0
$\begingroup$

How do we show the ring homomorphism for

$\phi :\mathbb F_p(\alpha) \rightarrow\mathbb F_p(\alpha)$ which is defined as $ \phi(\alpha)=\alpha +1$.

This is a very basic fact but I am unable to prove it by the definition of ring homomorphism. Same thing happens for ring homomorphism over $\phi :K[x] \rightarrow K[X]$ defined as $\phi (X)=X+1$. I have studied this earlier and assumed it as trivial but never tried to see the proof.

Thanks in advance.

$\endgroup$
  • $\begingroup$ You meant defined for $\;\alpha\;$ as you show and then generalized to $\;\Bbb F_p(\alpha)\;$ by the usual way. What is $\;\alpha\;$ , though? $\endgroup$ – DonAntonio Feb 21 '16 at 12:15
  • $\begingroup$ Here $\mathbb F_p (alpha)$ is the splitting field field of $x^p-x+a$ and $\alpha $ is a root of that polynomial $\endgroup$ – Germain Feb 21 '16 at 12:19
1
$\begingroup$

Ok, so $\;\Bbb F_p(\alpha)\;$ is a finite, and thus algebraic, extension of $\;\Bbb F_p\;$ , but also

$$\alpha^p-\alpha+a=0\implies (\alpha+1)^p-(\alpha+1)+a=\alpha^p+1-\alpha-a+a=0$$

so both $\;\alpha,\,\alpha+1\;$ are roots of the same polynomial irreducible polynomial over $\;\Bbb F_p[x]\;$ and thus there always exist a $\;\Bbb F_p$-automorphism of $\;\Bbb F_p(\alpha)\;$ mapping one of these roots to the other one, which is given by your map.

For example:

$$\phi(a_{p-1}\alpha^{p-1}+\ldots+a_1\alpha+a_0)=0\iff$$

$$0=a_{p-1}(\alpha+1)^{p-1}+\ldots+a_1(\alpha+1)+a_0=$$

$$=a_{p-1}\alpha^{p-1}+\ldots+a_1\alpha+\left(a_0+a_{p-1}+\ldots+a_1\right)\stackrel{\text{by lin. indep.}}\implies$$

$$\implies\begin{cases}a_{p-1}=0\\a_p=0\\...\\a_1=0\\a_0+\ldots+a_{p-1}=0\end{cases}\;\;\;\implies a_i=0\;\;\;\forall\;1=0,1,2,...,p-1$$

and $\;\phi\;$ is injective then. And this is enough, because $\;\phi\;$ is in particular a linear operator on a finite dimensional linear , so it is injective$\;\iff\;$ it is surjective $\;\iff\;$ it is bijective.

By the way, you could as well take $\;\alpha+x\;,\;\;\forall\,x\in\Bbb F_p\;$ . Any of these is a root of the above polynomial.

$\endgroup$
2
$\begingroup$

Let us start with your second question.

Let $K$ be a field, and $K[x]$ be the polynomial rings. Let $B$ be a commutative ring with unity containing $K$ as a subring, and let $\beta \in B$. Then there is a unique ring homomorphism $$ v_{\beta} : K[x] \to B $$ which satisfies $$\begin{cases} v_{\beta}: &a \mapsto a &\text{for $a \in K$},\\ & x \mapsto \beta.\\ \end{cases}$$ This is just evaluation of a polynomial for $x = \beta$. The statement, which allows several generalizations, can be described as the universal property of polynomial rings.


As to your first question (which originally was missing the part on the polynomial of which $\alpha$ is a root), let $f = x^{p} - x - b$, for some $b \ne 0$. Note that $b$ has additive period $p$. If $\alpha$ is a root of $f$, apply the Frobenius morphism $z \mapsto z^{p}$, then $\alpha^{p}$ is also a root, as $$ 0 = f(\alpha)^{p} = (\alpha^{p} - \alpha - b)^{p} = (\alpha^{p})^{p} - (\alpha^{p}) - b = f(\alpha^{p}). $$ And clearly $\alpha^{p} = \alpha + b$. Since $\mathbb F_p(\alpha) = \mathbb F_p(\alpha+b)$, you have found that the Frobenius map induces a ring isomorphism $\sigma$ on $\mathbb F_p(\alpha)$ which fixes $\mathbb F_p$ elementwise and maps $\alpha$ to $\alpha + b$.

Now note that $\sigma^{i}(\alpha) = \alpha + i b$. Since $b \ne 0$ has additive period $p$, you can choose $i_{0}$ so that $i_{0} b = 1$, and thus $$ \sigma^{i_{0}} (\alpha) = \alpha + 1. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.