Is $\phi: \mathbb{F_p}(X) \rightarrow \mathbb{F_p}(X), a \mapsto a$ for $a \in \mathbb{F_p}$ and $X \mapsto X+1$ a field homomorphism?

Question

Let $\mathbb{F_p}$ be a finite field of cardinality $p$, $p$ a prime. Let $\phi: \mathbb{F_p}(X) \rightarrow \mathbb{F_p}(X), a \mapsto a$ for $a \in \mathbb{F_p}$ and $X \mapsto X+1$, aka $\phi$ is the function from the field of rational functions with coefficients in $ \mathbb{F_p}$ to itself that reduces to the identity on $\mathbb{F_p}$ and maps $X$ to $X+1$. I want to know if $\phi$ is a field homomorphism, i.e. if it is multiplicative and additive and maps the $1$ to the $1$.

The $1$ to the $1$ follows from the definition,but the multiplicative and additive part I don't know how to prove or disprove.

First of all, is $\phi$ completely determined on all of $\mathbb{F_p}(X)$ through the information given? If so, what is say $\phi(X^2)?$ or $\phi(X^3+p)?$

My hunch for the solution (which may be completely off) is that the answer is no, $\phi$ is not a homomorphism, because it doesn't map zeros of certain minimal polynomials to other zeroes of that minimal polynomial. What I mean is that if I "plug in" an element $a$ that is algebraic over $ \mathbb{F_p}$ into $X$ I have essentially adjoined that Element to $ \mathbb{F_p}$and given the minimal Polynomial of $a$ over $ \mathbb{F_p}$ $m_{a, \mathbb{F_p}}$ $\phi(a)$ may not be another zero of $m_{a, \mathbb{F_p}}$.

Answer 1

You are right in that $\phi$ is not a homomorphism, YET. But, it can naturally be extended to one!

A standard way of descibing a homomorphism from one algebraic structure to another is to give its effect on the generators of the domain. My guess is that the first time you met this principle was on a course in linear algebra: a linear transformation is fully determined after you have given the images of the elements of some basis. The same thing happens with groups, but there you (hopefully) learned about the pitfall: the generators may satisfy some hidden relations, and as the homomorphism must respect those relations we can no longer choose the images of the generators freely.

OTOH, in the polynomial ring $K[X]$ the indeterminate $X$ doesn't have any hidden relations. We can map it relatively freely, and get a homomorphism of rings. The only thing we need to worry about is that $X$ is to commute (multiplicatively) with the constants. So if $R$ is any ring containing $K$ in its center, and $a\in R$ is ANY element, then we get a homomorphism of rings $ev_a:K[X]\to R$ by "evaluating polynomials at $a$": $$ ev_a(p(X))=p(a). $$

On we go. We can select here $R=K[X]$ and $a=X+1$. This gives us a homomorphism of rings $$ ev_{X+1}:p(X)\mapsto p(X+1). $$ Furthermore, this homomorphism is injective because its kernel is trivial (do you see why?). That means that it induces a homomorphism of the corresponding fields of fractions: $$ev_{X+1}:K(X)\to K(X), \frac{p(X)}{q(X)}\mapsto\frac{p(X+1)}{q(X+1)}. $$ And this way, logically extending step-by-step, we end up with the homomorphism your query is about.

It is a homomorphism of fields, when you extend it "naturally".

The term "naturally" here may be a bit loaded at first, but it becomes second nature quickly. That's why algebra texts often omit the description of all those steps.

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A few closing remarks are due.

• We could have also applied a homomorphism of the field of constants $K$ to all the constants. Your field $\Bbb{F}_p$ has no automorphisms, so this route is not open to us this time.
• This homomorphism $K(X)\to K(X)$ is actually an automorphism, i.e. a bijection. That's because it has as an inverse the homomorphism gotten by mapping $X\mapsto X-1$.
• We could also map a similar evaluation homomorphism from $K[X]$ to $K(X)$ by sending $X$ to any non-constant rational function, like $\alpha(X)/\beta(X)$, $\alpha,\beta\in K[X]$, and evaluating all the polynomials at $\alpha(X)/\beta(X)$. We again get a homomorphism from the field of fractions $K(X)\to K(X)$. In general that will no longer be an automorphism. A standard exercise is to show that it is an automorphism, iff $\alpha$ and $\beta$ are both at most linear polynomials in $X$ (such that their ratio is not a constant).

Answer 2

$\phi(X^2)$ must equal $\phi(X)^2=(X+1)^2$ etc. In general $\phi$ takes $f(X)/g(X)$ to $f(X+1)/g(X+1)$. One checks this map is well-defined and respects addition/multiplication etc.