So elements of this ring look like $$a+b\sqrt[3]{2}+c\sqrt[3]{4}$$ If I want to find the multiplicative inverse element for the above general element, then I'm trying to find $x,y,z\in\mathbb{Q}$ such that $$(a+b\sqrt[3]{2}+c\sqrt[3]{4})(x+y\sqrt[3]{2}+z\sqrt[3]{4})=1$$ I can see that expanding gives me the system $$ax+2cy+2bz=1$$ $$bx+ay+2cz=0$$ $$cx+by+az=0$$ I don't want to solve this using matrices because I know it will turn ugly. Is there a more elegant way to approach the inverse calculation to avoid the ugly calculation? The only thing I thought of was setting the bottom two equations equal to each other $$bx+ay+2cz=cx+by+az$$ Which seems to indicate that $$a=b, b=c, a=2c$$ but this would make me think $a=b=c=0$ and thus a multiplicative inverse does not exist.
-
"Which seems to indicate that ..." is of course wrong. – Martin Brandenburg May 25 '15 at 13:06
-
2You could represent your field by matrices – GPerez May 25 '15 at 13:10
-
I was sure that was not true, but a last minute effort before posting – Iceman May 25 '15 at 13:10
-
1Some posts that are related: http://math.stackexchange.com/questions/294993/show-mathbbq-sqrt32-is-a-field-by-rationalizing and http://math.stackexchange.com/questions/533655/is-mathbbq%CE%B1-ab%CE%B1c%CE%B12-a-b-c-%E2%88%88-mathbbq-with-%CE%B1-sqrt32-a-fiel – Martin Sleziak May 25 '15 at 14:44
-
@Martin Sleziak: duplicates, actually. – Martin Brandenburg May 25 '15 at 14:48
-
This post shows that denominator, which appears in the expressions given in the answers, is non-zero: I will just add that the denominator is non-zero: Does $a^3 + 2b^3 + 4c^3 = 6abc$ have solutions in $\mathbb{Q}$ – Martin Sleziak May 25 '15 at 14:49
-
Of course no additional proof is necessary for this, since we know that $x \mapsto x \cdot y$ is invertible for abstract reasons. – Martin Brandenburg May 25 '15 at 16:16
3 Answers
First let me remark that there is a general abstract argument which shows that if $L/K$ is a field extension and $a \in L$ is algebraic over $K$, then the $K$-algebra $K[a]$ is a field. Namely, $K[a]$ is an integral domain which is also a finite-dimensional vector space over $K$. This implies for $0 \neq x \in K[a]$ that the linear map $K[a] \to K[a]$, $y \mapsto x \cdot y$ is surjective, since it is injective, which means that $x$ is invertible. To get a constructive proof, we just have to sit down and make the linear algebra argument here explicit.
The underlying vector space of $\mathbb{Q}[\sqrt[3]{2}]$ has basis $1,\sqrt[3]{2},\sqrt[3]{4}$. For some fixed non-zero element $x=a + b \sqrt[3]{2} + c \sqrt[3]{4}$, let's write down the linear map $y \mapsto x \cdot y$ in terms of this basis:
$~~~\,1 \mapsto a + b \sqrt[3]{2} + c \sqrt[3]{4}$
$\sqrt[3]{2} \mapsto 2c + a \sqrt[3]{2} + b \sqrt[3]{4} $
$\sqrt[3]{4} \mapsto 2b+2c\sqrt[3]{2} + a \sqrt[3]{4} $
The corresponding matrix is:
$$\begin{pmatrix} a & 2c & 2b \\ b & a & 2c \\ c & b & a \end{pmatrix}$$
From linear algebra we know how to invert matrices, for example via Cramer's rule. In this case, we get:
$$\frac{1}{a^3 - 6abc + 2b^3 + 4c^3} \cdot \begin{pmatrix} a^2 - 2bc & 2b^2 - 2ac & 4c^2 - 2ab \\ 2c^2 - ab & a^2 - 2bc & 2b^2 - 2ac \\ b^2 - ac & 2c^2 - ab & a^2 - 2bc \end{pmatrix}$$
The determinant $a^3 - 6abc + 2b^3 + 4c^3$ has been computed via the Rule of Sarrus and the cofactors have been computed by the usual formula for $2 \times 2$-determinants.
This matrix represents the linear map $y \mapsto x^{-1} \cdot y$ with respect to our basis. Thus, to get $x^{-1}$, we just have to evaluate at $1$, and we get:
$$x^{-1} = \frac{1}{a^3 - 6abc + 2b^3 + 4c^3} \cdot ((a^2 - 2bc) + (2c^2 - ab) \sqrt[3]{2} + (b^2 - ac) \sqrt[3]{4})$$
Of course, this method works quite generally. For example, for $x=a + b \sqrt[3]{p} + c \sqrt[3]{p}^2$ we have:
$$x^{-1} = \frac{1}{a^3 - 3pabc + pb^3 + p^2 c^3} \cdot ((a^2 - pbc) + (pc^2 - ab) \sqrt[3]{p} + (b^2 - ac) \sqrt[3]{p}^2)$$
- 163,620
-
This looks great,...except I'm confused with where you say "evaluate at 1" . I'm assuming you just mean let $y=1=(1,0,0)^T$ in terms of vectors because the coefficients of the basis radicals is just the first column of $x^{-1}$? – Iceman May 25 '15 at 13:42
-
1Yes, that's correct. One evaluates the linear map at $1$, which means that we multiply the matrix with the corresponding vector $(1,0,0)$ (which I would not write as $1$, by the way) and then take the corresponding ring element of the vector. – Martin Brandenburg May 25 '15 at 13:44
-
This was a great answer. I understood everything very clearly. I had the matrix and seeing the invertible matrix argument using Cramer's rule was very insightful. Thanks for the help. – Iceman May 25 '15 at 13:47
- 2,419
-
Perhaps it would be better to remove the two previous posts that contain the same information... – Michael Burr May 25 '15 at 13:20
-
@Michael Burr \ Didn't realize that I had inadvertently done two comments. I chucked them finally. – Senex Ægypti Parvi May 25 '15 at 13:47
Not a great method if you'd like larger matrices, but I used Maple to get:
$$ x=\frac{a^2-2bc}{a^3-6abc+2b^3+4c^3}\qquad y=-\frac{ab-2c^2}{a^3-6abc+2b^3+4c^3}\qquad z=-\frac{ac-b^2}{a^3-6abc+2b^3+4c^3}. $$
I didn't check that I typed everything correctly, but it's not too bad.
- 32,867
-
3
-
I have mathematica and I was going to do that, but as Martin said, I wanted to see if there was a better way to solve the system instead of row reduction, which is terrible with three variables. – Iceman May 25 '15 at 13:39
-
@MartinBrandenburg Of course this isn't the way to understand the problem, but it depends on the application of interest. If one just needs the inverse formally for a specific case, it works. The OP was worried about an ugly calculation, so I pointed out that the calculation isn't too bad. – Michael Burr Jun 01 '15 at 22:41