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There are proofs that treats the cases of real and non-real $\chi$ on an equal footing. One proof is in Serre's Course in Arithmetic, which the answers by Pete and David are basically about. That method is using the (hidden) fact that the zeta-function of the $m$-th cyclotomic field has a simple pole at $s = 1$, just like the Riemann zeta-function. Here is another proof which focuses only on the $L$-function of the character $\chi$ under discussion, the $L$-function of the conjugate character, and the Riemann zeta-function.

Consider the product $$H(s) = \zeta(s)^2L(s,\chi)L(s,\overline{\chi}).$$ This function is analytic for $\sigma > 0$, with the possible exception of a pole at $s = 1$. (As usual I write $s = \sigma + it$.)

Assume $L(1,\chi) = 0$. Then also $L(1,\overline{\chi}) = 0$. So in the product defining $H(s)$, the double pole of $\zeta(s)^2$ at $s = 1$ is cancelled and $H(s)$ is therefore analytic throughout the half-plane $\sigma > 0$.

For $\sigma > 1$, we have the exponential representation $$H(s) = \exp\left(\sum_{p, k} \frac{2 + \chi(p^k) + \overline{\chi}(p^k)} {kp^{ks}}\right),$$ where the sum is over $k \geq 1$ and primes $p$. If $p$ does not divide $m$, then we write $\chi(p) = e^{i\theta_p}$ and find
$$\frac{2 + \chi(p^k) + \overline{\chi}(p^k)}{k} = \frac{2(1 + \cos(k\theta_p))}{k} \geq 0.$$
If $p$ divides $m$ then this sum is $2/k > 0$.
Either way, inside that exponential is a Dirichlet series with nonnegative coefficients, so when we exponentiate and rearrange terms (on the half-plane of abs. convergence, namely where $\sigma > 1$), we see that $H(s)$ is a Dirichlet series with nonnegative coefficients. A lemma of Landau on Dirichlet series with nonnegative coefficients then assures us that the Dirichlet series representation of $H(s)$ is valid on any half-plane where $H(s)$ can be analytically continued.

To get a contradiction at this point, here are several methods.

[Edit: In the answer by J.H.S., and due to Bateman, is the slickest argument I have seen, so let me put it here. The idea is to look at the coefficient of $1/p^{2s}$ in the Dirichlet series for $H(s)$. By multiplying out the $p$-part of the Euler product, the coefficient of $1/p^s$ is $2(\chi(p) 2 + \overline{\chi}(p))$, chi(p) + \overline{\chi}(p)$, which is nonnegative, but the coefficient of$1/p^{2s}$is$(\chi(p) + \overline{\chi}(p) + 1)^2 + 1$, which is not only nonnegative but in fact is greater than or equal to 1. Therefore if$H(s)$has an analytic continuation along the real line out to the number$\sigma$, then for real$s \geq \sigma$we have$H(s) \geq \sum_{p} 1/p^{2s}$. The hypothesis that$L(1,\chi) = 0$makes$H(s)$analytic for all complex numbers with positive real part, so we can take$s = 1/2$and get$H(1/2) \geq \sum_{p} 1/p$, which is absurd since that series over the primes diverges. QED!] 1. If you are willing to accept that$L(s,\chi)$(and therefore$L(s,\overline{\chi})$) has an analytic continuation to the whole plane, or at least out to the point$s = -2$, then$H(s)$extends to$s = -2$. The Dirichlet series representation of$H(s)$is convergent at$s = -2$by our analytic continuation hypothesis and it shows$H(-2) > 1$, or the exponential representation implies that at least$H(-2) \not= 0$. But$\zeta(-2) = 0$, so$H(-2) = 0$. Either way, we have a contradiction. 2. There is a similar argument, pointed out to me by Adrian Barbu, that does not require analytic continuation of$L(s,\chi)$beyond the half-plane$\sigma > 0$. If you are willing to accept that$\zeta(s)$has zeros in the critical strip$0 < \sigma < 1$(which is a region that the Dirichlet series and exponential representations of$H(s)$are both valid since$H(s)$is analytic on$\sigma > 0$), we can evaluate the exponential representation of$H(s)$at such a zero to get a contradiction. Of course the amount of analysis that lies behind this is more substantial than what is used to continue$L(s,\chi)$out to$s = -2$. 3. We consider$H(s)$as$s \rightarrow 0^{+}$. We need to accept that$H$is bounded as$s \rightarrow 0^{+}$. (It's even holomorphic there, but we don't quite need that.) For real$s > 0$and a fixed prime$p_0$(not dividing$m$, say), we can bound$H(s)$from below by the sum of the$p_0$-power terms in its Dirichlet series. The sum of these terms is exactly the$p_0$-Euler factor of$H(s)$, so we have the lower bound $$H(s) > \frac{1}{(1 - p_0^{-s})^2(1 - \chi(p_0)p_0^{-s})(1 - \overline{\chi}(p_0)p_0^{-s})} = \frac{1}{(1 - p_0^{-s})^2(1 - (\chi(p_0)+ \overline{\chi}(p_0))p_{0}^{-s} + p_0^{-2s})}$$ for real$s > 0$. The right side tends to$\infty$as$s \rightarrow 0^{+}$. We have a contradiction. QED These three arguments at some point use knowledge beyond the half-plane$\sigma > 0$or a nontrivial zero of the zeta-function. Granting any of those lets you see easily that$H(s)$can't vanish at$s = 1$, but that "granting" may seem overly technical. If you want a proof for the real and complex cases uniformly which does not go outside the region$\sigma > 0$, use the method in the answer by Pete or David [edit: or use the method I edited in as the first one in this answer]. 5 edited body There are proofs that treats the cases of real and non-real$\chi$on an equal footing. One proof is in Serre's Course in Arithmetic, which the answers by Pete and David are basically about. That method is using the (hidden) fact that the zeta-function of the$m$-th cyclotomic field has a simple pole at$s = 1$, just like the Riemann zeta-function. Here is another proof which focuses only on the$L$-function of the character$\chi$under discussion, the$L$-function of the conjugate character, and the Riemann zeta-function. Consider the product $$H(s) = \zeta(s)^2L(s,\chi)L(s,\overline{\chi}).$$ This function is analytic for$\sigma > 0$, with the possible exception of a pole at$s = 1$. (As usual I write$s = \sigma + it$.) Assume$L(1,\chi) = 0$. Then also$L(1,\overline{\chi}) = 0$. So in the product defining$H(s)$, the double pole of$\zeta(s)^2$at$s = 1$is cancelled and$H(s)$is therefore analytic throughout the half-plane$\sigma > 0$. For$\sigma > 1$, we have the exponential representation $$H(s) = \exp\left(\sum_{p, k} \frac{2 + \chi(p^k) + \overline{\chi}(p^k)} {kp^{ks}}\right),$$ where the sum is over$k \geq 1$and primes$p$. If$p$does not divide$m$, then we write$\chi(p) = e^{i\theta_p}$and find $$\frac{2 + \chi(p^k) + \overline{\chi}(p^k)}{k} = \frac{2(1 + \cos(k\theta_p))}{k} \geq 0.$$ If$p$divides$m$then this sum is$2/k > 0$. Either way, inside that exponential is a Dirichlet series with nonnegative coefficients, so when we exponentiate and rearrange terms (on the half-plane of abs. convergence, namely where$\sigma > 1$), we see that$H(s)$is a Dirichlet series with nonnegative coefficients. A lemma of Landau on Dirichlet series with nonnegative coefficients then assures us that the Dirichlet series representation of$H(s)$is valid on any half-plane where$H(s)$can be analytically continued. To get a contradiction at this point, here are several methods. [Edit: In the answer by J.S.E.J.H.S., and due to Bateman, is the slickest argument I have seen, so let me put it here. The idea is to look at the coefficient of$1/p^{2s}$in the Dirichlet series for$H(s)$. By multiplying out the$p$-part of the Euler product, the coefficient of$1/p^s$is$2(\chi(p) + \overline{\chi}(p))$, which is nonnegative, but the coefficient of$1/p^{2s}$is$(\chi(p) + \overline{\chi}(p) + 1)^2 + 1$, which is not only nonnegative but in fact is greater than or equal to 1. Therefore if$H(s)$has an analytic continuation along the real line out to the number$\sigma$, then for real$s \geq \sigma$we have$H(s) \geq \sum_{p} 1/p^{2s}$. The hypothesis that$L(1,\chi) = 0$makes$H(s)$analytic for all complex numbers with positive real part, so we can take$s = 1/2$and get$H(1/2) \geq \sum_{p} 1/p$, which is absurd since that series over the primes diverges. QED!] 1. If you are willing to accept that$L(s,\chi)$(and therefore$L(s,\overline{\chi})$) has an analytic continuation to the whole plane, or at least out to the point$s = -2$, then$H(s)$extends to$s = -2$. The Dirichlet series representation of$H(s)$is convergent at$s = -2$by our analytic continuation hypothesis and it shows$H(-2) > 1$, or the exponential representation implies that at least$H(-2) \not= 0$. But$\zeta(-2) = 0$, so$H(-2) = 0$. Either way, we have a contradiction. 2. There is a similar argument, pointed out to me by Adrian Barbu, that does not require analytic continuation of$L(s,\chi)$beyond the half-plane$\sigma > 0$. If you are willing to accept that$\zeta(s)$has zeros in the critical strip$0 < \sigma < 1$(which is a region that the Dirichlet series and exponential representations of$H(s)$are both valid since$H(s)$is analytic on$\sigma > 0$), we can evaluate the exponential representation of$H(s)$at such a zero to get a contradiction. Of course the amount of analysis that lies behind this is more substantial than what is used to continue$L(s,\chi)$out to$s = -2$. 3. We consider$H(s)$as$s \rightarrow 0^{+}$. We need to accept that$H$is bounded as$s \rightarrow 0^{+}$. (It's even holomorphic there, but we don't quite need that.) For real$s > 0$and a fixed prime$p_0$(not dividing$m$, say), we can bound$H(s)$from below by the sum of the$p_0$-power terms in its Dirichlet series. The sum of these terms is exactly the$p_0$-Euler factor of$H(s)$, so we have the lower bound $$H(s) > \frac{1}{(1 - p_0^{-s})^2(1 - \chi(p_0)p_0^{-s})(1 - \overline{\chi}(p_0)p_0^{-s})} = \frac{1}{(1 - p_0^{-s})^2(1 - (\chi(p_0)+ \overline{\chi}(p_0))p_{0}^{-s} + p_0^{-2s})}$$ for real$s > 0$. The right side tends to$\infty$as$s \rightarrow 0^{+}$. We have a contradiction. QED These three arguments at some point use knowledge beyond the half-plane$\sigma > 0$or a nontrivial zero of the zeta-function. Granting any of those lets you see easily that$H(s)$can't vanish at$s = 1$, but that "granting" may seem overly technical. If you want a proof for the real and complex cases uniformly which does not go outside the region$\sigma > 0$, use the method in the answer by Pete or David [edit: or use the method I edited in as the first one in this answer]. 4 deleted 25 characters in body three several methods. First [Edit: In the answer by J.S.E., and due to Bateman, is the slickest argument I have seen, so let me put it here. The idea is to look at the coefficient of$1/p^{2s}$in the Dirichlet series for$H(s)$. By multiplying out the$p$-part of the Euler product, the coefficient of$1/p^s$is$2(\chi(p) + \overline{\chi}(p))$, which is nonnegative, but the coefficient of$1/p^{2s}$is$(\chi(p) + \overline{\chi}(p) + 1)^2 + 1$, which is not only nonnegative but in fact is greater than or equal to 1. Therefore if$H(s)$has an analytic continuation along the real line out to the number$\sigma$, then for real$s \geq \sigma$we have$H(s) \geq \sum_{p} 1/p^{2s}$. The hypothesis that$L(1,\chi) = 0$makes$H(s)$analytic for all complex numbers with positive real part, so we can take$s = 1/2$and get$H(1/2) \geq \sum_{p} 1/p$, which is absurd since that series over the primes diverges. QED!] • If you are willing to accept that But$\zeta(-2) = 0$(ha!), , so$H(-2) = 0$. Either way, we have a contradiction. • get a contradiction. Since Of course the amount of continue$L(s,\chi)$out to$s = -2$, the first method is the simpler of the two so far. For a third method, we 2$.

• We consider $H(s)$ as $s \rightarrow 0^{+}$. We need to accept that $H$ is bounded as $s \rightarrow 0^{+}$. (It's even

All of these

• These three arguments at some point used something use knowledge beyond the half-plane $\sigma > 0$ or a nontrivial zero of the zeta-function. Granting any of those lets you see easily that $H(s)$ can't vanish at $s = 1$, but that "granting" may seem overly technical. If you want a proof for the real and complex cases uniformly which does not go outside the region $\sigma > 0$, the argument as in Pete's answer is use the way to go.

Edit: A fourth method along these lines, mentioned in the answer by J.S.E. and due to Bateman, is quite nice! The idea is to look at Pete or David [edit: or use the coefficient of $1/p^{2s}$ method I edited in $H(s)$. By multiplying out the $p$-part of the Euler product, the coefficient of $1/p^s$ is $2(\chi(p) + \overline{\chi}(p))$, which is nonnegative, but as the coefficient of $1/p^{2s}$ is $(\chi(p) + \overline{\chi}(p) + 1)^2 + 1$, which is not only nonnegative but first one in fact is greater than or equal to 1. (Note $\chi(p) + \overline{\chi}(p)$ is real.) Therefore if $H(s)$ has an analytic continuation along the real line out to the number $\sigma$, then for real $s \geq \sigma$ we have $H(s) \geq \sum_{p} 1/p^{2s}$. The hypothesis that $L(1,\chi) = 0$ makes $H(s)$ analytic for all complex numbers with positive real part, so we can take $s = 1/2$ and get $H(1/2) \geq \sum_{p} 1/p$, which is absurd since that series over the primes divergesthis answer].QED!

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