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Define the ring $\Bbb Z[x]/\langle x^2-3,2x+4\rangle$. Find an isomorphism between the given ring and some other ring.

Apparently the isomorphism is $\Bbb Z[x]/\langle x^2-3,2x+4\rangle \cong \Bbb Z_2[x]/\langle x^2-3\rangle$, but I don't see how they got $\Bbb Z_2[x]$ instead of $\Bbb Z[x]$ here.

What I did was that I noted $x^2-3=0 \implies x^2=3$ and that $2x+4 \implies 2x=-4$ so $6=2x^2 = 2x \cdot x = -4x$ so $$6+4x=0 \implies2(3+2x)=2(3-4)=-2 =0 \implies 2=0$$

so the generator becomes $\langle x^2-3,2x+4\rangle = \langle x^2-3, 0 \rangle = \langle x^2-3\rangle$.

Thus $$\Bbb Z[x]/\langle x^2-3,2x+4\rangle \cong \Bbb Z[x] /\langle x^2-3\rangle$$

but where does the $\Bbb Z_2[x]$ come from?

user26857
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Teemu
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  • I did find multiple questions regarding the same problem, but neither of them had answer for the question I was asking here? Is this common practice here to mark questions as duplicates even if they are asking about a different thing related to the same problem asked before? – Teemu Mar 27 '22 at 15:25
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    What you asked it is very well explained in the many answers to the original question. – user26857 Mar 27 '22 at 15:53
  • Btw, the equalities you used hold in the quotient ring, so $2=0$ in the quotient ring, too. Your mistake is to consider that this holds in the polynomial ring $\mathbb Z[x]$. Furthermore, $2=0$ in the quotient ring means that $2$ belongs to the ideal, so you can factorize by $2$, and here you are $\mathbb Z/2\mathbb Z$. – user26857 Mar 27 '22 at 15:57
  • Simpler: $,2(2!+!x)=0,\Rightarrow, 2 = 0,$ by cancelling the unit $,2!+!x,$ (by scaling by its inverse $,2!-!x;,$ intutively it's just $(2!+!\sqrt 3)(2!-!\sqrt 3) = 1,,$ cf. this Remark in the linked dupe). More generally many similar inferences follow by taking norms – Bill Dubuque Mar 27 '22 at 19:46

1 Answers1

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Note that $$2=2(x^2-3)-(x-2)(2x+4) $$ is in the ideal. Vice versa, $2x+4$ is a multiple of $2$, and $x^2-3\equiv x^2-1\pmod 2$. Hence the ring can also be written as $$\Bbb Z[x]/\langle2,x^2-1\rangle. $$ This is a ring with four elements $[0],[1],[x], [x+1]$.

Hagen von Eitzen
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