1

I am confused as to how to deal with basic polynomials in the sense of principal ideals.

For example, let $I=\lbrace p(x) \in \mathbb{Z}[X] : p(0)=7j \text{ for some } j\in \mathbb{Z} \rbrace$ be an ideal of $\mathbb{Z}[X]$ (the ring of all polynomials with integer coefficients). I am wanting to show that $I$ is not a principal ideal of $\mathbb{Z}[X]$. We know that $x+7\in I$ and $x+14\in I$, and also that these elements are not multiples of each other. What does not being multiples of each other mean in this context? In other words, how does this show that $I$ is not principal ideal of $\mathbb{Z}[X]$? Any help would be appreciated.

W. G.
  • 1,766
  • 2
    If $I=(f)$ then $f$ is a factor of both $x+7$ and $x+14$. – Angina Seng Apr 26 '18 at 18:17
  • @LordSharktheUnknown So, this means that $f g=x+7$ and $f h=x+14$ for some $g, h\in I$. But I am still stuck. I have watched videos/looked through other problems which explain how similar problems are a contradiction right at this step, and this is the part (the one right after this step) is what I am confused on. – W. G. Apr 26 '18 at 18:27
  • Like how does that tell you these aren't multiples of each other if that makes sense. I appreciate the help. – W. G. Apr 26 '18 at 18:31
  • 1
    Can you work out which elements of $\Bbb Z[x]$ are factors of, say, $x+7$? – Angina Seng Apr 26 '18 at 18:34
  • I don't know what the factors are. I really have been thinking about it. If it was just $x+7$ for $f$, then $g$ would be $1$, so it can't be that. – W. G. Apr 26 '18 at 18:58
  • Looking at the $x+7$, we know $f=...+c_1x+c_27$ and $g=...+c_3x+c_47$ where all the constants are integers. This means $fg=...+c_2c_4 49$. So, $c_2c_449=7$ which means the product of two integers is a fraction which is a contradiction. Thus, $x+7$ has no factors in $\mathbb{Z}[x]$ right? How does the $fh$ come in to play? I am still confused. – W. G. Apr 26 '18 at 19:21
  • You say $x+7$ has no factor in $\Bbb Z[x]$. Surely $x+7$ is a factor of $x+7$? – Angina Seng Apr 26 '18 at 19:35
  • Okay $x+7$ is a factor. It is the only one. – W. G. Apr 26 '18 at 19:39
  • To be honest, I don't know why it is a factor in the sense of it being a generator in the context to the ideal. I get that $x+7$ is a factor of $x+7$, same reason $x+2$ is a factor of $x^2-4$. But that is besides the point. – W. G. Apr 26 '18 at 19:45
  • So $x+7$ is the only factor of $x+7$? Then $1$ isn't a factor of $x+7$? – Angina Seng Apr 26 '18 at 19:52
  • Besides $1$. But $1\notin I$. – W. G. Apr 26 '18 at 20:03
  • Actually, I read this wrong. Factors of $\mathbb{Z}[X]$ for $x+7$ are $\pm 1, \pm (x+7)$; I was looking inside the ideal for factors. My apologies. – W. G. Apr 26 '18 at 22:32
  • @LordSharktheUnknown I got it! Thank you for your help!!! So, $f=\pm 1$ as the $\lbrace \pm 1, \pm (x+7)\rbrace \cap \lbrace \pm 1, \pm (x+14)\rbrace= \lbrace \pm 1 \rbrace$. Yet, $\pm 1 \notin I$. Thus, $I$ is not a principal ideal of $\mathbb{Z}[X]$. – W. G. Apr 26 '18 at 22:43

0 Answers0