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I have the following question on ring theory, I would like somebody to help me:

Let $R =\mathbb Z_2[x]$, and consider the ideal $I$ of $R$ generated by the irreducible polynomial $f(x)=x^2+x+1\in R$. Show that the factor ring $R/I$ is a field.

Here is the solution:

$R/I =\{f(x)+I\mid f(x)\in R\}$ so by division algorithm $f(x)=q(x)( x^2+x+1)+r(x)$, then $f(x)= q(x)( x^2+x+1)+ax+b$. Hence in $R/I$: $$f(x)+I=q(x)(x^2+x+1)+ax+b+I=ax+b+I$$ $$R/I=\{ax+b+I\mid a,b\in\mathbb Z_2\}=\{0+I,1+I,x+I,1+x+I\}$$ Every element in $R/I$ has a multiplicative inverse, hence is a field.

I'm not clear with this - somebody help me, thanks.

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    I have not gone through the question thoroughly! But, if every element in $R/I$ has an inverse, $0_{R/I}=1_{R/I}$ and hence, $R/I$ will be a zero ring! So, you probably wanted to say, every non-zero element has an inverse!2012-01-29
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    Hint: $\mathbb{Z}_2[x]/(x^2 + x + 1) = \mathbb{Z}/(2, x^2 + x + 1)$. To show that this is a field is equivalent to showing that $(2,x^2 + x + 1)$ is a maximal ideal in thepolynomial ring $\mathbb{Z}[x]$.2012-01-29
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    is the last line the part you don't follow?2012-01-29
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    $f\;$ is irreducible $\implies$ $I=(f)$ is maximal $\implies$ $R/I$ is a field...2012-01-29
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    Sorry bad mistake it should have been $\mathbb{Z}[x]/(2,x^2 + x + 1)$.2012-01-29
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    In the solution which you posted, they probably just meant this: Noticing that $(1+I)(1+I)=1+I$ and $(x+I)(1+x+I)=1+I$ you see that every non-zero element has an inverse. Thus it is a field. (Other properties, like commutativity and asociativity of $+$ and $\cdot$ etc., should be relatively easy to show.)2012-01-29
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    @BenjaminLim: True, but isn't it easier to show that $(x^2+x+1)$ is maximal in the polynomial ring $\mathbb{Z}[x]$? The latter is a Euclidean domain, whereas $\mathbb{Z}[x]$ is not. (And it's very easy to verify that $x^2+x+1$ is irreducible in $\mathbb{Z}_2$!)2012-01-29
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    @ArturoMagidin Sorry, which are you referring to being the Euclidean domain?2012-01-30
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    @Lim: Sorry, bad typing. $\mathbb{Z}_2[x]$ is a Euclidean domain (the ring of polynomials over a field is always a Euclidean domain, with "degree" being the Euclidean function).2012-01-30

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