This is a problem similar to one of my homework problems, but not on the homework. The problem states that:
Find a primitive root $\beta$ of $F_2[x]/(x^4+x^3+x^2+x+1)$.
Questions:
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I know what a primitive root of a prime number is, but what is a primitive root of a polynomial (or is it called something like a field extension here)?
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My book gives this hint: $[x]=\alpha$ doesn't work because $\alpha^5=1$. There are eight choices of $\beta$. I am basically lost on the hint. What is $\alpha$? Why doesn't it work? Why are there 8 choices for $\beta$? Wouldn't there be $2^4 =16$ choices? (or that's the number of polynomials in the field?)
Sorry for the long questions, since I am quite lost right now. Hopefully my questions make sense, and any help would be appreciated!
Best Answer
The multiplicative group of nonzero elements of a finite field is always cyclic. A "primitive root" of a finite field is a generator for the multiplicative group of nonzero elements.
Note that $x^4+x^3+x^2+x+1$ is irreducible over $\mathbb{F}_2$: it has no roots, and it is not the product of two irreducible quadratics (the quadratics are $x^2$, $x^2+1$, $x^2+x$, and $x^2+x+1$, and the only irreducible one is the latter; but $(x^2+x+1)^2 = x^4 + x^2 + 1$). So $\mathbb{F}_2[x]/(x^4+x^3+x^2+x+1$ is a field of degree $4$ over $\mathbb{F}_2$ (hence, of order $2^4 = 16$). So you are looking for an element in the field of $16$ elements whose multiplicative order is exactly $15$. The book is noting that even though this field equals $\mathbb{F}_2(\alpha)$ (where $\alpha$ is the class of $x$ in the quotient), $\alpha$ doesn't work because it is of order $5$. That is: while it is true that if $\beta$ is a primitive root for the field $GF(p^k)$, then $GF(p^k) = \mathbb{F}_p(\beta)$, the converse does not necessarily hold as this example shows.