4

(Confusion) Suppose $R$ is an integral domain with the distinct elements $\{0,1,a_1,a_2.......a_n\}$. If $p$ is a prime element belonging to $R$ then as per theorem $pR$ is a prime ideal.

Now I have a confusion here. As per the definition of prime ideal it is also a proper ideal. So $pR \neq R$. This implies that some of the elements of $pR$ are not distinct. Let us assume that $pa=pb$ ($a=a_i$, $b=a_j$ for some $i$ & $j$ where $i\neq j$ belongs to $R$) or, $p(a-b)=0 \Rightarrow a=b$ (as there is no zero divisor in $R$). This is a contradiction with the fact that $\{0,1,a_1,a_2........a_n\}$ is a set of distinct elements.

Please help.

egreg
  • 238,574
Sourav Chakraborty
  • 371
  • 1
    Your idea is correct but the articulation has a slight issue. Repetitions are not allowed in sets, so elements of $pR$ cannot be nondistinct by design. Rather, you want to say that the function $R\to pR$ given by multiplication by $p$ is not injective. Ultimately this is a proof by contradiction that there are no prime elements in a finite integral domain, which can be easily augmented to prove that all finite integral domains are fields. – anon Dec 07 '13 at 10:22

1 Answers1

3

You have accidentally stumbled upon the fact that finite integral domains are fields.

Fields don't have any proper nonzero ideals. In particular, no prime elements.

anon
  • 151,657
  • In my book I have a theorem stating precisely this " Let R be an integral domain.A non zero element p belonging to R is a prime element iff pR is a prime Ideal of R." So do you mean to say this theorem dose not hold true for finite integral domains but stands true to infinite domains? Proof given in the book also states pR==p => 1 does not belong to pR. Could you please explain this argument ? – Sourav Chakraborty Dec 07 '13 at 10:31
  • @SouravChakraborty No, "$p\in R$ prime iff $pR$ is a nonzero prime ideal" is still just as true as ever in fields, in particular in finite integral domains, because there no prime elements in fields and the only prime ideal is $(0)$. Also, indeed $I\ne R\Leftrightarrow 1\not\in I$ for any ideal $I\triangleleft R$, in particular for $I=pR$. It is logically equivalent to $I=R\Leftrightarrow 1\in R$: try to show both directions of implications by yourself. Can you say that $I=R\Rightarrow 1\in R$? Can you show $1\in I\Rightarrow I=R$? – anon Dec 07 '13 at 23:34
  • For your purposes, in order to show a finite integral domain $R$ is a field, you want to show every nonzero element $a\in R$ has an inverse, for which you can show that the multiplication-by-$a$ map is injective for all nonzero $a$, for which you can use the fact that $R$ is a domain. (Argue that non-injectivity of one of those multiplication-by-$a$ maps would imply $a$ is a zero divisor, impossible in a domain.) – anon Dec 07 '13 at 23:38