Prove that ideal is not principal

abstract-algebraidealsring-theory

Let $F$ be a field and $R$ be the subring of polynomials $F[x]$ such that coefficient of $x$ is zero. Let $I$ be the ideal of $R$ such that constant term is zero.
I have to prove that $I$ is not a principal ideal of $R$.

To show that I will have to prove that every element of $I$ is not generated by a single element of $R$. But to me it seems that every element of $I$, say, $f(x)= a_2x^2+a_3x^3+\dots+a_nx^n$ can be generated by an element of $R$, say, $g(x)=b_0+b_2x^2+\dots +b_mx^m$ by taking $b_2=b_3=\dots=b_n=0$, i.e.

$f(x)=b_0(\alpha_2x^2+\alpha_3x^3+\dots+\alpha_nx^n)$

where $b_0=g(x)$ and $\alpha_2x^2+\alpha_3x^3+\dots+\alpha_nx^n$ belongs to $R$. Thus it seems to me a principal ideal generated by $g(x)$. So what am I missing here? Please help.

Best Answer

Hint: look at $x^2$ and $x^3$. Can you find a non-unit $f\in R$ with $g,h\in R$ so that $fg=x^2$ and $fh=x^3$? There's a solution under the spoiler, but give yourself a chance before looking at it, please.

No - by examining the factorization $fg=x^2$ in $F[x]$, we find $f=cx^2$, but then writing $h=h_0+h_2x^2+\cdots$, we have that $fh=ch_0x^2+ch_2x^4+\cdots$.

Related Question