Show that $(X,X^2+1)$ generate $A[X]$ as $A[X]$-module but is not free.

Question

Let $A$ a commutative ring. Show that $(X,X^2+1)$ generate $A[X]$ as $A[X]$-module but is not free.

For the "generating" I tried by induction : if $a\in A$, then $$a=(X^2+1)-aX(X).$$ Then suppose it's true for degree $n$. Let $a_{n+1}X^{n+1}+a_nX^n+...+a_1X+a_0\in A[X].$ Since $a_{n+1}X^{n+1}=a_{n+1}X^{n}X\in (X,X^2+1)$ and that by induction $a_nX^n+...+a_1X+a_0\in (X,X^2+1)$, we have the claim.

Q1) Does it work ?

For the fact that it's not free, we have that $$X(X^2+1)-(X^2+1)X=0$$ and thus, it's not free.

Q2) Does it work for that ?

Q3) Don't have simply that $X$ generate $A[X]$ as $A[X]$-module ? And thus it's free since it's a basis ?

Answer 1

It's simpler to note that $(X^2+1) - X\cdot X = 1 \in (X, X^2 +1)$.

Yes, the fact that a linear combination of elements, $X(X^2+1) - (X^2+1)X$, is zero is sufficient to show that your set isn't a free generating set.

If it were the case that $X$ generated $A[X]$ as an $A[X]$-module we could write $1 = X\cdot f(X)$ for some polynomial $f$. Do you think this is possible?

Answer 2

Q1) If too complicated, in my opinion: $A[X]$ is an $A[X]$-module, with basis $1$ (this is true for any commutative ring). So, to prove $(X, X^2+1)$ fenerate $A[X]$, it is enough to prove you can obtain the basis. Indeed, $$1=X^2+1-X\cdot X.$$

Q2) is fine.

Q3) $X$ does not generate $A[X]$ since you can't obtain $1$ as a multiple of $X$ ($\deg(F\,X=\deg F+1\ge 1$).

Added: More generally, in any commutative ring, an ideal $I$ is free if and only if it is principal with a non-zero divisor as a generator.