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On page 41 in the proof of proposition 3.103.10., change

"i) $\implies$ ii) by (3.5) and (2.20)" to "i) $\implies$ ii) by (3.7) and (2.19)"

On page 52 in remark 1) at the bottom of the page, change

"(see Chapter 1, Exercise 25)" to "(see Chapter 1, Exercise 27)"

On page 65 at the end of the proof of proposition 5.18. the black square to denote end of proof is missing.

On page 66 we need to correct the proof of corollary 5.22., one correct version is the following: We start with the quotient map $\pi: A[x^{-1}] \to A[x^{-1}] /m$ where $m$ is a maximal ideal containing $x^{-1}$. We take an algebraic closure $\Omega$ of the field $A[x^{-1}] /m$ and consider the map $i \circ \pi: A[x^{-1}] \to \Omega$. Then by the previous theorem, (5.21), we can extend $i \circ \pi$ to some valuation ring $B$ of $K$ containing $A[x^{-1}]$: $g: B \to \Omega$ such that $g|_{A[x^{-1}]} = i \circ \pi$. Then $g(x^{-1}) = 0$. Hence $x^{-1} \in ker(g)$ and since the kernel is a proper ideal of $B$, $x^{-1}$ is not a unit in $B$ and hence $x$ is not in $B$. (also see math.SE)

On page 77 in the proof of proposition 6.7., change

"...a composition series, by ii);..." to "...a composition series, by i);..."

On page 41 in the proof of proposition 3.10, change

"i) $\implies$ ii) by (3.5) and (2.20)" to "i) $\implies$ ii) by (3.7) and (2.19)"

On page 52 in remark 1) at the bottom of the page, change

"(see Chapter 1, Exercise 25)" to "(see Chapter 1, Exercise 27)"

On page 65 at the end of the proof of proposition 5.18 the black square to denote end of proof is missing.

On page 66 we need to correct the proof of corollary 5.22., one correct version is the following: We start with the quotient map $\pi: A[x^{-1}] \to A[x^{-1}] /m$ where $m$ is a maximal ideal containing $x^{-1}$. We take an algebraic closure $\Omega$ of the field $A[x^{-1}] /m$ and consider the map $i \circ \pi: A[x^{-1}] \to \Omega$. Then by the previous theorem, (5.21), we can extend $i \circ \pi$ to some valuation ring $B$ of $K$ containing $A[x^{-1}]$: $g: B \to \Omega$ such that $g|_{A[x^{-1}]} = i \circ \pi$. Then $g(x^{-1}) = 0$. Hence $x^{-1} \in ker(g)$ and since the kernel is a proper ideal of $B$, $x^{-1}$ is not a unit in $B$ and hence $x$ is not in $B$. (also see math.SE)

3 added 112 characters in body

On page 41 in the proof of proposition 3.10, change

"i) $\implies$ ii) by (3.5) and (2.20)" to "i) $\implies$ ii) by (3.7) and (2.19)"

On page 52 in remark 1) at the bottom of the page, change

"(see Chapter 1, Exercise 25)" to "(see Chapter 1, Exercise 27)"

On page 65 at the end of the proof of proposition 5.18 the black square to denote end of proof is missing.

2 added 131 characters in body