Let $K$ be a field and let $S$ be the set of the irreducible polynomials in $K[x]$. Now let $A$ be the ring of the polynomials over $K$ in the variables $x_f$, one for every $f\in S$; denote with $I$ the ideal generated in $A$ by all the $f(x_f)$, for $f\in S$. How do I prove that $I$ is proper?
I should prove that there aren't $y_f\in A$, $f\in S$, such that $1=\sum_{f\in S}y_ff(x_f)$. It doesn't seem immediate to see from here, so maybe I should prove that $A/I$ is non-zero. I don't know if I can proceed by induction proving that $L[x_f]/f(x_f)$ is an algebraic extension of $L$ for any
$K\subseteq L$ algebraic, because $S$ is not necessarily countable. Any suggestion?
Best Answer
If $f\in K[x]$ is an irreducible polynomial then there is a field $K\subseteq E$ such that $f$ has a root in $E$. Just take $E=K[x]/(f)$, this is a field because the ideal $(f)\subseteq K[x]$ is maximal, and the element $x+(f)$ is a root. By induction it follows that if $f_1,...,f_r$ is a finite number of irreducible polynomials in $K[x]$ then there is an extension field $E$ where all these polynomials have a root.
So now, suppose we indeed have an equation of the form:
$1=\sum_{i=1}^r y_i f_i(x_{f_i}) \ \ \ \ (*)$
Where $y_i\in A$ and $f_i\in S$. Take an extension field $E$ of $K$ where all the polynomials $f_1, f_2,...,f_r$ (which are polynomials in one variable) have roots, say, $\alpha_1,\alpha_2,...,\alpha_r$. Now we can think of $(*)$ as an equation of polynomials over $E$. And now substitute $x_{f_1}=\alpha_1, x_{f_2}=\alpha_2,...,x_{f_r}=\alpha_r$ and $x_f=0$ for all other $f\in S$ into $(*)$. We obtain $1=0$, a contradiction.