I am working with the $\mathbb{Z}$-algebra generated by the two elements $\alpha$ and $\beta=(\alpha^3+1)/2$ of $\mathbb{Q}(\alpha)$ such that $$\alpha^4+5\alpha^3+12\alpha^2+21\alpha+21=0.$$ I guess I could also see it as $$\Omega\cong\frac{\mathbb{Z}[x,y]}{(x^4+5x^3+12x^2+21x+21,\ 2y-x^3-1)}.$$ Now I need to tensor it with $\mathbb{Z}/2\mathbb{Z}$: my first guess was $$\Omega\otimes_\mathbb{Z}\mathbb{Z}/2\mathbb{Z}\cong\frac{\mathbb{Z}/2\mathbb{Z}[x,y]}{(x^4+x^3+x+1,\ x^3+1)}$$ with the modulo two simply eating away all even occurrences in the quotient. However that cannot be as I lose any information about $y$ this way: for example, by calculation it follows that $\beta^2=-6\alpha^3-18\alpha^2-42\alpha-68$ which would then be zero modulo $2$.

What am I doing wrong? Is the first description of $\Omega$ valid? And how does that tensor product behave?


Perhaps the above description of $\Omega$ as quotient algebra is not sufficient, since it seems that it doesn't give enough information about the relationship between $\alpha$ and $\beta$: for example I can't find a way to show that $\alpha\beta=-8-10\alpha-6\alpha^2-5\beta$. I could simply go on by adding in the quotient the corresponding condition with $x$ and $y$, but again, what would grant that then it would be enough? Should I check that the operation tables built from the generators $\alpha$ and $\beta$, $x$ and $y$ are the same to conclude it?

  • 2
    $\begingroup$ Yes, you are right, your description of $\Omega$ is incorrect. $\endgroup$
    – user26857
    Nov 4, 2017 at 15:37
  • $\begingroup$ @user26857 Does the description in my answer work better? $\endgroup$ Nov 4, 2017 at 16:37

1 Answer 1


(I set $f(x)=x^4+5x^3+12x^2+21x+21$.) The description of $\Omega$ is not sufficient, since it doesn't give enough information about the relationship between $\alpha$ and $\beta$, for example that $\alpha\beta=-8-10\alpha-6\alpha^2-5\beta$. Adding that and $\beta^2=-62-42\alpha-18\alpha^2+12\beta$ a better version seems to be $$ \Omega\cong\frac{\mathbb{Z}[x,y]}{I}, $$ where $I=(f(x),\ 2y-x^3-1,\ xy+5y+8+10x+6x^2,\ y^2-12y+62+42x+18x^2)$, the conditions cited above. Elements of this quotient are of the form $a_1+a_2x+a_3x^2+a_4x^3+a_5y$, where $a_5\in\{-1,0,1\}$ and the other are integer coefficients.

Now, considering the evaluation map

$$\hat{\Psi}:\mathbb{Z}[x,y]\rightarrow\Omega,\ x\mapsto \alpha,\ y\mapsto \beta$$ by the conditions above $\hat{\Psi}(I)=0$ and so it induces a map $$\Psi:\frac{\mathbb{Z}[x,y]}{I}\rightarrow\Omega.$$ The set $\{1,\alpha,\alpha^2,\beta\}$ is a basis for $\Omega$ as $\mathbb{Z}$-module (it follows from the fact that $f(x)$ is actually the minimal polynomial for $\alpha$) and hence $\Psi$ is surjective. Now suppose $$\Psi(a_1+a_2x+a_3x^2+a_4x^3+a_5y)=0,\mbox{ i.e. } a_1+a_2\alpha+a_3\alpha^2+a_4\alpha^3+a_5\beta=0.$$ Since $\alpha^3=2\beta-1$ we get $$(a_1-a_4)+a_2\alpha+a_3\alpha^2+(2a_4+a_5)\beta=0$$ which immediately implies $a_2=a_3=0$. Since $2a_4=-a_5$ then $a_5=0$, hence $a_1=a_4=0$ and the map is injective too.

Then $$\Omega\otimes_\mathbb{Z}\frac{\mathbb{Z}}{2\ \mathbb{Z}}\cong\frac{\mathbb{Z/2\mathbb{Z}}[x,y]}{(x^4+x^3+x+1, x^3+1, xy+y,y^2)}\cong\frac{\mathbb{Z/2\mathbb{Z}}[x,y]}{((x+1)(x^2+x+1), y(x+1),y^2)}.$$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .