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I'm quite confused about the existence of prime elements in the ring $R=\mathbb{Q}.$ We know that $r \in R$ is a prime iff $r$ is a nonzero, nonunit of $R$ and $r|ab \implies r|a \ \text{or} \ \ r|b \ (a,b \in R).$ But given any nonzero $r\in R,$ $\frac{1}{r}$ will always exist, i.e. $r(\frac{1}{r})=(\frac{1}{r})r =1).$ Hence $r$ is a unit and $R$ has no prime elements?

Appreciate if someone can correct my misconceptions. Thank you.

glS
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Alexy Vincenzo
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    You are correct; $\mathbb{Q}$ is a field, and in a field every nonzero element is a unit. Hence there are no primes. – vadim123 Mar 25 '14 at 05:40

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Your reasoning is correct. Note that when you go from the integral domain $\mathbb{Z}$ to the field of fractions $\mathbb{Q}$, you lose all your prime elements.

In general, if $F$ is a field, then $F$ has no prime elements.

Caleb Stanford
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  • Thank you. To use Eisenstein's criterion to prove $$(x+1)^6+(x+1)^3+1=x^6+6x^5+15x^4+21x^3+18x^2+9x+3$$ is irreducible in $\mathbb{Q}[X],$ I need to find prime $p \in \mathbb{Q}$ such that $p$ divides the product of all the coefficients, $p^2$ does not divide $3$ and $p$ does not divide $1$ in $\mathbb{Q}.$ But how is this possible? – Alexy Vincenzo Mar 25 '14 at 05:44
  • And isn't $r|s, \ s|r, \forall r,s \in \mathbb{Q}^{*} \ ? $ – Alexy Vincenzo Mar 25 '14 at 05:46
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    You do not use Eisenstein's criterion directly in $\Bbb{Q}[x]$. Instead, you apply it in $\Bbb{Z}[x]$, and then use Gauss' Lemma (the one about primitive polynomials) to conclude that it is also irreducible in $\Bbb{Q}[x]$. – Ted Mar 25 '14 at 05:51