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 $K$ be a field and $R = K\times K$. Let $P = 0\times K$ be a prime ideal of $R$. If $\phi : K \rightarrow R$ is defined by $\phi(x) = (x, x)$ then we need to prove the induced map $\phi_{P} : K_{0} \rightarrow R_{P}$ is surjective, where $K_{0} = K_{\phi^{-1}(P)}$. Thanks.

share|improve this question

1 Answer 1

up vote 1 down vote accepted

$K_0=K$, so $\phi_P(x)=(x,x)/(1,1)$. Take $(a,b)/(s,t)\in R_P$, where $a,b,s,t\in K$ and $s\neq 0$. Then $(a,b)/(s,t)=(x,x)/(1,1)$ where $x=as^{-1}$, so $\phi$ is surjective.

Edit. $(a,b)/(s,t)=(x,x)/(1,1)\Leftrightarrow \exists(u,v)\in R-P$ such that $(u,v)[(a,b)(1,1)-(s,t)(x,x)]=(0,0)\Leftrightarrow \exists(u,v)\in R-P$ such that $(u,v)(a-sx,b-tx)=(0,0)$. Since $a-sx=0$ we can choose, for example, $(u,v)=(1,0)$.

share|improve this answer
    
Thanks YACP. This makes sense. –  Rajesh Dec 27 '12 at 17:38

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.