Let $A=\mathbb{Z}[\sqrt[3]{2}]$ and $I=(4+\sqrt[3]{2^2})$. Elements in $A$ have the form $a\cdot 1+b\cdot 2^{\frac{1}{3}}+c\cdot 2^{\frac{2}{3}} \Rightarrow$ elements in $I$ have the form $$ (a\cdot 1+b\cdot 2^{\frac{1}{3}}+c\cdot 2^{\frac{2}{3}})(4+2^{\frac{2}{3}}) = a(4+2^{\frac{2}{3}})+b(2+4\cdot 2^{\frac{1}{3}})+c(2\cdot 2^{\frac{1}{3}}+4\cdot 2^{\frac{2}{3}}) $$ Let $v_1=1, v_2= 2^{\frac{1}{3}}$, and $v_3=2^{\frac{2}{3}}$. Then elements in $I$ are linear combinations of $4v_1+v_3, 2v_1+4v_2$, and $2v_2+4v_3$. To find the number of elements in the group $A/I$, we can take the determinant of the matrix that represents these elements: $$ \begin{vmatrix} 4&0&1\\ 2&4&0 \\0&2&4 \end{vmatrix}=68 $$ I'd like to conclude that $A/I$ is an abelian group with 68 elements. However, when I was looking at this problem from another perspective, I noticed that the surjection $A\to A/I$ contains 34 in the kernel which makes me think that $A/I$ is actually a cyclic group with 34 elements. I want to know if the fact that 34 is in the kernel says anything about the structure of the group itself, or if it's just a coincidence that $34\cdot 2=68$ turned out to be the determinant calculated above.

  • $\begingroup$ You cannot conclude that $A/I$ is cyclic of order $34$ because $A$ is not cyclic generated by $1$; as an abelian group, $A\cong \mathbb{Z}\times\mathbb{Z}\times\mathbb{Z}$, and the fact that $34$ is in the kernel only tells you that $(34,0,0)$ lies in the kernel; that is, the factor group $A/I$ factors through $(\mathbb{Z}/34\mathbb{Z})\times\mathbb{Z}\times\mathbb{Z}$, but need not be equal to it. $\endgroup$ – Arturo Magidin Apr 4 '12 at 5:25
  • $\begingroup$ Dear Chris, Your computation with the determinant is correct. This doesn't quite answer your question, because there is more than one isomorphism class of abelian groups of order $68$. But there aren't so many. If you know some algebraic number theory (e.g. something about ideals and norms) this would provide the most natural perspective from which to sort out the remaining details. If not, you can still probably figure out exactly which group you get; the first step will be to work out the a priori possibilities. And of course it is no coincidence that $34$ is in the kernel. Regards, $\endgroup$ – Matt E Apr 4 '12 at 5:28

No, it's not (entirely) a coincidence.

As an abelian group, $A$ is isomorphic to a direct sum/product of three copies of $\mathbb{Z}$, $A\cong \mathbb{Z}\times\mathbb{Z}\times\mathbb{Z}$. The fact that $34\in I$ tells you that, again as abelian groups, the map $A\to A/I$ factors through $$\frac{\mathbb{Z}}{34\mathbb{Z}}\times \mathbb{Z}\times\mathbb{Z}.$$ So of course the $34$ is intimately connected to the size of the group.

As an abelian group, $A$ is generated by the elements $4 + \sqrt[3]{4}$, $-8+\sqrt[3]{2}$, and $1$; the ideal $I$ is generated, again as an abelian group, by $4+\sqrt[3]{4}$, $-16+2\sqrt[3]{2}$, and $34$. So, as an abelian group (I'm not taking the multiplicative structure into account here): $$\frac{A}{I}\cong \frac{\langle 4+\sqrt[3]{4}\rangle}{\langle 4+\sqrt[3]{4}\rangle} \times \frac{\langle -8+\sqrt[3]{2}\rangle}{2(-8+\sqrt[3]{2})} \times \frac{\langle 1\rangle}{\langle 34\rangle} \cong \frac{\mathbb{Z}}{\mathbb{Z}}\times\frac{\mathbb{Z}}{2\mathbb{Z}}\times \frac{\mathbb{Z}}{34\mathbb{Z}},$$ which tells you that as an abelian group $A/I$ is isomorphic to $C_2\oplus C_{34}$.

  • 1
    $\begingroup$ The fact that $32$ was in the kernel tells you that one cyclic factor of the quotient has order dividing $32$; if you knew that no positive integer smaller then $32$ lay in the kernel, you would know the quotient has one cyclic factor of order $32$, so that, together with your order computation, would tell you the isomorphism type of the group $A/I$. By the by, the computations can be done using the Smith Normal Form as well. $\endgroup$ – Arturo Magidin Apr 4 '12 at 5:39
  • $\begingroup$ Thank you. This is precisely the information I was looking for. $\endgroup$ – chris Apr 4 '12 at 5:45
  • $\begingroup$ @chris: (In the comment, "32" should be "34" throughout, of course); I take the comment partially back: you need a bit more information than just knowing that $34$ lies in the kernel and no smaller positive integer does to know the quotient has a factor of order exactly $34$; that information is to be found in your given basis, but it takes a bit more work than I implied in the previous comment. $\endgroup$ – Arturo Magidin Apr 4 '12 at 5:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.