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This problem is from a practice exam I was working on.

What is the cardinality of the quotient $\mathbb{Z}[x]/(x^2-3,2x+4)$ ?

Thoughts. If I find a ring that is easier to handle then this then I can go from there. So I think this is isomorphic to $\mathbb{Z}_2[x]/(x^2-3,x+2)$. And then I am stuck.

Am I correct so far? What should I be looking to do from here?

Thanks for your time and your answers.

Cousin
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  • Use long division of the elements of $\mathbb{Z}[x]$ by $x^3-3$. Notice that we can do this and stay in integer coefficients, since the leading coefficient of $x^3-3$ is $1$. You get representatives being the possible remainders, i.e. $ax^2+bx+c$. Now you need to separate in cases, depending on the parity of $a,b,c$ to 'divide' by $2x-4$ as much as possible, these remainders. The remainders in the different cases give you representatives for the elements of the quotients. – OR. Aug 15 '13 at 22:40
  • Thanks for your comment. But why divide by $x^3-3$. One of the factors are $x^2-3$? Am I missing something? – Cousin Aug 15 '13 at 22:53
  • Yes, divide by $x^2-3$. I misread the exponent. – OR. Aug 16 '13 at 02:56

3 Answers3

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Your start is good as long as you can show that $2\in(x^2-3,2x+4)$. Fortunately this is the case: $x(2x+4)-2(x^2-3)=4x+6$ and $2(2x+4)-(4x+6)=2$. (Actually one can show that $(x^2-3,2x+4)=(x^2+1,2)$.) So $$\mathbb{Z}[x]/(x^2-3,2x+4)\simeq \mathbb{Z}_2[x]/(x^2+1)$$ and this ring has $4$ elements.

  • thanks for the answer. So to show the equality of the two ideals I imply show that the respective generator is in the other ideal and visa versa and I'm done right? – Cousin Aug 15 '13 at 23:00
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A slight modification of YACP's answer:

If a ring contains an element $x$ satisfying $x^2=3$ and $2x=-4$, then we get $6=2x^2=-4x=8$, hence $2=0$, and the second relation becomes superflous. This shows that $\mathbb{Z}[x]/(x^2-3,2x+4)=\mathbb{Z}/2[x]/(x^2-3)$.

I just want to emphasize that you don't have to come up with clever linear combinations of the generators of the ideal. Just compute inside the quotient ring!

(In fact, as usual, you can also see this as an application of the Yoneda Lemma: The algebraic manipulations above show that $\hom(\mathbb{Z}[x]/(x^2-3,2x+4),-) \cong \hom(\mathbb{Z}/2[x]/(x^2-3),-)$.)

Martin Brandenburg
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  • Very nice. I like your answer. – Cousin Aug 21 '13 at 01:02
  • @RohanK More conceptually: $,2(2!+!x)=0,\Rightarrow, 2 = 0,$ by cancelling unit $,2!+!x,$ [by scaling by its inverse $,2!-!x;,$ intutively it's just $(2!+!\sqrt 3)(2!-!\sqrt 3) = 1,,$ cf. my answer] – Bill Dubuque Mar 27 '22 at 19:43
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Hint $\ $ Note $\,(2\!-\!x)(2\!+\!x) = (2\!-\!\sqrt 3)(2\!+\sqrt 3)=1\,$ in $\ \Bbb Z[x]/(\color{#0a0}{x^2\!-\!3})\cong \,\Bbb Z[\sqrt 3],\, $ therefore, since $\ 2\!+\!x\,$ in a unit $\!\bmod \color{#0a0}{x^2\!-\!3},\,$ we can cancel it from the generator $\,\color{#c00}2(2\!+\!x)\, $ of $I,\,$ yielding

$\qquad\ \ I = (\color{#0a0}{x^2\!-\!3},\,\color{#c00}2(2\!+\!x)) = (x^2\!-\!3,\color{#c00}2) = ((x\!+\!1)^2,2),\ $ so $\,\ \Bbb Z[x]/I = \Bbb Z_2[x]/((x\!+\!1)^2)$

Remark $\ $ Alternatively we can do the cancellation purely equationally as below

$\bmod I\!:\,\ \color{#0a0}{x^2\equiv 3},\,\ \color{}2(\color{}{2\!+\!x)\equiv 0} \,\Rightarrow\,0\equiv \color{}2(2\!+\!x)(2\!-\!x) \equiv 2(4\!-\!\color{#0a0}{x^2})\equiv 2(4\!-\!\color{#0a0}3) \equiv \color{#c00}2\ $

More generally when studying algebraic number rings we can often deduce much by taking norms (e.g. above we used: $\,\alpha = 2+\sqrt{3}\,$ is a unit since its norm $\,N(\alpha ) = \alpha \bar\alpha = 1)$.

The resulting quotient ring is $\,\cong \Bbb Z_2[\epsilon]/(\epsilon^2),\,$ the ring of dual numbers over $\,\Bbb Z_2.\,$ Such dual number rings prove useful for algebraically modelling derivatives, tangent spaces, etc.

The hint amounts to Euclid's Lemma in ideal form, i.e.

Lemma $\,\ \color{#0a0}{(a,b)=1}\,\Rightarrow\, (a,bc) = (a,c)\ $

Proof $\ \ \ (a,bc) = (a,ac,bc) = (a,\color{#0a0}{(a,b)}c) = (a,c)$

Bill Dubuque
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