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How can we show that $x^N - 1$ is an ideal of the ring $\mathbb{Z}[x]$?

I understand that $I$ is an ideal if $fg$ is in I for all $f$ in $\mathbb{Z}[x]$, all $g \in I$. But how do we see this for a value like $x^N - 1$? I feel that I am missing an insight here short of multiplying everything out to see why it is an ideal.

rschwieb
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Ymi
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    $x^N - 1$ is an element of the ring $\mathbb Z [x]$. An ideal is by definition a subset of the ring. Do you mean something else? – Sven-Ole Behrend Nov 22 '23 at 12:58
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    @Sven-OleBehrend Sorry I'm a bit confused. I sometimes see it notated as <x^N - 1>, I presume its the ring generated by the polynomial – Ymi Nov 22 '23 at 13:00
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    It is the other way around. The ideal $(f)$ contains all polynomial $gf$ for $g\in \Bbb Z[x]$. So it is a definition in fact, in your case with $f=x^N-1$, the ideal generated by $f$. First try it with ideals $(a)$ in $\Bbb Z$. – Dietrich Burde Nov 22 '23 at 13:00
  • Duplicate of your question yesterday. Please don't do that - it violates site policy. – Bill Dubuque Nov 22 '23 at 18:12
  • The element $f = x^N-1$ is not an ideal. What you mean is the ideal generated by $f$ in $\Bbb Z[x],,$ i.e. the ideal $(f) := f\Bbb Z[x],,$ see the linked dupe and the comments on your prior question. – Bill Dubuque Nov 22 '23 at 18:17

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Given any subset $S \subseteq R$ of any (commutative) ring, we can define the generated ideal of $S$, denoted by $\langle S\rangle$ or sometimes also $(S)$, to be smallest ideal, which contains $S$. If $S = \{r\}$ contains only a single element we sometimes also write $\langle r\rangle$ instead of $\langle \{r\}\rangle$.

So $\langle x^N - 1\rangle$ is an ideal simply by the definition. If you want to express it as a set, you have $\langle x^N - 1\rangle = \{ f \cdot (x^N - 1) | f \in \mathbb Z [x]\}$. One can show that the right side is an ideal by showing that it is an additive subgroup and closed under multiplication by arbitrary ring elements.

rschwieb
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Sven-Ole Behrend
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  • Does this mean that any arbitrary <f(x)> in Z[x] can be an ideal? Also if S has more than one element, how do we generate the ideal elements? Do we multiply elements of Z[x] separately to elements in S and union them in the ideal together? – Ymi Nov 22 '23 at 13:17
  • Yes, every polynomial $f(x) \in \mathbb Z [x]$ generates an ideal in the exact same way. If $S$ has multiple elements, the description is a little bit more complicated. You first take all multiples of elements of $S$ and consider their union. But now, this is not necessarily closed under addition. So you have to consider finite sums of multiples of elements in $S$, this works. For example, $< 2x^2 , x^3>$ contains $2x^2 + x^3$, which is neither a multple of $2x^2$ nor of $x^3$ (at least in $\mathbb Z[x]$). – Sven-Ole Behrend Nov 22 '23 at 14:21
  • protip: the latex you were looking for was \langle ... \rangle rather than < ... >. Good answer btw – rschwieb Nov 22 '23 at 14:46
  • Please strive not to post more (dupe) answers to dupes of FAQs. This is enforced site policy, see here. – Bill Dubuque Nov 22 '23 at 18:15