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I am wondering if anyone can validate my proof for this problem I am working on.

Question: Let $f(x)$ be a non-constant polynomial in $\mathbb Z[x].$ Prove that $\langle f(x)\rangle$ is not maximal in $\mathbb Z[x]$.

Clearly $\frac{\mathbb Z[x]}{\langle f(x)\rangle}$ is a field iff $\langle f(x)\rangle$ is a maximal ideal.

Proof: $\langle f(x)\rangle = \{g(x)f(x) \mid g(x) \in \mathbb Z[x]\}$. Thus $\langle f(x)\rangle$ is just all non-constant polynomials.

Let $a \in \frac{\mathbb Z[x]}{\langle f(x)\rangle} = \{ a_0 + \langle f(x)\rangle \mid a_0 \in \mathbb Z\}$. Thus $a_0$ is a constant and is not 'absorbed' by $\langle f(x)\rangle$.

Does there exists an element $b \in \frac{\mathbb Z[x]}{\langle f(x)\rangle} = \{ b_0 + \langle f(x) \rangle \mid b_0 \in \mathbb Z[x]\}$ s.t. $ab = 1 + \langle f(x)\rangle$ (the identity).

But $\langle f(x)\rangle$ absorbs all non-constant polynomials, thus $a_0$, $b_0$ must be non-constant, and $a_0 b_0 \in \mathbb Z$.

Clearly, not all elements of $\mathbb Z$ have multiplicative inverses, and there does not exist elements

$a = \{a_0 + \langle f(x)\rangle \mid a_0 \in \mathbb Z\}$, $b = \{b_0 + \langle f(x) \rangle \mid b_0 \in \mathbb Z[x]\}$ where $a_0 b_0 = 1 + \langle f(x)\rangle$.

Therefore, not all elements of $\frac{\mathbb Z[x]}{<f(x)>}$ are units, $\frac{\mathbb Z[x]}{\langle f(x)\rangle}$ is not a field, $\langle f(x)\rangle$ is not a maximal ideal of $\mathbb Z[x]$.

user26857
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Stack
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    No, no, $f(x)$ is one particular non-constant polynomial. So the ideal it generates is not all non-constant polynomials. Back to the drawing board. – Ted Shifrin Oct 08 '16 at 23:28
  • The claim is true, and you indeed have to show that $\mathbb{Z} [X] / (f (X))$ is not a field. However, $(f (X))$ does not consist of all non-constant polynomials. –  Oct 08 '16 at 23:29
  • At first I felt tempted to remark that writing $<f(x)>$ instead of $\langle f(x) \rangle$ is a frequent typesetting error here, and it's disturbingly seemingly impossible to eradicate. But that's only one of the problems: the list of corrections I did in copy-editing was quite extensive. In particular, notice that in an expression like $$\langle f(x)\rangle = {g(x) f(x)\mid g(x)\in Z[x]},$$you should not keep alternating in and out of MathJax, but you should remain within MathJax until the end. Not doing that results in lack of proper spacing and alignment and mismatches in$,\ldots\quad$ – Michael Hardy Oct 08 '16 at 23:31
  • $\ldots,$font sizes, as when you write $a$ + $b$ instead of $a+b. \qquad$ – Michael Hardy Oct 08 '16 at 23:33
  • I am not sure then what $\langle f(x)\rangle$ is, but I know that the ideal cannot generate a constant polynomial, is that correct? – Stack Oct 08 '16 at 23:53
  • @Stack Let's say that if $;f(x);$ is any polynomial of positive degree, then the only constant polynomial in it is the zero polynomial. – DonAntonio Oct 09 '16 at 00:11

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Hint: if $R = \Bbb{Z}[x]/(f(x))$ were a field, then for each positive integer $n$ there would be an element $t_n \in R$ such that $nt_n = 1$ (because $n$ must be invertible in $R$). So for every positive integer $n$, there would be $h_n(x) \in \Bbb{Z}[x]$ representing $t_n \in R$ and $g_n(x) \in \Bbb{Z}[x]$ such that $nh_n(x) = 1 + g_n(x)f(x)$. Is that possible?

user26857
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Rob Arthan
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