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It is known that $L(V,s)$ is analytic and non-vanishing in $\Re(s)\geq 1$. Equivalently, $\log L(V,s)$ is analytic in $\Re(s)\geq 1$. Note that $\log L(V,s)$ is given by an absolutely convergent Dirichlet series $D(V,s)$ in $\Re(s)>1$. This Dirichlet series $D(V,s)$ has bounded coefficients, it is supported on prime powers, and the contribution of a given non-archimedean place $v$ equals $\log L_v(V,s)$. In particular, $$\log L(V,s)=\sum_v\log L_v(V,s),\qquad\Re(s)>1.\tag{1}$$ By Ingham's Tauberian theorem (for which Newman gave a simple proof in 1980), $D(V,s)$ converges in $\Re(s)\geq 1$. Hence, by Theorem 1.1 in Montgomery-Vaughan: Multiplicative number theory I (CUP, 2006), $D(V,s)$ is continuous in $\Re(s)\geq 1$. It follows that $$\log L(V,s)=D(V,s),\qquad \Re(s)\geq 1.\tag{2}$$ On the right-hand sides of $(1)$ and $(2)$, the contributions of prime powers that are not primes converges absolutely for $\Re(s)>1/2$, hence they are equal, while the contributions of primes are identical Dirichlet series. Hence in fact $$\log L(V,s)=\sum_v\log L_v(V,s),\qquad\Re(s)\geq 1.$$ Equivalently, $$L(V,s)=\prod_v L_v(V,s),\qquad\Re(s)\geq 1.$$

It is known that $L(V,s)$ is analytic and non-vanishing in $\Re(s)\geq 1$. Equivalently, $\log L(V,s)$ is analytic in $\Re(s)\geq 1$. Note that $\log L(V,s)$ is given by an absolutely convergent Dirichlet series $D(V,s)$ in $\Re(s)>1$. This Dirichlet series $D(V,s)$ has bounded coefficients, it is supported on prime powers, and the contribution of a given non-archimedean place $v$ equals $\log L_v(V,s)$. In particular, $$\log L(V,s)=\sum_v\log L_v(V,s),\qquad\Re(s)>1.\tag{1}$$ By Ingham's Tauberian theorem (for which Newman gave a simple proof in 1980), $D(V,s)$ converges in $\Re(s)\geq 1$. Hence, by Theorem 1.1 in Montgomery-Vaughan: Multiplicative number theory I (CUP, 2006), $D(V,s)$ is continuous in $\Re(s)\geq 1$. It follows that $$\log L(V,s)=D(V,s),\qquad \Re(s)\geq 1.\tag{2}$$ On the right-hand sides of $(1)$ and $(2)$, the contributions of prime powers that are not primes converge absolutely for $\Re(s)>1/2$, hence they are equal, while the contributions of primes are identical Dirichlet series. Hence in fact $$\log L(V,s)=\sum_v\log L_v(V,s),\qquad\Re(s)\geq 1.$$ Equivalently, $$L(V,s)=\prod_v L_v(V,s),\qquad\Re(s)\geq 1.$$

It is known that $L(V,s)$ is analytic and non-vanishing in $\Re(s)\geq 1$. Equivalently, $\log L(V,s)$ is analytic in $\Re(s)\geq 1$. Note that $\log L(V,s)$ is given by an absolutely convergent Dirichlet series $D(V,s)$ in $\Re(s)>1$. This Dirichlet series $D(V,s)$ has bounded coefficients, it is supported on prime powers, and the contribution of a given prime $v$ equals $\log L_v(V,s)$. In particular, $$\log L(V,s)=\sum_v\log L_v(V,s),\qquad\Re(s)>1.\tag{1}$$ By Ingham's Tauberian theorem (for which Newman gave a simple proof in 1980), $D(V,s)$ converges in $\Re(s)\geq 1$. Hence, by Theorem 1.1 in Montgomery-Vaughan: Multiplicative number theory I (CUP, 2006), $D(V,s)$ is continuous in $\Re(s)\geq 1$. It follows that $$\log L(V,s)=D(V,s),\qquad \Re(s)\geq 1.\tag{2}$$ Comparing the right-hand sides of $(1)$ and $(2)$, it follows that $$\log L(V,s)=\sum_v\log L_v(V,s),\qquad\Re(s)\geq 1.$$ Equivalently, $$L(V,s)=\prod_v L_v(V,s),\qquad\Re(s)\geq 1.$$

It is known that $L(V,s)$ is analytic and non-vanishing in $\Re(s)\geq 1$. Equivalently, $\log L(V,s)$ is analytic in $\Re(s)\geq 1$. Note that $\log L(V,s)$ is given by an absolutely convergent Dirichlet series $D(V,s)$ in $\Re(s)>1$. This Dirichlet series $D(V,s)$ has bounded coefficients, it is supported on prime powers, and the contribution of a given non-archimedean place $v$ equals $\log L_v(V,s)$. In particular, $$\log L(V,s)=\sum_v\log L_v(V,s),\qquad\Re(s)>1.\tag{1}$$ By Ingham's Tauberian theorem (for which Newman gave a simple proof in 1980), $D(V,s)$ converges in $\Re(s)\geq 1$. Hence, by Theorem 1.1 in Montgomery-Vaughan: Multiplicative number theory I (CUP, 2006), $D(V,s)$ is continuous in $\Re(s)\geq 1$. It follows that $$\log L(V,s)=D(V,s),\qquad \Re(s)\geq 1.\tag{2}$$ On the right-hand sides of $(1)$ and $(2)$, the contributions of prime powers that are not primes converges absolutely for $\Re(s)>1/2$, hence they are equal, while the contributions of primes are identical Dirichlet series. Hence in fact $$\log L(V,s)=\sum_v\log L_v(V,s),\qquad\Re(s)\geq 1.$$ Equivalently, $$L(V,s)=\prod_v L_v(V,s),\qquad\Re(s)\geq 1.$$

It is known that $L(V,s)$ is analytic and non-vanishing in $\Re(s)\geq 1$. Equivalently, $\log L(V,s)$ is analytic in $\Re(s)\geq 1$. Note that $\log L(V,s)$ is given by an absolutely convergent Dirichlet series $D(V,s)$ in $\Re(s)>1$. This Dirichlet series $D(V,s)$ has bounded coefficients, it is supported on prime powers, and the contribution of a given prime $v$ equals $\log L_v(V,s)$. In particular, $$\log L(V,s)=\sum_v\log L_v(V,s),\qquad\Re(s)>1.$$ By Ingham's Tauberian theorem (for which Newman gave a simple proof in 1980), it follows that $D(V,s)$ converges in $\Re(s)\geq 1$. By a basic theorem on Dirichlet series, $D(V,s)$ is continuous in $\Re(s)\geq 1$, hence it equals $\log L(V,s)$ there. Comparing $D(V,s)$ to the Dirichlet series of the various terms $\log L_v(V,s)$, it actually follows that $$\log L(V,s)=\sum_v\log L_v(V,s),\qquad\Re(s)\geq 1.$$ Equivalently, $$L(V,s)=\prod_v L_v(V,s),\qquad\Re(s)\geq 1.$$

It is known that $L(V,s)$ is analytic and non-vanishing in $\Re(s)\geq 1$. Equivalently, $\log L(V,s)$ is analytic in $\Re(s)\geq 1$. Note that $\log L(V,s)$ is given by an absolutely convergent Dirichlet series $D(V,s)$ in $\Re(s)>1$. This Dirichlet series $D(V,s)$ has bounded coefficients, it is supported on prime powers, and the contribution of a given prime $v$ equals $\log L_v(V,s)$. In particular, $$\log L(V,s)=\sum_v\log L_v(V,s),\qquad\Re(s)>1.\tag{1}$$ By Ingham's Tauberian theorem (for which Newman gave a simple proof in 1980), $D(V,s)$ converges in $\Re(s)\geq 1$. Hence, by Theorem 1.1 in Montgomery-Vaughan: Multiplicative number theory I (CUP, 2006), $D(V,s)$ is continuous in $\Re(s)\geq 1$. It follows that $$\log L(V,s)=D(V,s),\qquad \Re(s)\geq 1.\tag{2}$$ Comparing the right-hand sides of $(1)$ and $(2)$, it follows that $$\log L(V,s)=\sum_v\log L_v(V,s),\qquad\Re(s)\geq 1.$$ Equivalently, $$L(V,s)=\prod_v L_v(V,s),\qquad\Re(s)\geq 1.$$