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Problem: Prove that if $R$ is a commutative ring, then $R[x]$ is never a field.

My attempt: We prove that by contradiction. Suppose $R[x]$ is a field, then we have $a_0 + a_1x + \dots +a_nx^n \in R[x]$, then $x(a_0 + a_1x + \dots +a_nx^n) = 1$. So we have $a_0x + a_1x^2 + a_2x^3 + \dots +a_nx^{n+1} = 1 \Rightarrow 0 + a_0x + a_1x^2 + a_2x^3 + \dots +a_nx^{n+1} = 1 + 0x + 0x^2 + \dots +0x^{n+1}$. Identities coefficients we have $a_n=0, \dots, a_2=0,a_1=0,0=1$, which means $R$ is a trivial ring $\Rightarrow$ contradiction. Q.E.D

Is that right?? Thank all!!!

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  • $\begingroup$ Well, $\:\Bbb Q[\sqrt2]\;$ is a field....you mean $\;x\;$ is an indeterminate (or transcendental) element, don't you? $\endgroup$
    – DonAntonio
    Mar 10, 2019 at 13:53
  • $\begingroup$ The idea is correct, but you should mention what you do : You show that $x$ cannot have a multiplicate inverse. To show this you only need to consider the degrees of the polynomials and that the leading cofficient of $x$ is $1$ $\endgroup$
    – Peter
    Mar 10, 2019 at 13:54
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    $\begingroup$ You should be more precise. It seems that you argue that if $R[x]$ is a field, then $x$ must have an inverse which has the form $a_0 + a_1x + \dots +a_nx^n$. But this point is not really clear in your question. $\endgroup$
    – Paul Frost
    Mar 10, 2019 at 13:57
  • $\begingroup$ I think the question is clear because in a field, every element has inverse. $\endgroup$
    – Minh
    Mar 10, 2019 at 13:58

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