Show that $\mathbb{F}_9 \not \subset \mathbb{F}_{27}$

Question

The usual answer will go like this:

Since $2 \not | \ 3$ and $\mathbb{F}_{p^r} \subset \mathbb{F}_{p^s}$ if and only if $r | s$, then $\mathbb{F}_9 \not\subset \mathbb{F}_{27}$.

However, I'm interested in a proof that does not require any use of finite fields other than the fact that the quotient of a field $\mathbb{F}_p$ with an irreducible polynomial of degree $n$ in $\mathbb{F}_p[t]$ results in a field of order $p^n$, and that if two sets, $A$ and $B$ are such that $A \subset B$, then there exists and injection $g:A \to B$ (the identity).

So first I constructed explicitly two fields of $9$ and $27$ elements by taking the quotient of $\mathbb{Z}_3$ with $f_1(x)=x^2+1$ and $f_2(x)=x^3+2x^2+1$ respectively. Now I think I should prove somehow that there is no injection between those fields, but I'm stuck on this part, so amy help will be highly appreciate. Thanks in advance!

Answer 1

If $\Bbb F_9 \subset \Bbb F_{27}$, then the groups of invertible elements should be in the same relation, i.e. $\Bbb F_9 ^* \subset \Bbb F_{27} ^*$. But $\Bbb F_9 ^*$ has $8$ elements, while $\Bbb F_{27} ^*$ has $26$, and $8 \nmid 26$, so $\Bbb F_9 \not\subset \Bbb F_{27}$.

In fact, the argument can be extended: if $\Bbb F_{p^m} \subset \Bbb F_{p^n}$, then their corresponding groups of invertible elements should be in the relation $\Bbb F_{p^m} ^* \subset \Bbb F_{p^n} ^*$, so their orders should be in the relation $p^m - 1 \mid p^n - 1$. Let $n = km + r$ with $k \in \Bbb N$ and $0 \le r < m$. Then

$$p^n - 1 = (p^m)^k p^r - 1 = [(p^m - 1) + 1]^k p^r -1 \equiv p^r - 1 \pmod {p^m - 1} .$$

Since $p^m - 1 \mid p^n - 1$, we must have $p^r - 1 \equiv 0 \pmod {p^m - 1}$. Since $0 \le r < m$, we must have $p^r - 1 = 0$, so $r = 0$ and thus $m \mid n$.

The converse is clear: if $m \mid n$, then $\Bbb F_{p^m} \subset \Bbb F_{p^n}$.

This is probably the shortest proof of "$\Bbb F_{p^m} \subset \Bbb F_{p^n} \iff m \mid n$".

Answer 2

Does this argument satisfy your needs?

If $\mathbb{F}_{9}\subset\mathbb{F}_{27}$ then $\mathbb{F}_{27}$ would be a vector space over the field $\mathbb{F}_{9}$ (since it is an abelian group with compatible multiplication of $\mathbb{F}_9$). If $\{a_1,\dots,a_n \}$ is a basis for this vector space, then every element in $\mathbb{F}_{27}$ can be written uniquely as $$\lambda_1 a_1+\dots+\lambda_n a_n$$ with $\lambda_i \in \mathbb{F}_9$. So there are $(9)^n$ elements in $\mathbb{F}_{27}$, which can never equal $27$.

It seems to me that a direct argument of showing that there exists no injection between the two will not be very easy: you can maybe try to look where a generator gets mapped to and examine all the possibilities (since this is a finite problem) but that will get pretty technical.

Answer 3

There are many standard ways of showing the general result. Jef Laga's is a textbook method for proving one direction. Here's something specifice to your construction.

Let use denote by $\alpha$ the coset $x+\langle f_1(x)\rangle\in\Bbb{F}_3[x]/\langle f_1(x)\rangle$. Because $\alpha$ is a zero of $f_1$, we see that $\alpha^2=\alpha^2-f_1(\alpha)=-1$. Consequently $\alpha^4=(-1)^2=1$, so $\alpha$ is of order four.

On the other hand the multiplicative group of $\Bbb{F}_3[x]/\langle f_2(x)\rangle$ is of order $3^3-1=26$. Because $4\nmid 26$ Lagrange's theorem from elementary group theory tells us that there are no elements of order fours in this field.

But an injective homomorphism preserves orders of elements, so....

Answer 4

Here is a very elementary solution which shows the power given to us by all those nice results the other solutions use.

Any morphism of unitary rings $$ f\colon \mathbf F_3[x]/(x^2+1)\rightarrow \mathbf F_3[x]/(x^3+2x^2+1) $$ is dominated by a morphism $$ g\colon \mathbf F_3[x]\rightarrow \mathbf F_3[x] $$ in the following sense: the diagram $$ \begin{array}{ccc} \mathbf F_3[x] & \overset{g}{\rightarrow} & \mathbf F_3[x] \\ \downarrow & & \downarrow \\ \mathbf F_3[x]/(x^2+1) & \overset{f}{\rightarrow} & \mathbf F_3[x]/(x^3+2x^2+1) \end{array} $$ commutes, where the vertical maps are the quotient maps. Indeed, denoting by $\pi$ the left quotient map and by $\rho$ the right one, choose any $\rho$-preimage $p\in\mathbf F_3[x]$ of $f(\pi(x))$. Of course, we may choose $p$ of degree $\leq 2$, by euclidean division. Define $g$ to be the unique morphism of unitary rings with $g(x)=p$. Then one has indeed $\rho\circ g=f\circ \pi$. In particular, $\rho(g(x^2+1))=f(\pi(x^2+1))=f(0)=0$. This means that $$ g(x^2+1)=g(x)^2+1=p^2+1\in(x^3+2x^2+1). $$ Write $p=ax^2+bx+c$ with $a,b,c\in\mathbf F_3$. Then $$ p^2+1=(dx+e)(x^3+2x^2+1) $$ for some $d,e\in\mathbf F_3$, that is $$ a^2x^4+2abx^3+(2ac+b^2)x^2+2bcx+c^2+1=dx^4+(2d+e)x^3+2ex^2+dx+e. $$ It follows that, looking at trailing coefficients, $e=c^2+1$. Since $c^2=0$ or $c^2=1$, this means that $e=\pm1$. If $e=1$ then $c=0$ and $d=2bc=0$, leading to $a=0$ and $e=0$. Contradiction. If $e=-1$ then $c=\pm1$. Looking at the coefficient of $x^1$, $\mp b=d$. Then $\mp a+d^2=2ac+b^2=2e=1$, $\pm ad=2ab=2d+e=-d-1$ and $a^2=d$. In particular, $d=0$ or $d=1$. Now, $d$ cannot be zero otherwise $0=\pm ad=-d-1=-1$. Hence $d=1$. Then $\mp a+ 1=\mp a +d^2=1$ and $a=0$. Then $-d-1=0$ and $d=-1$. Contradiction. Therefore, there is no morphism $f$.