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This is my first question on StackExchange. I'm taking a second semester course of Abstract Algebra. I have a general understanding of Principal Ideal Domains, but I am a bit confused on proving that a specific ring is a PID.

Could I get an example of showing that a specific ring is a PID? Or possibly an example of a ring that is not a PID?

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    $\begingroup$ The ring $\mathbb{Z}$ is a PID. The ring $\mathbb{Z}[X]$ is not, since the ideal generated by the polynomials $2$ and $X$ (that is, $(2,X)$) is not a principal ideal. $\endgroup$ – Sigur Oct 28 '13 at 19:43
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    $\begingroup$ To prove that a general ring is a PID, you must (1) prove the ring is an integral domain (i.e. no zero divisors), and (2) prove that every ideal is principal. $\endgroup$ – vadim123 Oct 28 '13 at 19:45
  • $\begingroup$ This can be hard. Sometimes you just try to prove it. Sometimes you get lucky and can try other routes, like first showing it is a Euclidean Domain, since EDs are PIDs. $\endgroup$ – davidlowryduda Oct 28 '13 at 19:46
  • $\begingroup$ @vadim123 How do I prove that every ideal is principle? This is what I am stuck on. $\endgroup$ – Pubbie Oct 28 '13 at 19:47
  • $\begingroup$ See this answer (and others) for more details on Sigur's comment. $\endgroup$ – Ayman Hourieh Oct 28 '13 at 19:47
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The polynomial ring $\;\Bbb F[x]\;$ is a PID iff $\;\Bbb F\;$ is a field, thus the usual polynomial rings $\;\Bbb R[x]\,,\,\Bbb Q[x]\,,\,\Bbb C[x]\,,\,\Bbb F_q[x]\;$ are PID's , whereas the nice ring $\;\Bbb Z[x]\;$ , for example, isn't: the ideal $\;\langle 2,x\rangle\;$ (can you characterize this ideal's elements?) is not principal.

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Here's an example from number theory. In the ring $R = \mathbb{Z}\lbrack\sqrt{-5}\rbrack = \{a + b\sqrt{-5} : a,b \in \mathbb{Z}\}$ is not a PID! Specifically, the ideal $I = (2,1+\sqrt{-5})$ is not principal.

The easiest way to see this is to use the norm map $N: \mathbb{Z}\lbrack\sqrt{-5}\rbrack \to \mathbb{Z}$ defined by $N(x+y\sqrt{-5}) = (x+y\sqrt{-5})(x-y\sqrt{-5}) = x^2 + 5y^2$. This map is well-defined and multiplicative ($N(rs) = N(r)N(s)$ for $r,s \in \mathbb{Z}\lbrack\sqrt{-5}\rbrack$). Multiplicativity implies that if $z = xy$ then $N(z) = N(x)N(y)$ and so $N(x), N(y) \mid N(z)$ (since $N(x)$, $N(y)$, and $N(z)$ are all integers).

Now onto the proof. First, notice that $1 \not\in I$, and so $I$ is a proper ideal of $R$. By way of contradiction, assume that $I = (\alpha)$ for some $\alpha \in R$. Then $\alpha \mid 2$, $\alpha \mid (1+\sqrt{-5})$. Hence by the above observation about the norm map we know $N(\alpha) \mid N(2) = 4$ and $N(\alpha) \mid N(1+\sqrt{-5}) = 6$. Hence $N(\alpha) \mid \text{gcd}(4,6) = 2$, which (since $N(\alpha) \geq 0$) makes $N(\alpha) \in \{1,2\}$. But we know $\alpha = a + b\sqrt{-5}$ for integers $a,b$ and $N(\alpha) = a^2+5b^2$. The equation $a^2 + 5b^2 = 2$ has no solutions $(a,b)$ in $\mathbb{Z}^2$, and the only solutions $(a,b)$ in $\mathbb{Z}^2$ to $a^2 + 5b^2 = 1$ are $(a,b) = (\pm1,0)$. Thus $\alpha \in\{\pm 1\}$ is a unit, which means $I = (\alpha) = R$. But we've already noted above (and it's easy to very directly) that $1 \not\in I$, and so this is a contradiction.

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