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Let $R$ be a UFD and $P$ a prime ideal. Here we are defining a UFD with primes and not irreducibles.

Is the following true and what is the justification?

If $a$ in $R$ is prime, then $(a+P)$ is prime in $R/P$.

Nikolaj-K
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The bijection between ideals of $\,R/I\,$ and ideals of $R$ containing $I$ restricts to prime ideals, hence $\,(a+P)\,$ is prime in $\,R/P\,$ iff $\,I = (a)+P\,$ is prime in $\,R.\,$ But generally this is not true, e.g. $\,I=1\,$ for $\,a\nmid P\,$ primes in $\,\Bbb Z,\,$ or $\,I = (4\!-\!x)+(x) = (4,x)\,$ is not prime in $\,\Bbb Z[x],\,$ by $\,2^2\in I\,$ but $\,2\not\in I.$

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Bill Dubuque
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There are many cases that I'm tempted to call degenerate: $R/P$ could be a field, for example. For concreteness, consider $R = \mathbf Z$, $P = 2\mathbf Z$, and $a = 3$.

Maybe it's helpful to think geometrically: if $k$ is a field of characteristic $\neq 2$ we can describe a parabola in the $k$-plane via the ring $A = k[x, y]/(y - x^2)$. The element $y - 1$ is certainly prime in $k[x, y]$, but \[ A/(y - 1)A \approx k[x]/(1 - x^2) \approx k^2 \] is not a domain, so the image of $y - 1$ in $A$ is not prime. The picture is that the line $y = 1$ and the parabola $y = x^2$ intersect in two points.

Dylan Moreland
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  • Of course, this does not use unique factorization in a very deep way, so I'd be interested in seeing an answer that does. – Dylan Moreland Jun 20 '12 at 20:44
  • Also, it's very possible that you haven't thought about describing geometry using algebra before and that's perfectly alright — you can treat this example purely formally. But in the long run I think that thinking about the geometry of algebraic facts is worth it. I should add later that another thing to remember is that sums of prime ideals need not be prime. – Dylan Moreland Jun 20 '12 at 21:01
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I think thats wrong. If $ a \in P $ holds, then $ a + P = 0 + P \in R/P$ and therefore $a+P$ not prime.

sebigu
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    $(a+P)=P$ if a is prime, true, thanks. But I'm uncertain about the terminology of $P$ in $R/P$, which I feel is, to draw a comparison, like the the ideal in the integers modulo that ideal, i.e. it's like $0$, right? – Nikolaj-K Jun 20 '12 at 20:21
  • I think you have got it right. In $R/P$, all elements of $P$ become $0$ and two elements $a,b \in R$ become equal if $a-b \in P$. If you take $\mathbb{Z}$ as $R$ and $<2>$ as $P$, then in $R/P$ all even numbers become $0$, therefore $R/P$ only consists of two elements, the residue classes of $0$ and $1$. – sebigu Jun 20 '12 at 20:28
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    @sebigu But $0$ is a prime ideal in a domain. – Bill Dubuque Jun 20 '12 at 20:32
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    I guess it's confusing, because normally in UFDs we would talk about irreducibles and the convention is usually that $0$ is not an irreducible element. [Edit: I guess prime elements can't be zero either. The more you learn!] – Dylan Moreland Jun 20 '12 at 20:36
  • As I recall it, a prime element need to be nonzero. Guess the definition is from Eisenbud, Commutative Algebra, but I'm not shure. Also Wikipedia says so ;). $0$ might generate a prime ideal, and you have this correspondece you mentionend, but 0 is not prime by definition, so you have no correspondence by prime elements without letting zero, and I guess then also every invertible element be a prime. – sebigu Jun 20 '12 at 20:40
  • Also zero is not the generator of the zero ideal, the empty set is a generating set. $0$ is just an element. – sebigu Jun 20 '12 at 20:42
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    @sebigu I don't understand the objection. $0$ is a fine generator for $(0)$, isn't it? It's not like generating sets are unique :) – Dylan Moreland Jun 20 '12 at 20:43
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    @Dylan I read the question as asking if the ideal $\rm: (a + P):$ is prime. – Bill Dubuque Jun 20 '12 at 20:43
  • @DylanMoreland The set ${2,4}$ generates $<2>$, but 4 is not prime either. So you can not call $0$ a prime by this ;). – sebigu Jun 20 '12 at 20:44
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    @BillDubuque Ah, good eye. There are parentheses there. – Dylan Moreland Jun 20 '12 at 20:45
  • @BillDubuque Guess you are right, very good! – sebigu Jun 20 '12 at 20:47
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    @sebigu $\ (0)$ is prime in $\rm:R\iff R:$ is a domain. Indeed $\rm:0:|:ab\iff 0:|:a\ $ or $\rm:0:|:b:$ is equivalent to $\rm:0=ab\iff 0=a:$ or $\rm:0=b.:$ It's the special case $\rm:P=0:$ of $\rm:R/P:$ is a domain iff $\rm:P:$ is prime. $\ $ $\ $ – Bill Dubuque Jun 20 '12 at 20:50
  • @BillDubuque You are right. Also had a look at the book of Eisenbud, and in a Domain $0$ is said to be prime there, too. Thank you very much. – sebigu Jun 20 '12 at 20:58
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    @sebigu In domains, some authors do, for convenience, exclude $,0,$ as a prime element, but most authors do consider $(0)$ to be a prime ideal. It boils down to convention, and authors often choose the convention that proves most convenient in their context. – Bill Dubuque Jun 20 '12 at 21:13