11
$\begingroup$

I had to calculate the number of elements of this quotient ring: $$R = \mathbb{Z}[X]/(X^2-3, 2X+4).$$

This is what I've got by myself and by using an internet source:

Writing the ring $R = \mathbb{Z}[X]/(X^2−3, 2X+4)$ as $\mathbb{Z}[X]/I$, we note that $I $ contains $2(X^2−3)−(X−2)(2X+4) = 2$. We then note that the generator $2X + 4$ is actually superfluous since $2X + 4 = 2(X + 2)$.

Now we can write $R = \mathbb{Z}[X]/(X^2 − 3, 2) = \mathbb{Z}[x]/((X+1)^2, 2)$, because $(X+1)^2=X^2+2X+1=X^2-3+2X+4$. The internet source states now the following:

$$ R \cong (\mathbb{Z}/2\mathbb{Z})[\alpha] \quad \text{with} \quad \alpha = X+1$$ I guess that $\mathbb{Z}/2\mathbb{Z}[\alpha]$ represents the set of dual numbers of the field $\mathbb{Z}/2\mathbb{Z}$. I see that $\alpha^2=0$, but what exactly implies the isomorphism? And does this mean that the quotient ring $R$ contains four elements?

If necessary, you can take at the site I used. (It's about page 5, exercise 4.3a.) http://www.math.umn.edu/~musiker/5286H/Sol1.pdf

I thank you in advance for your answers.

$\endgroup$
1
  • $\begingroup$ Yeah, ignore my previous comment. What do you mean "dual number of the field..." $\endgroup$ Jul 30 '13 at 13:23
5
$\begingroup$

Here is an alternative approach. Remember, for ideals $I,J \trianglelefteq R$ we have $R/(I+J) \cong (R/I)/\overline{J}$ where $\overline{J}$ denotes the embedding of $J$ in $R/I$.

In this case we have $\mathbb{Z}[X]/(X^2-3,2X+4) \cong (\mathbb{Z}[X]/(X^2-3))/(2 \overline{X}+4) \cong \mathbb{Z}[\sqrt{3}]/(2 (\sqrt{3}+2))$. In $\mathbb{Z}[\sqrt{3}]$ the element $2+\sqrt{3}$ is a unit (with inverse $2-\sqrt{3}$), so $\mathbb{Z}[\sqrt{3}]/(2 (\sqrt{3}+2)) = \mathbb{Z}[\sqrt{3}]/(2) \cong \mathbb{Z}[X]/(X^2-3,2) \cong \mathbb{F}_2[X]/(X^2+1)$.

The latter is obviously a two-dimensional $\mathbb{F}_2$-vector-space so the number of elements is clear.

Of course, instead of considering the first line of isomorphisms you could also show the equality $(X^2-3,2X+4) = (X^2-3,2)$ directly. Nevertheless the correspondence to $\mathbb{Z}[\sqrt{3}]$ is useful, because it gives us an idea how to show this.

$\endgroup$
4
  • $\begingroup$ Sorry for my late reaction. Thank you for your answer, but what you are telling me right now is slightly ambitious for me, because I have knowledge about groups, and I know just a little about rings. I don't know terms as "embidding" that well, I've only heard that word once or twice in my life. esides, I don't get the link with this and vector spaces. It will take me some time to figure out what you wrote down. (Anekdote: My last name "Duin" is Dutch for "Dune." $\endgroup$ Jul 31 '13 at 10:01
  • 1
    $\begingroup$ Don't mind the word 'embedding'. What I wrote in the first line is nothing but the third isomorphism theorem for rings. That $\mathbb{F}_2[X]/(X^2+1)$ is a two dimensional space, you can see as follows: if $K$ is a field and $p \in K[X]$ a polynomial of degree $n$ then $K[X]/p$ is a $K$-vector-space with basis $\{1, X, X^2, \dots, X^{n-1}\}$ - in particular it is $n$-dimensional. I hope this helps a bit. :) $\endgroup$
    – Dune
    Jul 31 '13 at 10:43
  • $\begingroup$ How can you show $R/(I+J) \cong (R/I)/\overline{J}$? Just by routine arguments or can we use some isomorphism theorems? $\endgroup$
    – izimath
    Apr 13 '21 at 2:56
  • 1
    $\begingroup$ @izimath This is one of the standard isomorphism theorems of rings. It is proven by applying the first isomorphism theorem to the canonical homomorphism $R/I \to R/(I+J)$. $\endgroup$
    – Dune
    Apr 13 '21 at 7:06
2
$\begingroup$

You almost have it. To get representatives of the classes in the quotient ring divide by $X^2-3$ and take the remainders as those representatives. Notice you can do long division by $X^2-3$ over the integers because the leading coefficient is $1$. What remainders do you get? Then the mod $2$ reduces the list further, from an infinite list to a finite list.

$\endgroup$
0

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.