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Suppose we are given the field $\mathbb{F}_5$ and $p(X) = X^2-2 \in \mathbb{F}_5[X]$, an irreducible polynomial over $\mathbb{F}_5$. Let $\mathbb{K}$ denote the extension of $\mathbb{F}_5$ in which $p(X)$ has a root $\alpha$. $\mathbb{K}$ is an extension of degree $2$ and of cardinality $5^2=25$.

It is easily verified that $\alpha$ is not a square in $\mathbb{K}$, so $q(Y) = Y^2 - \alpha \in \mathbb{K}[Y]$ is an irreducible polynomial over $\mathbb{K}$. We can therefore form an extension of $\mathbb{K}$ in which $q(Y)$ has a root $\beta$, i.e. an element such that $\beta^2=\alpha$. I will call this extension $\mathbb{L}$. $\mathbb{L}$ is an extension of degree $2$ over $\mathbb{K}$ and of cardinality $5^4 = 625$.

I am trying to find the minimal polynomial of $\beta$ over $\mathbb{F}_5$ which gives rise to the same extension $\mathbb{L}$, i.e. the polyomial $s(Z) \in \mathbb{F}_5[Z]$ such that $s(\beta)=0$.

$s(Z)$ should be of degree $4$ and $\mathbb{L}$, being the splitting field of $s(Z)$, will contain all its roots which are given by $\beta, \beta^5, \beta^{25}, \beta^{125}$. So, one way to find $s(Z)$ is to compute and simpify the product $s(Z) = (Z-\beta)(Z-\beta^5)(Z-\beta^{25})(Z-\beta^{125})$. I got $s(Z)=Z^4+3$.

My question: is there another, more intelligent, method to find $s(Z)$ that would work even when the involved fields are of greater cardinality?

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    $\begingroup$ Because the minimal polynomial of $\alpha$ is $f(x)=x^2-2$ and $\beta^4=\alpha^2=2$ then the minimal polynomial of $\beta$ is $g(x)=f(x^2)=x^4-2=x^4+3$. $\endgroup$ – Harto Saarinen Oct 23 '14 at 21:36
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You're adding in $\sqrt{2}$ and then $\sqrt{\sqrt{2}}$. So intuitively it's obvious that really you're just adding a root of $x^4-2$.

This is the same polynomial as you got.

More formally speaking, you could verify the isomorphism of rings:

$\mathbb{F}_5[x,y]/\langle x^2 - 2, y^2 -x \rangle \cong \mathbb{F_5}[y]/\langle y^4-2 \rangle$

This amounts to checking that the equality of the ideals

$\langle x^2 - 2, y^2 -x \rangle = \langle y^4-2, y^2 - x\rangle$

The only non-trivial step there is that $x^2-2 \in \langle y^4-2, y^2 - x\rangle$.

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    $\begingroup$ $$x^2 - 2 = (y^2 - x) \cdot (-y^2 - x) + (y^4 - 2) \cdot 1$$ $\endgroup$ – user14972 Oct 23 '14 at 21:50
  • $\begingroup$ Thank you for your answers! My example was rather obvious, indeed. But what happens if we extend $\mathbb{K}$ with the (suppose) irreducible polynomial $q(Y) = Y^3 + (1+3\alpha)Y + 4\alpha$ which has root $\beta$. How would one then find the irreducible polynomial of $\beta$ over $\mathbb{F}_5$? $\endgroup$ – Nocturne Oct 23 '14 at 22:04
  • $\begingroup$ I don't know. You should ask a new question, if you want answer. $\endgroup$ – user40167 Oct 23 '14 at 22:12
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In general, suppose you have a field extension $K(\beta) / K$, and $\alpha \in K(\beta)$, and suppose that $f(x)$ is the minimal polynomial of $\beta$ over $K(\alpha)$.

Recall that there is a norm function $K(\alpha) \to K$. There are a few different ways to express this e.g. if $L/K$ is the Galois closure of $K(\alpha) / K$, then the norm of $\xi$ is the product of all of the conjugates of $\xi$. i.e. if $m$ is the minimal polynomial of $\xi$ then

$$ N_{K(\alpha) / K}(\xi) = \prod_{\substack{\rho \in L \\ m(\rho) = 0}} \rho $$

There are other approaches involving resultants or determinants too; which one we use isn't especially important....

But the key idea is that we can extend the norm function to polynomials and rational functions: e.g. $K(x, \alpha) / K(x)$ is a field extension, and there is a norm function sending $K(x, \alpha) \to K(x)$. This extends the norm on $K(\alpha) / K$.

And so, we can set $g(x) = N_{K(x,\alpha)/K(x)}(f(x))$. Then $g(x)$ is a polynomial over $K$, $\beta$ is one of its roots, and

$$\deg g(x) = (\deg f(x)) \cdot [K(\alpha) : K] = [K(\beta) : K(\alpha)] \cdot [K(\alpha) : K] = [K(\beta) : K]$$

Consequently, $g(x)$ must be the minimal polynomial of $\beta$ over $K$,

e.g. in your example, the norm of $Y^2 - \alpha$ is

$$ (Y^2 - \alpha)(Y^2 - \bar{\alpha}) = Y^4 - (\alpha + \bar{\alpha})Y^2 + \alpha \bar{\alpha} = Y^4 - 2$$

where $\bar{\alpha} = -\alpha$ is the other square root of $2$. The fact $\alpha + \bar{\alpha} = 0$ can be obtained by looking at the minimal polynomial of $\alpha$: in particular, it is the negation of the coefficient on $x$ (which is zero). The same with $\alpha \bar{\alpha} = -2$.

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  • $\begingroup$ As an aside, writing this answer made me fully realize something I was only dimly aware of before: while I knew that norms made sense on the polynomials, this is the first time I explicitly realized that it could be expressed as the norm on the extension of function fields. I wish I could +1 the original question again for helping me realize that! $\endgroup$ – user14972 Oct 26 '14 at 0:29
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$\{ 1, \alpha, \beta, \alpha \beta \}$ is a $\mathbf{F}_5$-basis for $\mathbf{F}_5(\alpha, \beta)$, and so there must be a linear combination of $1, \beta, \beta^2, \beta^3, \beta^4$, which gives us the minimal polynomial. We can compute these:

$$ \beta^0 = 1$$ $$ \beta^1 = \beta$$ $$ \beta^2 = \alpha$$ $$ \beta^3 = \alpha \beta $$ $$ \beta^4 = 2 $$

and we can do linear algebra to solve. Of course, in this case we can do it by inspection, and see that

$$ \beta^4 - 2 = 0$$

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  • $\begingroup$ Thanks! So, in general, we might express the powers of $\beta$ in terms of the basis vectors and solve a system of linear equations to find the linear combination which gives the zero vector. $\endgroup$ – Nocturne Oct 23 '14 at 23:00

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