0

Let $R_1$ be the ring of polynomials (over $\mathbb{C}$) and $R_2$ a finite free module over $R_1$ (generated by elements $b_1,\ldots,b_m$). Now I consider an ideal $I\subset R_2$ and the ideal $J=I\cap R_1$ and consider the corresponding quotients $Q_1=R_1/J$ and $Q_2=R_2/ I$. Is $Q_2$ now a free $Q_1$ module?

J. W. Tanner
  • 60,406
user382144
  • 145

1 Answers1

1

Consider $R_2 = \mathbb{C}[x, y]/(y^2) = \mathbb{C}[x, y], y^2 = 0$. Then let $I = (xy)$ so $J = (0)$. Then $Q_2$ cannot be free because $y$ would be a torsion element on $Q_2$ over $Q_1$.

On the other hand, by results on modules over PIDs (see Theorem 6.7 of Roman's Advanced Linear Algebra), if $I$ is an ideal in $R_2$, it is a free submodule and we can choose its basis s.t. $b_1, \dots, b_n$ is a basis for $R_2$ over $R_1$ and $\{c_1b_1, \dots, c_m b_m \}$ is a basis for $I$. Now, if $I$ is prime, either $c_i$ or $b_i$ is in $I$ for $i \leq m$. Remove all $b_i$ s.t. $b_i \in I$, and consider $\{ b_1, \dots, b_j \}$. Then clearly this is a generator set, and furthermore if $\sum a_i b_i \in I$, then $\sum a_i b_i = \sum r_i c_i b_i$, so by freeness $a_i = r_i c_i$, but by our choice of indexes $c_i \in I$ so $r_i c_i \in J$

interregno
  • 474