## Is there another proof for Dirichlet’s theorem? [closed]

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Is a “non-analytic” proof of Dirichlet’s theorem on primes known or possible?

Dirichlet's theorem on primes in arithmetic progression states that there are infinitely many primes of the form $kn+h$ given that $k$ and $h$ are coprime. Is there a short proof for this?

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short answer: no. – KConrad Jun 14 2010 at 20:30

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"It is rash to assert that a particular theorem cannot proved in a particular way." Thought you were an endorser of that viewpoint, Professor K. C. – J. H. S. Jun 14 2010 at 20:40

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Every proof I've even seen takes the same form up until the final steps. (1) Introduce the character group of the unit group of $Z/N$. (2) Consider the sum $\sum \chi(k)/k$, where $chi$ is a character of $Z/N$. Notice that this sum is much larger for $\chi$ trivial than for $\chi$ nontrivial. (3) Use the multiplicativity of $\chi$, and step 2, to deduce that $\sum 1/p$ grows much faster than $\sum \chi(p)/p$. (4) THE HARD STEP: Deduce somehow that $\sum \chi(k)/k \neq 0$, so $- \sum \chi(p)/p$ is also small. (5) Deduce the theorem. – David Speyer Jun 14 2010 at 20:52

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J.H.S.: I didn't mean there can't be a short proof, but rather that (right now) there isn't one. That's what it seemed he was asking about, not some meta-mathematical query on the possible existence of a short proof. – KConrad Jun 14 2010 at 21:04

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I have voted to close, but not for the reason that others have (they said "exact duplicate"). Rather, I think that the question "does there exist a short proof of Theorem X?" is inherently vague and subjective and could well lead to arguments of the form "Proof X which assumes Y and takes Z pages is / is not short." Please clarify what you actually want to know. There are proofs of Dirichlet's theorem which avoid complex or even real analysis, but I am not aware of a proof which could be given in an undergraduate course in less than a week of lectures. – Pete L. Clark Jun 15 2010 at 3:12
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## 1 Answer

Well, there are short proofs of particular instances of the result. For example, emulating the Euclidean assault on the infinitude of primes, one can establish, almost effortlessly, that there are infinitely many primes of the form 4k+3. Nevertheless, you have to be warned that there is no way to strenghten this technique in order to get the result for every arithmetic progression. You may want to take a look at [1]. In that note, Professor Murty mentions that it was I. Schur the one who first derived a sufficient condition for the existence of an "Euclidean" proof for the infinitude of primes in the arithmetic progression {$mk+a$}$_{k \in \mathbb{N}}$.

Edit: As David Speyer mentioned above one on the main ingredients in the proof is a certain non-vanishing result for L-series. Hence, a way in which one might shorten the proof is by spotting the shortest demonstration for the corresponding non-vanishing theorem. I higly recommend that you take a look at the thread in [2] if you wish to learn more about this particular matter.

References

[1] M. R. Murty, Primes in certain arithmetic progressions, J. Madras. Univ. (1988), 161-169.

[2] Shortest/Most elegant proof for the non-vanishing of $L(1, \chi)$ :

http://mathoverflow.net/questions/25794/shortest-most-elegant-proof-for-l1-chi-neq-0/25815#25815

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Oh, that's a different matter! The person who asked the question should definitely clarify if he was asking for a short proof of the general theorem or if he'd be content with a short proof of some special cases (which usually don't give a flavor of the general proof). – KConrad Jun 14 2010 at 21:01
Point taken, Sire. Nonetheless, I decided to enter the reply because I thought it'd help to disseminate the fact that there are specific arithmetic progressions for which the related proofs are as short as possible. – J. H. S. Jun 14 2010 at 21:44

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There's also a short proof for $4k+1$. Assume there were a finite number of such primes, take their product, square it, and add one. Then there is a prime $p$ that divides this result. This forces $-1$ to be a square in $F_p$, hence $p$ is congruent to 1 modulo 4, contradiction. – David Carchedi Jun 14 2010 at 22:09
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