0

In examples 1 and 2 p. 300 of Dummit's book it is written that the ideal $(2, x)$ is not principal in the ring $\mathbb{Z} [x]$ but it is principal in the ring $\mathbb{Q} [x]$.

To understand this by example I can see that no single element of $\mathbb{Z} [x]$ can generate both of $2x^2$ and $2x^3$ but what single element of $\mathbb{Q} [x]$ can generate both of $2x^2$ and $2x^3$?

Bill Dubuque
  • 282,220

2 Answers2

1

I don't know exactly what's confusing you, since whether you are working in $\mathbb Z[x]$ or $\mathbb Q[x]$, you have $(2x^3) \subseteq (2x^2)$ (because $2x^3 = x \cdot 2x^2$). So for instance, $x^2$ generates $(2x^2)$ and $(2x^3)$ since $2x^2 = 2 \cdot x^2$ and $2x^3 = 2x \cdot x^2$ (in both rings).

Hope that helps,

Patrick Da Silva
  • 42,548
  • well yeah, but the ideal $(2,x)$ is certainly not principal in $\mathbb Z[x]$ – Asinomás Aug 28 '16 at 13:38
  • 1
    So? Why is that an issue? – Patrick Da Silva Aug 28 '16 at 13:38
  • I didn't say it was an issue, but the first paragraph talks about principality. Every subset of a ring is generated by $1$. – Asinomás Aug 28 '16 at 13:41
  • @CarryonSmiling : the only ideal that $1$ generates is the proper ideal. Do not confuse "containment" with "equality". For instance, $2x^2 \in (x)$, but $x$ does not generate $(2x^2)$, it generates $(x)$. It is true though that $2x^2$ lies in the ideal $(x)$, which is the notion that you seem to call "generate". We say that a subset $S$ generates the ideal $I$ when $(S) = I$, not when $(S) \supseteq I$. The latter is a useful notion of course but it is not named as you say. – Patrick Da Silva Aug 28 '16 at 13:43
  • (2,x) in Z[x] means all xr+2s such that r and s are in Z[x]. By that, is there any example of a polynomial that shows that (2,x) is not principle in Z[x]? –  Aug 28 '16 at 13:47
  • @CarryonSmiling : I confused you with the OP for a second, sorry if I sounded condescendant. – Patrick Da Silva Aug 28 '16 at 13:58
  • @PatrickDaSilva not at all, don't worry about it. – Asinomás Aug 28 '16 at 13:59
  • 1
    @L.G. : If you think about it long enough, you will see that as a set, the ideal $(2,x)$ is the set of polynomials in $\mathbb Z[x]$ whose constant term is even, i.e. the set of $f(x) = \sum_{i=0}^n a_i x^i$ such that $2$ divides $a_0$. (Hint : compute modulo that ideal and see what conditions it imposes on the polynomial to lie in $(2,x)$, i.e. to be zero modulo $(2,x)$. ) So $\mathbb Z[x] \setminus (2,x) = 1 + 2 \mathbb Z$, the set of odd integers. – Patrick Da Silva Aug 28 '16 at 13:59
0

The ideal $(2,x)$ in $\mathbb Q[x]$ is $\mathbb Q[x]$. Why? because $2$ is invertible.

Asinomás
  • 107,565
  • So $\dfrac12 \in (2,x)$? If so, how $\dfrac13 \in (2,x)$? –  Aug 28 '16 at 13:37
  • because $\frac{1}{3}=\frac{1}{2}\times \frac{2}{3}$ – Asinomás Aug 28 '16 at 13:37
  • the ideal $(2,x)$ must contain all the polynomials of the form $2P$, with $P\in \mathbb Q[x]$. but this is all of the polynomials! because $ \frac{P}{2}\in \mathbb Q[x]$ always! – Asinomás Aug 28 '16 at 13:40
  • Is there any example of a polynomial that shows that (2,x) is not principle in Z[x]? –  Aug 28 '16 at 13:43
  • no there isn't, but it is easy to argument $(2,x)$ is not principle. If a polynomial $P$ generated $(2,x)$ it would have to divide $2$. So the only options are $P=-2,-1,1,2$. And none works. – Asinomás Aug 28 '16 at 13:50
  • (2,x) in Z[x] means all xr+2s such that r and s are in Z[x]. s can be 0 also why it needs to always divide 2? –  Aug 28 '16 at 13:55
  • 1
    becuse you want $P$ to generate $(2,x)$. So the polynomial $2$ must be of the form $PQ$ for some $Q\in \mathbb Z[x]$ – Asinomás Aug 28 '16 at 13:57
  • I got it! But still I can't find any $P\in \mathbb Q(x)$ such that $(P)=(2,x)$? –  Aug 28 '16 at 14:04
  • 1
    the polynomial $1$ works. Because $(2,x)=\mathbb Q[x]$. (an invertible polynomial automatically generates the whole ring, and $2$ is invertible in $\mathbb Q[x]$) – Asinomás Aug 28 '16 at 14:05