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Assume we have a finite field $\mathbb F_p$, an irreducible polynomial $f(x)$ of degree $m$ over $\mathbb F_p$, and an irreducible polynomial $g(y)$ of degree $n$ over $\mathbb F_p[x]/(f(x))$. Then $\bigl(\mathbb F_p[x]/(f(x))\bigr)[y]/(g(y))$ is a finite field with $p^{mn}$ elements.

Now, let's consider $\alpha\colon \mathbb F_p[z]\to \bigl(\mathbb F_p[x]/(f(x))\bigr)[y]/(g(y))$, acting as constant on elements $\mathbb F_p$, and mapping $z^{im+j}$ to $x^j y^i$ (or maybe it maps $z^{in+j}$ to $x^i y^j$?). It's a surjective homomorphism onto a field, so its kernel has to be generated by an irreducible polynomial $h(z)$ of degree $mn$.

How do I obtain this $h(z)$ from $f(x)$ and $g(y)$?

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As @walcher observed, your map $\alpha$ is not a morphism of rings.

Your question about determining an irreducible polynomial $h\in\mathbf F_p[z]$ such that $$ \mathbf F_p[z]/(h)\cong \left(\mathbf F_p[x]/(f)\right)[y]/(g) $$ is nice and has a neat answer.

Note that $\mathbf F_p[x]/(f)$ is the finite field $\mathbf F_q$ with $q=p^m$. Let $\phi$ be the Fobenius morphism from $\mathbf F_q$ into itself: $$ \phi(x)=x^p $$ for all $x\in\mathbf F_q$. Since $\mathbf F_q$ is of characteristic $p$, the map $\phi$ is indeed a ring morphism from $\mathbf F_q$ into itself. Since $\mathbf F_q$ is a field, $\phi$ is necessarily injective. Since $\mathbf F_q$ is finite, $\phi$ is an automorphism of $\mathbf F_q$.

If you know a little bit Galois Theory then you know that $\phi$ is of order $m$, i.e., composing $\phi$ $m$ times with itself $\phi^m=\phi\circ\cdots\circ\phi$ is equal to the identity map on $\mathbf F_q$. However, you do not need to know Galois Theory to see that. It suffices to apply Lagrange's Theorem to the multiplicative group $\mathbf F_q^\times$ in order to see that $x^{q-1}=1$ for all $x\in\mathbf F_q^\times$. Multiplying by $x$, one has $x^q=x$, which now holds for all elements of the field $\mathbf F_q$. Then, indeed $$ \phi^m(x)=x^{p^m}=x^q=x $$ for all $x\in\mathbf F_q$, and $\phi^m$ is the identity on $\mathbf F_q$.

The interesting thing about $\phi$ is that its set of fixed points is exactly equal to the subfield $\mathbf F_p$ of $\mathbf F_q$. Indeed, by Fermat's Little Theorem, one has $x^p=x$ for all $x\in \mathbf F_p$. Moreover, since the polynomial $x^p-x$ has degree $p$, it cannot have more roots than the elements of $\mathbf F_p$. Hence, the elements $x$ of the complement $\mathbf F_q\setminus\mathbf F_p$ satisfy $x^p\neq x$. This proves that $\phi(x)=x$ iff $x\in\mathbf F_p$, for all $x\in\mathbf F_q$.

We need a little more notation. If $\sigma$ is an automorphism of $\mathbf F_q$ and $r\in\mathbf F_q[y]$ is a polynomial, then $r^\sigma$ will denote the polynomial one obtains from $r$ by letting act $\sigma$ on its coefficients. If $s$ is another polynomial in $\mathbf F_q[y]$, then $$ (rs)^\sigma=r^\sigma \cdot s^\sigma. $$ If $\tau$ is another automorphism of $\mathbf F_q$, then $$ (r^\sigma)^\tau=r^{\tau\sigma} $$ as one applies $\sigma$ first to the coefficients, and $\tau$ next. Note also that $r^{\mathrm{id}}=r$, where $\mathrm{id}$ is the identity of $\mathbf F_q$.

Now, let us prove that $$ h=g\cdot g^\phi\cdot g^{\phi^2} \cdots g^{\phi^{m-1}} $$ is the polynomial you were after. Let us check that $h\in\mathbf F_p[y]$. It suffices to check that $h^\phi=h$ by what has been said above. And indeed, $$ h^\phi=\left(g\cdot g^\phi\cdot g^{\phi^2} \cdots g^{\phi^{m-1}}\right)^\phi= g^{\phi}\cdot g^{\phi^2}\cdot g^{\phi^3} \cdots g^{\phi^{m}}=g^{\phi}\cdot g^{\phi^2}\cdot g^{\phi^3} \cdots g=h $$ since $\phi^m$ is the identity. It follows that $h\in\mathbf F_p[y]$. Since $g$ has a root in $\mathbf F_{p^{mn}}$ that generates the field extension of $\mathbf F_{p^{mn}}$ over $\mathbf F_p$, the polynomial $h$ has such a root as well. Since $h$ is of degree $mn$, it is the minimal polynomial of such a root. In particular, $h$ is irreducible and $$ \mathbf F_p[z]/(h(z)=\mathbf F_{p^{mn}}=\mathbf F_{q^n}=\left(\mathbf F_p[x]/(f)\right)[y]/(g) $$ as required.

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    $\begingroup$ Huh. So I just find the polynomial whose splitting field I already know... now it seems really obvious and that means your answer is excellent. Thank you! $\endgroup$ – Joker_vD May 13 '16 at 12:44
  • $\begingroup$ What a great answer, Johannes! I hope it will get the upvotes it deserves. $\endgroup$ – Georges Elencwajg May 15 '16 at 8:06
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Your map $\alpha$ is not a homomorphism unless $g(y)=y^n-x \;(\text{mod}\;f(x))$. In fact: $x=\alpha(z^n)=\alpha(z\cdot z^{n-1})$, $y^n=y\cdot y^{n-1}=\alpha(z)\cdot\alpha(z^{n-1})$. If $g(y)=y^n-x$, then $h(z)=f(z^n)$.

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  • $\begingroup$ Hm, that's true. But I really need to construct an irreducible polynomial $h(z)$ from $f(x)$ and $g(y)$, so that the corresponding homomorphism is easily computible. Do you have any ideas? $\endgroup$ – Joker_vD May 23 '14 at 19:02

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