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I know that if $F[x]$ is a PID then $F$ is a field. Now $F[x]$ being a field implies that $F[x]$ is a PID, so $F$ is a field. Anyway, I tried to prove that $F$ is a field right away and the following argument came to mind:

Let $A[x]$ be a field, and let $p(x)= a + bx$, then there exists $q(x) = d x^n + ... + r$ where $b$ and $d$ are nonzero, such that $p(x)q(x) = 1.$ Carrying out the multiplication and equating the coefficient of $x^{n+1}$ to zero yields $bd = 0.$ Hence b and d are zero divisors and A is not even an integral domain.

What is wrong with this argument? And is $R[x]$ never a field?

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The ring $R[x]$ is never a field, since $x\neq0$ and it has no inverse.

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    $\begingroup$ Minor nitpick: you can have $x=0$, but only in the special case that $R$ is a zero ring. But in that case $R[x]$ is also a zero ring, and thus not a field. $\endgroup$ – user14972 Apr 2 '18 at 23:23
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A field has only the unproper ideals $(0),(1)$, but $(x)\subseteq R[x]$ is a proper ideal. Therefore $R[x]$ cannot be a field.


You could also look at the evaluation homomorphism $$\phi_0:R[x]\rightarrow R$$ which maps $f(x)\mapsto f(0)$. Now suppose $xf(x)=1$ for some $f\in R[x]$. Then $$1=\phi_0(xf(x))=0$$ which only happens if $R=\{0\}$. So, is $R[x]$ a field when $R=\{0\}$? Some would say no, because part of the definition of a field is that $0\neq 1$.

From Wikipedia:

"there exist two different elements $0$ and $1$ (...)"

Relevant links:

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    $\begingroup$ The notion of "field with one element" has absolutely nothing to do with the zero ring. It refers to a hypothetical object of a sort that generalizes the notion of "field", or more generally it relates to the sorts of things we would do with such an object if we knew what it was. $\endgroup$ – user14972 Apr 2 '18 at 23:26
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$R[x]$ is never a field, because, for example, $p(x) = x$ has no inverse. If a $q(x)$ had existed which $p(x)q(x) = 1$ we would have the degree of $p(x)q(x)$ equals to $0$ and this can never happen because $p(x)$ has degree $1$.

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