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I saw this question but I still do not understand:

What is the difference between ideal and principal ideal?

At my homework I had to prove to things about $I=\left\{p(x)\in \mathbb{Z}\left[X\right]:2\mid p(0)\right\}$:
1. That $I$ is an ideal. I prove it...
2. That $I$ is not a principal ideal.

Now how can I prove 2.? Why $I$ isn't a principal ideal?

Thank you!

Community
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CS1
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2 Answers2

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Can you show your ideal equals $(2,X)$? Can you show that cannot be principal? Hint If it were, say $(p)=(2,X)$ then we'd have $p\mid 2$ and $p\mid X$.

What is the difference between ideal and principal ideal?

If $R$ is a ring, a (double sided) ideal $I$ of $R$ is a subgroup of the additive group of $R$ that is closed under multiplication by elements of $R$ both on the right and on the left, that is, $RI=\{ri:r\in R,i\in I\}$ and $IR=\{ir:r\in R,i\in I\}$ are both contained in $I$.

A principal ideal is one that is generated by single element of your ring, that is, it consists of all $R$ multiples of a fixed element, for example $2\Bbb Z=\{2n:n\in\Bbb Z\}$ is a principal ideal in $\Bbb Z$.

Pedro
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  • What is $(2,X)$? (I don't this sign...) Thank you! – CS1 Jan 07 '14 at 08:35
  • @YoavFridman The ideal generated by $2$ and $X$. – Pedro Jan 07 '14 at 08:36
  • What makes it principal? I think that I didn't understand the definition... – CS1 Jan 07 '14 at 08:37
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    @YoavFridman You want to show it is not principal, don't you? – Pedro Jan 07 '14 at 08:38
  • I know that $\left<2\right>$ is ideal... I don't understand what is ideal that generated by $2$ and $X$... – CS1 Jan 07 '14 at 08:39
  • @YoavFridman If you know what it means for an ideal to not be principal it's important you know how to notate such ideals. The notation $(a_1,a_2,\cdots)$ means the set of all $R$-coefficient combinations of the elements $a_1,a_2,\cdots$. – anon Jan 07 '14 at 08:40
  • Yes, but what is the definition of principal, I think this is what I don't understand... – CS1 Jan 07 '14 at 08:40
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    @YoavFridman if that is the problem, then you should look up the definition where you got the problem from. – Tobias Kildetoft Jan 07 '14 at 08:41
  • @anon, this is what I don't know/understand: "what it means for an ideal to not be principal".... – CS1 Jan 07 '14 at 08:42
  • @TobiasKildetoft, I saw the definition but I still not understand, this is why I'm asking you.... – CS1 Jan 07 '14 at 08:42
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    @YoavFridman If you read the definition and don't understand it, then you need to formulate precisely what it is about it you don't understand, or there is no way anyone can help you. – Tobias Kildetoft Jan 07 '14 at 08:44
  • @TobiasKildetoft, can you give me a link to place that shows the definition with examples? – CS1 Jan 07 '14 at 08:44
  • @PedroTamaroff, Thank you for your edit, but how can I prove that $I$ is not made by a single element? – CS1 Jan 07 '14 at 08:49
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    @YoavFridman I added something. – Pedro Jan 07 '14 at 08:52
  • @PedroTamaroff, but how can I show it? what is the way to prove that is not principal? – CS1 Jan 07 '14 at 08:53
  • By example? why $(2,X)$ not makes $I$? – CS1 Jan 07 '14 at 08:54
  • @YoavFridman I am saying $(2,X)$ equals $I$. Can you try to be clearer? – Pedro Jan 07 '14 at 08:55
  • @PedroTamaroff, yes, sorry, I try to understand how can I show that $I$ is not principle, you say that $(2,X)$ equals $I$, so why $I$ is not principle?? – CS1 Jan 07 '14 at 08:56
  • @YoavFridman I have given you a hint in the answer. – Pedro Jan 07 '14 at 08:59
  • @PedroTamaroff, I'm confuse - If $(2,X)=I$ so $(2,X)$ makes $I$ so $(2,X)$ is principle, no?? – CS1 Jan 07 '14 at 09:03
  • @YoavFridman Please try to understand what a principal ideal is. – anon Jan 07 '14 at 09:05
  • @anon, I'm try, but still I think I miss something at the definition that is not clear to me... I still can't understand why $(2,X)$ shows as that $I$ is not principal... – CS1 Jan 07 '14 at 09:07
  • @PedroTamaroff, can you add please one more example at your answer, but that example will of something that is not principal? that I will see an opposite example? Thank you! – CS1 Jan 07 '14 at 09:09
  • @TobiasKildetoft, how can I show that something is not a principal ideal? example is enough? Thank you! – CS1 Jan 07 '14 at 09:25
  • @YoavFridman You might want to know that your ideal is also a maximal ideal. – Robert M Jan 07 '14 at 09:43
  • @RobertM, why $I$ is maximal? and how does it helps me to prove that $I$ is not principal? – CS1 Jan 07 '14 at 09:47
  • @YoavFridman You don't need maximality of $I$. That was only an extra information about $I$. Meanwhile $I$ is maximal because when you divide $\Bbb{Z}[x]$ by $I$ you get $\Bbb{Z}_2$ which is a field. But if you are not familiar with this concept just ignore my comments. – Robert M Jan 07 '14 at 09:50
  • @RobertM, oh, thank you! I know what is maximal ideal, but I didn't understand how does it helps me to prove is not principal... sorry. is because I really want to understand how to prove this... – CS1 Jan 07 '14 at 09:54
  • @Yoav: ${\bf Z}[X]$ is two-dimensional (since $\bf Z$ is one-dimensional and taking polynomials adds a dimension). Maximal ideals correspond to points, hence have dimension zero. Principal ideals on the other are generated by a single element, hence correspond to one-dimensional curves. This is an intuition for why a maximal ideal is not principal (it would be a full proof in ${\bf C}[X, Y]$, say, since maximal ideals correspond to points in $\bf C$ and principal ideals to curves). But these are just aside comments, you have complete answer given, so try to understand it before asking more. – Marek Jan 07 '14 at 11:18
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This would be a comment but it is too long for that. I am going to show that $ I= (2,x)$ is not principal. Suppose to the contrary that $I$ is principal so there would be $f \in \Bbb{Z}[x]$ such that $I = (f)$. Since $2 \in I$ we have $2 \in (f)$ and hence there is a polynomial $g$ such that $2 = fg$. This means that $f$ is a constant polynomial lets say $m$. But on the other hand $x\in I$ and hence $x \in (f)$ therefore there is a polynomial $h$ such that $x = mh$ if we compare the coefficients of $x$ we get that $m = 1$ which means that $(f)$ (and hence $I$) is equal the whole ring $\Bbb{Z}[x]$. A contradiction. Pleaes let me know if you need some more explanation.

Robert M
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  • You explain very well. If there be questions I'll ask you! Thank you! – CS1 Jan 07 '14 at 10:16
  • So, if I'm understand it right, if there is element at $I$ that makes all the ring is shows that $I$ is not principal, I'm right? – CS1 Jan 07 '14 at 10:31