# Questions about following proof regarding why $\mathbb{Z}[x]$ is not a principal ideal domain

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?

• 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 $x$). Jun 8, 2021 at 19:28

If we assume that $$I = \left$$ 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.

• Thank you for the answer, only one question, why then the part "but xf(x) + 2g(x) has even constant term." Jun 8, 2021 at 19:25
• 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. Jun 8, 2021 at 19:28
• What exactly do we mean by the constant term? Jun 8, 2021 at 19:30
• 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.) Jun 8, 2021 at 19:31
• 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). Jun 8, 2021 at 19:45

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?