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Problem Let $f(x)$ be a non-constant element in $\mathbb{Z}[x]$ . Prove that $\langle f(x) \rangle$ is not maximal in $\mathbb{Z}[x]$

Attempt $\langle f(x) \rangle \subset \langle 1 \rangle \subset \mathbb{Z}$ . Is this wrong?

Scientifica
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Bluey
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  • I edited your post just to clean up the $\LaTeX$ a little. Remember your "$" signs! Cheers! – Robert Lewis Oct 06 '18 at 06:10
  • So, can you name a non-constant element of $\mathbb{Z}[x]$ that could be the $f(x)$ in your attempt? In particular, can you exhibit a non-constant $f(x)$ satisfying $f(x) \in \langle 1 \rangle$? – Eric Towers Oct 06 '18 at 06:13
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    @RobertLewis Thanks a lot. – Bluey Oct 06 '18 at 06:15
  • @blueboy: my pleasure, sir. By the way, did you take your handle from the famous painting? – Robert Lewis Oct 06 '18 at 06:16
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    all the prime ideals and maximal ideals of Z[x] are sure https://math.stackexchange.com/q/174595/453628 – Jian Oct 06 '18 at 13:32
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    @RobertLewis sorry i don't know, it was set at random. Anyways my knowledge in this field (Art) is abysmal. Only painters i have heard of are Picasso and Van Gogh. It is sad. :) – Bluey Oct 06 '18 at 17:31

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Consider the leading coefficient of $f$. If $(f(x))$ is maximal, then $\Bbb{Z}[x]/(f(x))$ is a field $K$. Since no constant element is in $(f(x))$, $K$ has characteristic $0$, i.e., $p\ne 0$ in $K$ for all primes $p=1+\cdots +1$ ($p$ times). In $K$ each such $p$ must be invertible. Consider $2$. Multiplying by $2^{-1}$ in $K$ shows that to any element $g+(f)$ in $K$ there corresponds an $h+(f) = 2^{-1}g+(f)$, i.e., for any $g$ there is an $h$ such that $f$ divides $2h-g$. In particular, there is an $h$ such that $f$ divides $2h-1$. Since $f$ is not constant, this means its leading coefficient is divisible by $2$. (Do the long division, and you'll see this.) Repeating this argument for every prime, the leading coefficient of $f$ is divisible by every prime, a contradiction.

C Monsour
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