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I'm making exercises to prepare for my ring theory exam:

How many elements does the ring $ℤ[X]/(X^2-3,2X-4)$ have? Describe the structure of the this ring.

I find it always difficult if the ideal is generated by two elements.

I thought about someting like this. $X=2$ and $X^2=3$. Therefore $4=3$ therefore $1=0$. As $2=1$ then $X=0$. And then everyting seems to become $0$.

I'm not sure if I'm allowed to see it this way, but this is the first thing that comes to my mind.

Kasper
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    @user Did you notice that the ring there has $2x+4$ and not $2x-4$? – rschwieb Oct 18 '13 at 14:42
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    @rschwieb Did you notice that this doesn't matter and the problem is basically the same? –  Oct 18 '13 at 15:23
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    Dear @user Sure they are superficially very similar, and it makes a good "related" link. It changes the final answer, though. If this were closed as a duplicate, then we may as well close every question $Z[x]/(p(x),q(x))$ as a duplicate. Let's not be too tough posters like that. Regards – rschwieb Oct 18 '13 at 16:30
  • @rschwieb I won't call $(x^2-3,2x-4)=(x^2-3,2x+4)$ "superficially" very similar. (I think you missed this little detail.) –  Oct 18 '13 at 17:03
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    Dear @user : I did, and that means a lot of other readers will too. What is superficial/obvious is in the eye of the beholder. Perhaps you can replace the petty antagonism in your posts with more useful like "notice since $(2x-4)^2=-4$, the ideal here is the same as at this other post." That way we can all see the connection without partaking in your wonderful demeanor. Kind regards: – rschwieb Oct 18 '13 at 17:36
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    More clear: They are related by the isomorphism $x\mapsto -x$. As a grader, I declare that you should never say that things are "obviously true" or "basically equal" or "can easily be shown to hold", blah blah blah. Instead, practice the art of showing things in few words. If it is simple, then it is simply explained. When saying it's simple takes as many words as actually giving the explanation, then you might as well do the latter—just as when, if a loved one asks if you care, you smile and do something nice for them, instead of saying, "Duh, isn't it completely obvious that I care?" – Andrew Dudzik Oct 18 '13 at 23:49
  • @user A good reason to not say stuff like "oh, you can show that" is you get stuff like my 'answer'. – Henry Swanson Oct 19 '13 at 00:32
  • @user, indeed. But in this way I am able to also make a point. – Mariano Suárez-Álvarez Oct 27 '13 at 19:09

2 Answers2

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Here is another way to compute this.

In the quotient ring, we have $X^2 = 3$ (I willl continue to write $X$ to denote the image of $X$ in the quotient)). Thus $(X-2)(X+2) = X^2 - 4 = - 1,$ and so in the quotient we have $X-2$ is a unit. Thus the equation $2(X-2) = 0$ simplifies to $2 = 0$. Thus the quotient is equal to $(\mathbb Z/2\mathbb Z)[X]/(X^2 -3)$.

Now $\mathbb Z/2\mathbb Z$ is a field of two elements, and in particular in this field $(a+b)^2 = a^2 + b^2$. Using this, we see that $X^2 - 3 = X^2 + 1 = (X+1)^2$. If we make a change of variables $T = X+ 1$, then we can write the quotient as $(\mathbb Z/2\mathbb Z)[T]/(T^2).$

Matt E
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  • I love this proof. Really elegant. But I don't understand all your steps. I understand that $2X-4=0$ and $X^2=3$ imply that $2=0$ in this ring. And I understand that this imply that you have $ℤ/2ℤ$ as coefficients. But I don't see why this implies that $ℤ[X]/(X^2-3,2X-4)=ℤ/2ℤ[X]/(X^2-3)$. Why can we leave out $2X-4$? – Kasper Oct 19 '13 at 00:37
  • Or is this because adding the equation $2X=4$ is the same thing as saying $2X=0$, and in $ℤ/2ℤ[X]$ this is already true, therefore $(X^2-3,2X-4)=(X^2-3)$ in this ring ? – Kasper Oct 19 '13 at 00:49
  • One other note, I would think you get $(ℤ/2ℤ)[T-1]/(T^2)$. I don't see why you can write $(ℤ/2ℤ)[T]/(T^2)$. – Kasper Oct 19 '13 at 00:51
  • @Kasper: Dear Kasper, Since $2 = 0$ in $(\mathbb Z/2\mathbb Z)[X]$, the relation $2 X -4$ holds automatically, so we can omit it. Also, polynomials in $T-1$ can always be rewritten as polynomials in $T$ (and vice versa), so $R[T-1] = R[T]$ for any ring $T$. Regards, – Matt E Oct 19 '13 at 03:39
  • @user: Dear user, Thanks for pointing out the typo. As for the first isomorphism: we are in a ring generated over $\mathbb Z$ by $X$, with relations $X^2 = 3$ and $2(X-2) = 0$. The first relation implies that $(X-2)$ is a unit in this ring, and so the relations can be rewritten as $X^2 = 3$ and $2 = 0$. I'm not sure what more you would want to be said. Regards, – Matt E Oct 19 '13 at 12:03
  • @MattE: I understand that $R[T-1] \cong R[T]$. But are you saying that isomorphism takes $T^2$ to $T^2$? That I do not quite get; I thought it wasn't the case. Could you please clarify? –  Oct 19 '13 at 15:56
  • @Doldrums: Dear Doldrums, $R[T-1]$ is literally equal to $R[T]$; they are the same ring. In particular, the isomorphism I have in mind (which is simple equality) is the one that takes $T$ to $T$, and so takes $T^2$ to $T^2$. Regards, – Matt E Oct 19 '13 at 20:43
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Henry's answer does a good job of explaining what the congruence classes look like, except for a misstep about $b$. By using the Euclidean algorithm on $x^2-3$ and $2x-4$ in $\Bbb Q[x]$, we can find that $2(x^2-3)-(2x-4)(x+2)=2$, so $2$ is in that ideal, and hence $(x^2-3,2x-4)=(x^2+1,2)$. That further reduces the possibilities for $b$ to $0,1$ also.

I'm going to add a bit about the structure of your ring (which I'll call $R$).

Clearly it is a commutative finite ring: $\{0,1,x,x+1\}$, and you can use an isomorphism theorem to show it is isomorphic to $\Bbb F_2[x]/(x+1)^2$

Kasper
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rschwieb
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  • Do correct answers get less votes then good answers here ? ;) – Kasper Oct 18 '13 at 23:42
  • Ah, since $\mathbb{Q}$ is a field, $\mathbb{Q}[x]$ is Euclidean, so now we can find a GCD. I have a question about the method though; what happens when you don't end up in $\mathbb{Z}[x]$ at the end? I tried it with $3x^2+1$ and $2x-1$, I get quotient $3/2x - 3/4$ and remainder $1/4$. I think I'm looking at $(3x^2+1)p(x) + (2x-1)q(x) = k$, and then minimize $k$ while keeping $p(x),q(x) \in \mathbb{Z}[x]$? – Henry Swanson Oct 18 '13 at 23:50
  • @HenrySwanson When your polynomials don't end up in $\Bbb Z[x]$, you can multiply by the LCM of the denominators of the quotients to make it into something in $\Bbb Z[x]$ :) – rschwieb Oct 19 '13 at 00:37
  • Oh wow, "least" [integer] common multiple and "least" [integer] linear combination are the same least. Thanks! :D – Henry Swanson Oct 19 '13 at 00:57