I need to describe a ring $R=\mathbb{Z}[X]/(X^2-3,2X+4)$
I know that its element would be of the form $\{f(x)+(X^2-3,2X+4)|f(x)\in \mathbb{Z}[X]\}$ After that will i divide $f(x)$ by $X^2-3$ first and then by $2X+4$ to reduce the elements in $R$?
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rschwieb
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Mathronaut
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2First, perhaps, it would help to better understand $I=(x^2-3,2x+4)\leq\mathbb{Z}[x]$. – Jonathan Y. Aug 20 '13 at 18:57
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how do we reduce $f(x)$? – Mathronaut Aug 20 '13 at 18:59
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See also the answers to this duplicate question, whicg give further details and reamrks. – Bill Dubuque Dec 20 '13 at 16:46
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it looks to me like it is $\mathbb Z_2[X]/\left<X^2+1\right>$.
$2$ is in your ideal because:
$$2 = 2(X^2-3)-(2X+4)(X-2)$$
Also $X^2+1$ is in your ideal because: $$\begin{align}X^2+1 &= (X^2-3) + 4 = (X^2-3) + 4(X^2-3) - (2X+4)(2X-4) \\&= 5(X^2-3) - (2X+4)(2X-4) \end{align}$$
So we know that $\left<2,X^2+1\right>$ is contained in your ideal.
But it is obvious that the generators for your ideal are in $\left<2,X^2+1\right>$.
So we are done.
Note: Given that $(X+1)^2=X^2+1$ in $\mathbb Z_2$, this can be rewritten as isomorphic to $\mathbb Z_2[Y]/\left<Y^2\right>$ with $Y$ corresponding to $X+1$.
Thomas Andrews
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I actually started with $2x^2$ and write it as both $2(x^2-3) + 6$ and $(2x+4)x - 8x$. @rschwieb – Thomas Andrews Aug 20 '13 at 19:10
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The ideal is also $\left<2,X^2-1\right>$, and so the quotient ring is $\mathbb Z_2[u]$, with $u^2=1$. – lhf Aug 20 '13 at 19:16
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1$u^2=-1$ or $u^2=1$ doesn't really matter in $\mathbb Z_2$, @lhf. I think the reduction to $\mathbb Z_2[Y]/\left<Y^2\right>$ is probably best. – Thomas Andrews Aug 20 '13 at 19:19
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1Right, and that can be written as $\mathbb Z_2[w]$ with $w^2=0$, which may be more concrete for the OP. – lhf Aug 20 '13 at 19:20
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Perhaps. I always translate "$\mathbb R[w]$ with $P(w)=0$" as $\mathbb R[X]/\left<P(x)\right>$ in my head, @lhf. : ) – Thomas Andrews Aug 20 '13 at 19:22
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And, of course $2x^2 = (2x+4)x-4x$. My previous comment said $-8x$. – Thomas Andrews Aug 20 '13 at 21:42
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I have another answer. $\mathbb Z[x]/(x^2-3,2x+4) = (\mathbb Z[x]/(x^2-3))/(x^2-3,2x+4)/(x^2-3)$(by third isomorphism theorem) now define a map $\mathbb Z[x]$ to $\mathbb Z[\sqrt{3}], $ $F(f(x))=f(\sqrt{3})$( evaluated at root $3)$ then $F$ is surjective homomorphism.By 1st isomorphism theorem $\mathbb Z[x]/(x^2-3)=\mathbb Z[\sqrt{3}]$ and $(x^2-3,2x+4)/(x^2-3)=(2x+4+(x^2-3))$ and when it undergoes $F$,resultant ideal will be $(2\sqrt{3}+4)$ so finally description will be $\mathbb Z[\sqrt{3}]/(2\sqrt{3}+4)$
ANJAN SAMANTA
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