An exercise about field automorphisms and ideals.

Question

Consider a field $K$ and the $K$-algebra $K[x_1,\ldots,x_n]$ of polynomials in $n$ variables; $\mathfrak a$ is an ideal of $K[x_1,\ldots,x_n]$ and suppose that there exists a field $L\subseteq K$ with the following properties:

1. $\mathfrak a$ has basis in $L$, namely $\mathfrak a$ is generated by a set of polyomials $\{f_0,\ldots,f_m\}$ all with coefficients in $L\subset K$.
2. If $F$ is another subfield of $K$ such that $\mathfrak a$ has basis in $F$, then $L\subseteq F$

I have to prove the following proposition:

If $\sigma\in\text{Aut} K$ fixes pointwise $L$, then $\sigma(\mathfrak a)=\mathfrak a$

Do you have any hint for the solution?

Answer 1

Hint:

Figure out what $\sigma$ applied to a polynomial means and consider $\sigma(f_i)$ if all coefficients of $f_i$ are fixed by $\sigma$.