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In the ring $\mathbb{Z}[x]$, I want to prove that the principal ideal $I=(x-2)$ is not maximal.

Surely I just need to find an ideal $J$ that contains $I$ plus some other elements of $\mathbb{Z}[x]$. My instinct is to just say $J = (x-2) + (x)$ (not confident this is correct), or any other principal ideal that is not $(x-2)$ for that matter. However, I'm not sure how I would prove $I$ is contained in $J$ without them being equal even if I am correct.

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  • $\begingroup$ Your $J$ here actually works. Indeed, we have $\langle x-2 \rangle \subset J \subset \mathbb{Z}[x]$. To prove both of these containments are proper, notice $I$ cannot contain the polynomial $x$ (which $J$ does), and $J$ cannot contain $1$ (which $\mathbb{Z}[x]$ does). $\endgroup$ – Kaj Hansen Nov 3 '16 at 4:52
  • $\begingroup$ @Kaj Hansen Hmm, interesting. Perhaps I'm being particularly naive here, but it seems for most principal ideals I we can simply find another principle ideal, and add it to I to form a new ideal J which contains I and at least a few other elements. Is there any clear principal ideals which are maximal? It doesn't seem totally clear to me that there is. $\endgroup$ – SenoritaChad Nov 3 '16 at 5:43
  • $\begingroup$ Not in $\mathbb{Z}[x]$ there aren't. The only ideals in this ring that are maximal can be written in the form $\langle p, f(x) \rangle$, where $p$ is prime and $f$ is irreducible modulo $p$. $\endgroup$ – Kaj Hansen Nov 3 '16 at 5:45
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Hint:

Notice that $f(x)(x-2)$ cannot be a nonzero constant polynomial for any $f(x) \in \mathbb{Z}[x]$. Given this, can you think of an ideal that properly contains $\langle x-2 \rangle$ that is also a proper subset of $\mathbb{Z}[x]$? It'll be easiest to construct this new ideal in terms of a set of generators (which will include $x-2$ of course), and note that there may be more than one generator as $\mathbb{Z}[x]$ is not a principal ideal domain.


In addition to explicitly constructing an ideal that properly contains $\langle x-2 \rangle$, there is another approach. One can show that an ideal $I \subset R$ is maximal $\iff R/I$ is a field. I claim that $\mathbb{Z}[x] / \langle x-2 \rangle \cong \mathbb{Z}$ (not a field!). To prove this, find a surjective homomorphism $\phi:\mathbb{Z}[x] \rightarrow \mathbb{Z}$ with the appropriate kernel. The result would follow from the isomorphism theorem.

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  • $\begingroup$ If I simply take the homomorphism ϕ(f(x)) = f(2), that will yield my ideal as kerϕ, I'm unsure how to connect that to proving my ideal is not maximal? $\endgroup$ – SenoritaChad Nov 3 '16 at 3:28
  • $\begingroup$ To answer your second question: that's exactly right. The reason this is sufficient is due to this result: proofwiki.org/wiki/Maximal_Ideal_iff_Quotient_Ring_is_Field $\endgroup$ – Kaj Hansen Nov 3 '16 at 3:40
  • $\begingroup$ Sorry not sure I follow. So we have the quotient ring is isomorphic to the integers, but this theorem states J is maximal if and only if R/J is a field, which can't be true since the integers are not a field? $\endgroup$ – SenoritaChad Nov 3 '16 at 3:43
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    $\begingroup$ Glad I could help! $\endgroup$ – Kaj Hansen Nov 3 '16 at 3:54
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    $\begingroup$ I made a small error in a comment @SenoritaChad. To get a proper ideal containing $\langle x-2 \rangle$, you can add to the ideal all the $\mathbb{Z}[x]$-multiples of any set of constant polynomials so long as $\gcd(x, y) \neq 1$ for every pair. This is necessary due to Bezout's identity. Otherwise, we'd have $1$ as an element of the ideal, and thus the whole ring as well, meaning the ideal we're trying to construct is no longer proper. $\endgroup$ – Kaj Hansen Nov 3 '16 at 4:09
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The fact that $\mathbb{Z}$ is not a field is key here. For example $(x-2,2)$ is a proper ideal containing yours.

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    $\begingroup$ No, @KajHansen, every polynomial in that ideal has constant term divisible by $2$. $\endgroup$ – Lubin Nov 3 '16 at 3:56
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    $\begingroup$ @KajHansen How do you get $1=xf +2g$ ? $\endgroup$ – Rene Schipperus Nov 3 '16 at 3:56
  • $\begingroup$ Oops, my mistake; you are correct. $\endgroup$ – Kaj Hansen Nov 3 '16 at 4:00
  • $\begingroup$ Sorry not familiar with the notation, is this, as above, just adding the multiples of 2 as constant polynomials to the ideal (x-2)? $\endgroup$ – SenoritaChad Nov 3 '16 at 4:04
  • $\begingroup$ Yeah. The notation $\langle x, y \rangle$ means the ideal $I = \{rx + sy \ | \ r, s \in R \}$ $\endgroup$ – Kaj Hansen Nov 3 '16 at 4:15

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