# How to prove that ideal is not principal

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.

• @mrtaurho Ideal generated by generator $g(x)$ thus will contain polynomials containing constant terms also? And since $I$ does not contain any polynomials having constant terms, does that make I non-principal ideal? May 26 at 8:37
• That's what I was referring to! You've to be careful when choosing generators to not accidentally get too much. For example, the unit element $1$ "generates" every ideal in any ring by your argument. But it will never really generate any proper ideal. (I re-added my comment below) May 26 at 8:45
• You are correct in so far that the ideal generated by $g(x)$ (or any constant for that matter) contains $I$. But as the constants are units in $R[x]$ the principal ideal generated by $g(x)$ will be the whole ring instead of only $I$ which is a proper subset of $R[x]$. May 26 at 8:46

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 $$fh=ch_0x^2+ch_2x^4+\cdots$$.