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Let $A$ be a Krull ring and $p$ a prime ideal of height 1. Then $A_p$ is a DVR with corresponding valuation $v$ on the field of fractions $K$ of $A$.

Question: Can we extend this valuation to an additive valuation of $K(X)$?

Remark: Matsumura in his Commutative Ring Theory p.89, says that this can be done by defining $v(a_0+a_1x+\cdots+a_nx^n)=\min_{i}v(a_i)$. However, this extension of $v$ does not seem to me to be additive, i.e. $v(f(x)g(x)) \neq v(f(x))+v(g(x))$. What am i missing?

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This is easy: consider $f=a_0+a_1x+\cdots+a_mx^m$ and $g=b_0+b_1+\cdots+b_nx^n$. Denote $v(f)=r$, $v(g)=s$ and take $i$, respectively $j$ minimal with the property that $v(a_i)=r$, respectively $v(b_j)=s$. It's obvious that $v(fg)\ge r+s$, and it remains to prove that the equality holds. Set $c=\sum_{k+l=i+j}a_kb_l$. We have $v(a_kb_l)=v(a_k)+v(b_l)>v(a_i)+v(b_j)=v(a_ib_j)$ $=$ $r+s$ for $(k,l)\neq(i,j)$. This shows that $v(c)=r+s$. (Here I've used the following well known property of valuations: if $v(a)>v(b)$, the $v(a+b)=v(b)$.)

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  • $\begingroup$ Can you please explain what happens if two of $v(a_k b_l)$ are equal...? $\endgroup$ – Subhadip Majumder Aug 4 at 14:58

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