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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.

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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, $(u,v)=(1,0)$.

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Thanks YACP. This makes sense. – Rajesh Dec 27 '12 at 17:38

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