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I am not sure how to go about this. Could someone please check my work and see if I am doing something wrong?

If $F$ is a field, show that $F[x]$ is never a field.

Let $p(x)=x\in F[x]$. Then if $F[x]$ is a field, $g(x)=x^{-1}\in F[x]$ such that $p(x)$ has a multiplicative inverse where $p(x)g(x)=xx^{-1}=1\in F[x]$. However, $g(x)$ is not a polynomial, and so $p(x)$ is not unit. Therefore, $F[x]$ cannot be a field.

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    This https://math.stackexchange.com/questions/2514/why-cant-the-polynomial-ring-be-a-field?rq=1 ? – Martin R Apr 08 '18 at 20:18
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    It is basically correct, except you need to make the sentence "$g(x)$ is not a polynomial" more clear. True, your choice is not a polynomial, but why couldn't it be possible that another choice (of a polynomial, this time) also works? – Arnaud Mortier Apr 08 '18 at 20:20

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If $F[x]/(x)\cong F$ is a field and $F[x]$ is a field then $x$ has a multiplicative inverse so $(x)=F[x]$. But this forces $F=\{0\}$, which is not a field.

cansomeonehelpmeout
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  • Thank you for taking the time to reply! Could you explain further that the ideal forces $F=\left {0 \right }$? I don't really understand. –  Apr 08 '18 at 20:38
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    Sure! $(x)\subseteq F[x]$ is an ideal, and $F[x]/(x)=F$ (this holds for any polynomial ring). Now, if $F[x]$ were a field then $x\neq 0$ and so it would have an inverse, but all the ideals in a field is either $(0)$ or $(1)$ (the entire ring/field). This forces us to have $(x)=(1)=F[x]$. Looking at the quotient $F[x]/(x)\cong F[x]/F[x]\cong {0}\cong F$, so $F={0}$. ${0}$ is the "zero-ring", and is by definition not a field, since $0=1$ in ${0}$. (For reference: https://en.wikipedia.org/wiki/Zero_ring) – cansomeonehelpmeout Apr 08 '18 at 20:44
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    Thank you for your clear explanation. It is very useful to me. :) –  Apr 08 '18 at 20:46
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Actually, in principle, it could be the case that $x^{-1}$ equals some polynomial. So, you have to check that given $p(x)=x\in F[x]$, there's no polynomial $q(x)\in F[x]$ such that $p(x)q(x)=1$.

I think you'll be able to do the details using the properties of the product and sum of polynomials with respect to the degree of the polynomials involved.

Javi
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