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$2 + \langle x^3-x+5\rangle$ and $x^2 + \langle x^3-x+5\rangle$ are distinct elements of $\mathbb Z[x]/\langle x^3-x+5\rangle$ Is this true or false?

I would be greateful if someone can explain the procedure to find the elements.

3 Answers 3

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This should give you a start. Note that $2+\langle x^3-x+5\rangle $ and $x^2+\langle x^3-x+5\rangle$ are the same if and only if their difference, $(x^2-2)+\langle x^3-x+5\rangle,$ is the zero element, namely $0+\langle x^3-x+5\rangle$. Now think about what it means for a polynomial $f$ to have the property that $f+\langle x^3-x+5\rangle$ is equal to $0+\langle x^3-x+5\rangle$, or equivalently, for $f\equiv 0\mod x^3-x+5.$ Does $x^2-2$ have that property?

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    What I understand is (x^3-x+5)=0 and x^2-2 doesn't seem to have this property. Is this enough to show that these two elements are distinct?2012-04-05
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You need to review the definitions of ring, ideal, and quotient ring. The ideal generated by $x^3-x+5$ is by definition the set of polynomials $I=\{f\cdot(x^3-x+5)\;|\;f\in \mathbb{Z}[x]\}$.

For $g\in \mathbb{Z}[x]$, consider the coset $g+I=\{g+f\;|\;f\in I\}$. In particular, $0+I$ is the set $I$ itself, $1+I$ consists of all elements of $I$ shifted by 1, etc. The quotient $\mathbb{Z}[x]/I$ is the set $\{g+I\;|\;g\in \mathbb{Z}[x]\}$, i.e. it is a set of cosets. You should now go and review how this set is endowed with addition and multiplication to make it a ring.

Yes, I mean now.

Ok, so now you should be able to answer some basic questions:

  1. What is the 0 element in this ring?

  2. Why are addition and multiplication well-defined (remember, you are adding and multiplying sets; cosets, to be precise)? Where exactly is it used that $I$ is an ideal, rather than just a subring?

If you are happy with all this, you should have no trouble with your question. Is $x^2-2+I$ the 0 element in $\mathbb{Z}[x]$? If you still don't know the answer, then you have not understood the definitions yet. Feel free to comment, detailing your thoughts.

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    @faisal $x^3-x+5$ is not the 0 element. Indeed, it is not an element of the quotient at all. The elements of the quotient are _sets_, and the 0 element is the set $\langle x^3-x+5\rangle$ which consists of all multiples of this poly. The set $x^2-2+I$ is therefore 0 if and only if it is the same set. But clearly $x^2-2$ is not a multiple of a cubic polynomial.2012-04-05
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Hint $\ $ In $\rm\:R/I,\ \ a+I\: =\: b+I\iff a\!-\!b+I\: =\: 0+I\iff a\!-\!b\in I.\:$

In mod language, $\rm\ mod\ I\!:\:\ a\equiv b\iff a\!-\!b\equiv 0\iff a\!-\!b\in I.$

Further, when $\rm\:I\: = \langle d\rangle = d\:\!R\:$ is principal, we have $\rm\:c\in d\:\!R\iff d\ |\ c\:$ in $\rm R$.

Finally, $\rm\:deg\ f < 3\:$ and $\rm\:x^3\!-x+5\ |\ f\:$ in $\rm\:\mathbb Z[x]\:$ $\Rightarrow$ $\rm\:f= 0.$

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    Thanks a ton. This is what I was asking for.2012-04-05