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I'm using Michael Artin's book "Algebra". I understand what $\mathbb Z[x]$ is. I just don't understand how to think of $\mathbb Z[x]/(x^2+7)$ or things like $\mathbb Z[x]/(x^2-3, 2x-4)$. I understand $(x^2+7)$ represents an ideal.

I also don't understand how to think of $(x^2-3, 2x-4)$ (considered in $\mathbb Z[x]$), another student said that it's the set of all polynomials that can be divded by both of those polynomials, if this is the case: what does $(x^2-3, 2x-4)$ look like written out formally?

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    $\begingroup$ To understand quotients (at least by ideals generated by a single element), remember that what we are doing is setting the generator of the ideal to $0$. So when the ideal is generated by $x^2 + 7$ what we do is say "work as usual with polynomials, but whenever you have a $x^2$, replace it by $-7$ (which then means that all our elements are represented by polynomials of degree at most $1$). $\endgroup$ – Tobias Kildetoft Jan 31 '18 at 19:35
  • $\begingroup$ @TobiasKildetoft in that situation, will we ever get $x^3$, if so, what happens? $\endgroup$ – Mario Jan 31 '18 at 19:44
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    $\begingroup$ Well, $x^3 = x^2x$, so...? $\endgroup$ – Tobias Kildetoft Jan 31 '18 at 19:46
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    $\begingroup$ $(x^2-3,2x-4)=\{(x^2-3)f(x)+(2x-4)g(x); f(x),g(x) \in \Bbb Z[x] \}$ $\endgroup$ – Mustafa Jan 31 '18 at 20:12
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    $\begingroup$ Have a look at similar questions. They have good answers and might be helpful; e.g., see here. $\endgroup$ – Dietrich Burde Jan 31 '18 at 20:17
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We can explain $\Bbb Z[x]/(x^2-3,2x-4) \cong \Bbb Z_2 \times \Bbb Z_2$:

we have $2\in (x^2-3,2x-4); 2=2(x^2-3)-(x+2)(2x-4)$, so $(x^2-3,2x-4,2)=(x^2-3,2)$ and $\Bbb Z[x]/(2,x^2-3)\cong \Bbb Z_2[x]/(x^2-1) \cong \Bbb Z_2[x]/(x+1)(x-1)\cong \Bbb Z_2 \times \Bbb Z_2 $.

where $\Bbb Z_2[x]/(x^2-1)=\{a_0+a_1x+a_2x^2+a_3x^3+...+a_nx^n+(x^2-1); a_0,...,a_n\in \Bbb Z_2 \}=\{b_0+b_1x+I; b_0,b_1\in \Bbb Z_2 \}; x^2=1, x^3=x,...,I=(x^2-1)=\{0+I,1+I,x+I,(1+x)+I \}$

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  • $\begingroup$ Why are we allowed to omit $2x-4$ when we discover that $2$ is in $(x^2-3, 2x-4)$? Also how is $\mathbb Z_2[x]/(x+1)(x-1) \cong \mathbb Z_2 \times\mathbb Z_2$? $\endgroup$ – Mario Feb 1 '18 at 1:52
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    $\begingroup$ $2x-4=2(x-2)$. And by Chinese remainder theorem $\Bbb Z_2[x]/(x-1)(x+1) \cong \Bbb Z_2[x]/(x-1) \times \Bbb Z_2[x]/(x+1) \cong \Bbb Z_2 \times \Bbb Z_2 $ $\endgroup$ – Mustafa Feb 1 '18 at 5:23

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