Equality of ideals with two generators

Asked Jan 14 '17 at 15:45

Active Jan 14 '17 at 15:45

Viewed 170 times

Let $R$ be a commutative ring, $a, b \in R$, and $c, d \in (a, b)$. Under which conditions is $(a, b) = (c, d)$?

Let $c = x'\cdot a + y'\cdot b$ and $d = x''\cdot a + y''\cdot b$. Then

$$ \begin{align} (c, d) &= \{ xc + yd \mid x, y \in R \} \\ &= \{ x(x'\cdot a + y'\cdot b) + y(x''\cdot a + y''\cdot b) \mid x, y \in R \} \\ &= \{ (xx' + yx'')a + (xy' + yy'')b \mid x, y \in R \}, \end{align} $$

so $(a, b) = (c, d)$ iff $(x', x'') = (y', y'') = (1)$. Is this correct? Can I state something better?

asked Jan 14 '17 at 15:45

bbrot

1

See my remark here on unimodular linear transformations. – Bill Dubuque Jan 14 '17 at 16:02
2

If the matrix $\left(\begin{array}{cc}x' & y'\x'' & y''\end{array}\right)$ is invertible, that is, $x'y''-x''y'$ is invertible in $R$, then the ideals are equal. – user26857 Jan 15 '17 at 11:22
Thank you for your comments. user26857, is your condition equivalent to mine? – bbrot Jan 16 '17 at 12:44

Equality of ideals with two generators

0 Answers0