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.

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

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.

  • $\begingroup$ 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
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    $\begingroup$ @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

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.


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