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So I have to prove that there are infinitely many prime elements in $R = F[x]$ where $F$ is a field.s I can assume that $R$ is a UFD and therefore an irreducible element is same as a prime element.

This is my approach:

Assume the contrary that is there are finite prime numbers, let them be $(p_1 , p_2 , p_3\ldots p_n)$

Let there be a polynomial $(p_1.p_2.p_3\ldots p_n) \cdot x + 1$

Now since it belongs to the ring, it can either be reducible or irreducible.

Case 1 : Polynomial is irreducible:

If it is irreducible then we have found a new prime element in $R[F(x)]$ , so contradiction.

Case 2: I am not able to get a contradiction in the case where the polynomial is reducible..

Is my approach even correct??

FishDrowned
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Ansh
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    You are expected to understand the analogy between this result and Euclid's proof of infinitude of primes. How does case 2 go in that case? – Jyrki Lahtonen Nov 23 '23 at 07:21
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    @Ok So I read the article , can you confirm if I understood it correctly: We form a new prime from our previously finitely assumed primes , (p1.p2.p3...pn)+1, now we know this is not divisible by any primes in our list since every one of them leave a remainder of 1 , so this number is either a prime or a product of prime , if it is prime then we found a new prime , if it is a product of prime then also we found new primes since its factor can't be in our list as they leave remainder 1, so we have proved that there are infinite primes? – Ansh Nov 23 '23 at 07:29
  • Exactly that. Euclid used exactly this idea for the prime numbers. It can easily be used also in this case. – Peter Nov 23 '23 at 07:55
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    I added a different proof under the first occurence of this question as linked above. – Andrea Mori Nov 23 '23 at 08:49

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