectors of this action correspond to homomorphisms of algebras $\lambda : B \to \Omega$ that extend the homomorphism $\nu$. Deduce that the set of such homomorphisms $\lambda : B \to \Omega$ is finite and non-empty.

We can assume that $B = Ab_1 + ... + Ab_n$ where $b_1,...,b_n \in B$. The chose $v(b_i)=\omega_i \in \Omega-\{0\}$ and then define $$ v( a_1 b_1 + ... + a_n b_n ) = \nu(a_1) v(b_1) + ... + \nu(a_n) v_n = \nu(a_1) \omega_1 + ... + \nu(a_n) \omega_n \ .$$ This is an $A$-module homomorphism. Am I correct?

For the second part, that $V$ is not-zero follows from the first part. That it is finite dimensional follows from the fact that $v_i(b_j)=\delta_{i,j}$ (Kronecker's delta) with $i=1,...,n$ spans $V$ and hence $\dim V \le n < \infty$.

I define the action of $B$ over $V$ to be $$b \bullet v = \left( x \mapsto v(b)v(x) \right) \mbox{ for all } b,x \in B \ .$$ This seems to be the natural action to define. Am I correct?

However, I haven't managed to solved the rest (the eigenvector stuff and the relation to $\nu$).

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Yes, you are right, $A$ is not necessarily an integral domain. But I think the $S^{-1}A$ is not neccesary for the solution. Also, the algebras are over $\mathbb{Z}$ and can be considered commutative rings with a unit. –  LinAlgMan Nov 20 '12 at 16:23
    
What do you mean by an eigenvector of an action? –  user26857 Nov 20 '12 at 23:24
    
In eigenvector I mean $V \ni v : B \to \Omega$ such that the action $b \bullet v = \omega v$ with $\omega \in \Omega$. However, the professor did not defined this terms (both the action and the eigenvector). –  LinAlgMan Nov 21 '12 at 11:54
    
The question makes more sense if we instead consider the action of $A$ over $V$ by $$ a \bullet v = \left( x \mapsto v(ax)=\nu(a)v(x) \right) \ , $$ then $v$ extends $\nu$. –  LinAlgMan Nov 21 '12 at 14:47

2 Answers 2

up vote 2 down vote accepted

Let $A \subset B$ be a finite ring extension. Fix a ring homomorphism $\sigma : A \to \Omega$, where $\Omega$ is an algebraically closed field. Prove that there is at least one, but finitely many ring homomorphisms $\overline{\sigma}: B \to \Omega$ that extend $\sigma$.

Existence. We don't need to assume that the ring extension is finite. It is enough to be integral (see exercise 2, Chapter 5, from Atiyah and MacDonald). Let $\mathfrak p=\ker\sigma$. This is a prime ideal of $A$ and there is a prime ideal $P$ of $B$ lying over $\mathfrak p$, that is, $P\cap A=\mathfrak p$. So we have an integral extension of integral domains $A/\mathfrak p\subset B/P$. Since $A/\mathfrak p\to\Omega$ is injective and every nonzero element of $A/\mathfrak p$ is invertible in $\Omega$ we can extend this to a homomorphism $Q(A/\mathfrak p)\to\Omega$. But $Q(A/\mathfrak p)\subset Q(B/P)$ is an algebraic field extension and then we can extend $Q(A/\mathfrak p)\to\Omega $ to $Q(B/P)\to\Omega$. The extension of $\sigma$ we are looking for is $B\to B/P\to Q(B/P)\to\Omega$.

Finiteness. Now only two observations are enough to show that there are only finitely many extensions of $\sigma$ when $A\subset B$ is finite. The first is the following: if $A\subset B$ is finite, then there are only finitely many prime ideals of $B$ lying over $\mathfrak p$. The second comes from field theory and says that for a finite field extension $K\subset L$ and a homomorphism $\sigma:K\to\Omega$, where $\Omega$ is an algebraic closed field, there are only finitely many extensions of $\sigma$ to $L$.

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I talk with the professor and here is a partial answer: we first construct $v : B \to \Omega$ as an $A$-linear map ($v(ab)=\nu(a)v(b)$ and consider the vector space of all such maps. Then we define the action $b \bullet v = ( x \mapsto v(bx) )$ and since all the $b$'s are commuting operators then there are common eigenvectors. We write $b \bullet v = \lambda(b) v$ and then $\lambda$ is algebras homomorphism that satisfies $\lambda(bb') = \lambda(b) \lambda(b')$ and extends $\nu$. –  LinAlgMan Nov 26 '12 at 10:33
    
Okay. This makes sense, although I don't know how to prove the existence of an $A$-linear map $v:B\to\Omega$. If you don't mind I'd include your comment in my answer in order to have (another) complete answer. –  user26857 Nov 26 '12 at 10:41
    
OK, you can include my comment in your answer. –  LinAlgMan Nov 26 '12 at 11:33

$1.$ You don't need to assume that the ring extension is finite; it is enough to be integral. In this case your question is exercise 2, Chapter 5, from Atiyah and MacDonald, and you get a ring homomorphism $v:B\to\Omega$ that extends $\nu$.

Note that a homomorphism of $A$-modules $v:B\to\Omega$ satisfies the rule $v(ab)=\nu(a)v(b)$ for all $a\in A$ and $b\in B$. In particular, a ring homomorphism that extends $\nu$ does it, too.

Remark. At this moment I can't see a direct way to prove the existence of a homomorphism of $A$-modules $B\to\Omega$.

$2.$ $V$ is a finite dimensional $\Omega$-vectorspace: take $b_1,\dots,b_n\in B$ that generates $B$ as an $A$-module. Then define $\phi:V\to\Omega^n$, $\phi(v)=(v(b_1),\dots,v(b_n))$. This is an injective homomorphism of $\Omega$-vectorspaces, so $\dim_{\Omega}V\le n$.

Edit. The OP came up with some new explanations and these allow me to finish the proof.

Define an action of $B$ over $V$ (as he did) by $b \bullet v = v_b, \mbox{ where } v_b(x)=v(bx) \mbox{ for all } b,x \in B$. This gives raise to a family $(\varphi_b)_{b\in B}$ of commuting endomorphisms of the $\Omega$-vectorspace $V$, where $\varphi_b(v)= b \bullet v$. From linear algebra we know that a family of commuting endomorphisms of a finite dimensional vectorspace over an algebraically closed field has a common eigenvector. So there exists $v\in V$ such that $\varphi_b(v)=\lambda_b v$ for any $b\in B$, where $\lambda_b\in\Omega$ is a scalar depending on $b$. From $\varphi_b(v)=\lambda_b v$ we get $v(bx)=\lambda_bv(x)$ for all $b,x\in B$ and taking $x=1$ we obtain $\lambda_b=v(1)^{-1}v(b)$, and therefore $v(bx)=v(1)^{-1}v(b)v(x)$ for all $b,x\in B$. (Note that $v(1)\neq 0$, otherwise $v=0$ and this is false!) Although $v$ is not a ring homomorphism, after rescaling we can get one: define $w:B\to\Omega$ by $w(b)=v(1)^{-1}v(b)$ and check that $w$ is a ring homomorphism extending $\nu$.

Conversely, any ring homomorphism $B\to\Omega$ that extends $\nu$ is a common eigenvector of the family $(\varphi_b)_{b\in B}$.

So we have at least one, but finitely many ring homomorphisms $B\to\Omega$ that extend $\nu$.

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Thank you very much. I still need to think about the eigenvector stuff, because if the action is as I defined, it seems that everything in $V$ is eigenvector. –  LinAlgMan Nov 21 '12 at 11:55

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