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So in class we had an exercise that was:

Show that $\mathcal{GL}(4)$ is not the splitting field for $x^3+x+1$

Now in the lecture this was done by noting that if $\alpha$ was a root then $Z_2[\alpha]\subset \mathcal{GL}(4)$ but then $|Z_2[\alpha]|=8$ so this is not the case.

However could we not simply notice that if $\mathcal{GL}(4)$ was a splitting field then we would have $x^3+x+1=(x-a)(x-b)(x-c)$- so it would have $3$ roots in $\mathcal{GL}(4)$ but then we can simply check for the elements $0,1,2,3$ that this is not the case? Or have I misunderstood something badly?

Thanks for any help

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GL(4) isn't the same as integers mod 4, so checking 0,1,2,3 is likely not the way. –  coffeemath Nov 22 '12 at 14:45
    
Yeah I was not meaning $\mathbb{Z}_4$, i just used 0,1,2,3 to represent the elements of the field, which looking back would be some very misleading notation. Thanks –  hmmmm Nov 22 '12 at 15:14
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1 Answer

up vote 3 down vote accepted

As you said, GF(4) is not a splitting field of $x^3 + x + 1$, because any field over $\mathbb{Z}_2$ containing the roots of $x^3 + x + 1$ must contain at least eight elements.

About the actual question: yes, you can do it by calculating values of the polynomial for all elements of GF(4). The other solution saves you the work of actually construing GF(4) though.

As mentioned in the comments, you might be confusing GF(4) with $\mathbb{Z}_4$, the ring of integers modulo $4$. The ring $\mathbb{Z}_4$ is not even a field. See here for the addition/multiplication table of GF(4) (also denoted by $\mathbb{F}_4$) and more information.

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