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Prove that $\Bbb{Z[x]}$ is not a principal ideal domain.

Proof: Consider the ideal $I=\langle x,2\rangle$. We’ll show that this ideal is not principal. First note that I is not equal to $\Bbb{Z[x]}$ because 1 is not in I. If it were then $1=xf(x)+2g(x)$ hx, 2i because if it were then $1 = xf(x) + 2g(x)$ for f(x), g(x) ∈ Z[x], but xf(x) + 2g(x) has even constant term. Then suppose $I=\langle p(x) \rangle$ for some p(x) ∈ Z[x]. Then we must have x = p(x)f(x) and 2 = p(x)g(x) for some f(x), g(x) ∈ Z[x]. But the second implies that p(x) must be a constant polynomial, specifically p(x) = −2, −1, 1 or 2. We can’t have p(x) = ±1 because then I = Z[x] so p(x) = ±2. But then x = ±2f(x), a contradiction since ±2f(x) has even coefficients.

I don't understand this part But then x = ±2f(x), a contradiction since ±2f(x) has even coefficients. What is the thing about the even coefficients, and why xf(x) + 2g(x) has even constant term in the first part? How do they lead us to the contradiction?

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  • $\begingroup$ The constant term of $xf(x)+2g(x)$ is twice the constant term of $g$ (note that $xf(x)$ cannot have a constant term, because it is a multiple of $x$). $\endgroup$ Jun 8, 2021 at 19:28

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If we assume that $I = \left<p(x)\right>$ for some $p(x) \in \Bbb{Z}[x]$, that means that we can write $\textit{any}$ element of $I$ as a product of $p(x)$ and another element of $\Bbb{Z}[x]$. In particular, we then could write the two polynomials $x$ and $2$ in this way, i.e.,

$$x = f(x)p(x)$$ $$2 = g(x)p(x)$$

for some $f(x),g(x) \in \Bbb{Z}[x]$. Let us first analyze the second equality. $2 = g(x)p(x)$ implies that $p(x)$ must be a constant polynomial, since otherwise the degree would not match. Further, we can also see that (depending on what $g(x)$ would be) $p(x)$ could in any case only take the values $\pm1, \pm2$.

We already know that it cant be $\pm1$ since then we had $I = \Bbb{Z}[x]$, which you have already excluded. So let us now check the first of the above equations when we assume that $p(x) = \pm2$. If that is the case, then $f(x)p(x)$ for sure has even coefficients, no matter what $f(x)$ is. But that's a problem, since we have that $x = f(x)p(x)$, and $x$ doesn't have even coefficients! Here is where the contradiction arises.

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  • $\begingroup$ Thank you for the answer, only one question, why then the part "but xf(x) + 2g(x) has even constant term." $\endgroup$
    – Annalisa
    Jun 8, 2021 at 19:25
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    $\begingroup$ Sure! That refers to the fact that we can't have $1 = xf(x) + 2g(x)$. When we look at the right hand side, we see that the constant term is even for sure: It could be 0 (if the constant term of $g(x)$ is 0), or $2c$ where $c$ is the constant term of $g(x)$. So in any case, it will never be 1, so this equation can't hold. $\endgroup$ Jun 8, 2021 at 19:28
  • $\begingroup$ What exactly do we mean by the constant term? $\endgroup$
    – Annalisa
    Jun 8, 2021 at 19:30
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    $\begingroup$ Example: Consider the polynomial $3x^2 + 2x +5$, then the constant term is 5. Does this answer your question? (The other terms also have "nicknames", $2x$ could be called the "linear term", $3x^2$ could be called the "quadratic term" etc.) $\endgroup$ Jun 8, 2021 at 19:31
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    $\begingroup$ If we have $2 = g(x)p(x)$, we know that $p(x)$ must be just a constant term, otherwise we would have an $x$ also on the left hand side. It's not 0 of course, and if we had that $|p(x)| > 2$ then the left hand side could not be 2, it would be greater than 2. (Or $0$ if $g(x) = 0$, but that's of course also impossible). $\endgroup$ Jun 8, 2021 at 19:45
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The proof shows that $x=\pm 2f(x)$. But no matter what $f(x)$ is, after multiplying by $\pm 2$, all of the coefficients of $f(x)$ will be even (as a multiple of $2$). But the coefficient of $x$ is $1$, which is odd. So if $x=\pm 2f(x)$, then $1$ is even, a contradiction. Does this make sense?

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