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At this question I asked about specific one...

But I think that I don't understand the basic:

If I have an ideal $I$, what avoid it to be principal?

I think that I really need a good example of some ideal $I$ that is not(!!) principal.

At all the books and the examples I saw that $\left<m\right>=m\cdot \mathbb{Z}$ is a principal ideal, but I look for an opposite example, i.e. an ideal that is not principal and why this ideal is not principal...

I open another Q, because I really want to understand it but not via my example...

Thank you!

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CS1
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2 Answers2

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You will not find an example in $\mathbb{Z}$, since all of its ideals are principal.

Since you don't like $\mathbb{Z}[X]$, consider the other simple example, $k[X,Y]$, where $k$ is any field.

Take the set of elements with zero constant term. This is an ideal generated by $X$ and $Y$. But it's not principal. Why? Because if it were generated by a polynomial $p(X,Y)$, then $p(X,Y)$ would have to be a factor of $X$, and also a factor of $Y$. The only such polynomials are the non-zero constants, which don't belong to our ideal.

Intuitively, we don't expect the ideal generated by $X$ and $Y$ to be principal, because $X$ and $Y$ are abstract symbols with no relation between them.

Andrew Dudzik
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  • What is $k[X,Y]$? – CS1 Jan 07 '14 at 10:29
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    @YoavFridman The polynomial ring in two variables over $k$. $k$ could be $\mathbb{Q}$ or $\mathbb{R}$ or $\mathbb{C}$ or anything; it doesn't really matter. – Andrew Dudzik Jan 07 '14 at 10:31
  • Oh, we didn't learn it yet...:-). B.T.W, I'm ok with $\mathbb{Z}[X]$ LOL, I just want general example... – CS1 Jan 07 '14 at 10:34
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    @YoavFridman This is the most general example there is, since we've just taken a general ring with two abstract symbols and no relationship between them. Any other example will look pretty much like this—for instance, in your $\mathbb{Z}[X]$ example, we have $Y=2$... – Andrew Dudzik Jan 07 '14 at 10:36
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    @YoavFridman Also, you know about polynomial rings in two variables as long as you know about polynomial rings in one variable, since $k[X,Y] = (k[X])[Y] = (k[Y])[X]$. – Andrew Dudzik Jan 07 '14 at 10:38
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Consider the ideal $(x,y) \subset k[x,y]$.

By definition an ideal is principal iff there's an element $a \in I$ such that $I=(a)$

Now, our ideal $(x,y)$ is not principal. Suppose there's a polynomial $P(x,y)$ such that $(x,y)=(P(x,y))$, then there must exist polynomials $R(x,y)$ and $Q(x,y)$ such than $x=P(x,y)R(x,y)$ and $y=P(x,y)Q(x,y)$.

Now we obtain, looking at the degrees of $P$ and $R$ in the variables $x$ and $y$ that:

  • $deg_xP+deg_xR=deg_xx=1$
  • $deg_yP+deg_yR=deg_yx=0$

This implies:

  • $deg_yP=deg_yR=0$, follows that $P$ and $R$ are constant in $y$ and then have the only variable $x$
  • Two ptions about the degree in x: $deg_xP=1$ and $deg_xR=0$ or $deg_xP=0$ and $deg_xR=1$

Now we come to an absurd in a lot of ways, for exaple: If $P$ generates $(x,y)$ and it is a polynomial in the only variable $x$, how can be $y$ be a multiple of $P$?

Sabino Di Trani
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