# Prove that if $R$ is a commutative ring, then $R[x]$ is never a field [duplicate]

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!!!

• Well, $\:\Bbb Q[\sqrt2]\;$ is a field....you mean $\;x\;$ is an indeterminate (or transcendental) element, don't you? Mar 10, 2019 at 13:53
• 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 $1$ Mar 10, 2019 at 13:54
• 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. Mar 10, 2019 at 13:57
• I think the question is clear because in a field, every element has inverse.
– Minh
Mar 10, 2019 at 13:58