Consider $f(x) := x^3+2x+2$ and the field $\mathbb{Z_3}$. $f(x)$ is obviously irreducible over $\mathbb{Z_3}$. Let $a$ be a root in an extension field of $\mathbb{Z_3}$, then why is it that $[\mathbb{Z_3}(a):\mathbb{Z_3}] = 3$? What is the basis of $\mathbb{Z_3}(a)$ over $\mathbb{Z_3}$?

I know that $\mathbb{Z_3}(a) \simeq \mathbb{Z_3}[x]/<f(x)>$ and since $f(x)$ is irreducible in $\mathbb{Z_3}$, any polynomial in $\mathbb{Z_3}[x]$ can have degree atmost 2. But I don't understand how that ties to $[\mathbb{Z_3}(a):\mathbb{Z_3}] = 3$? And how does that imply $\mathbb{Z_3}(a)\simeq GF(3^3)$? Thanks.

  • $\begingroup$ See "Field and Galois Theory, Patrick Morandi, chapter 1, proposition 1.15". $\endgroup$ – Lucas Corrêa Apr 21 at 21:44

In general, the degree of $F(\alpha)$ over $F$ is the degree of the minimal polynomial of $\alpha$. In this case, the minimal polynomial is $f(x)=x^3+2x+2$ which has degree $3$. The basis is $\{1,\alpha,\alpha^2\}$.

Think of it this way: $F(\alpha)$ should consist of elements of the form $p(\alpha)/q(\alpha)$, where $p,q$ are polynomials. But using the relation $\alpha^3=-2\alpha-2$, you can see that every polynomial in $\alpha$ can be written as a linear combinations of $1,\alpha,\alpha^2$. And even $\alpha^{-1}$ can be written as such. That means every element of $F(\alpha)$ is a linear combination of $1,\alpha,\alpha^2$.

Let $K=\mathbb{F}_3(\alpha)$. To see why $K\simeq \mathbb{F}_9$, it's just a cardinality argument: since $K$ is a $3$-dimensional $\mathbb{F}_3$-vector space, we know from linear algebra that $K\simeq \mathbb{F}_3^3$ as vector spaces. The right hand side has 27 elements. So $K$ is the field of 27 elements.

  • $\begingroup$ I see, thanks. Why is $f(x)$ the minimal polynomial for $\alpha$? Why can't we have a polynomial of degree, say 2, whose zero is $\alpha$? $\endgroup$ – manifolded Apr 21 at 21:49
  • $\begingroup$ Thanks @egreg, my arithmetic is suspect. $\endgroup$ – Ehsaan Apr 21 at 21:58
  • $\begingroup$ @manifolded: The minimal polynomial of $\alpha$ has the property that it generates the ideal of all polynomials which vanish at $\alpha$. It is the unique (monic) irreducible polynomial with $\alpha$ as a root. $\endgroup$ – Ehsaan Apr 21 at 21:59

Look at the situation from a more abstract point of view. Let $F$ be a field and $f(x)\in F[x]$ an irreducible monic polynomial.

If $a$ is a root of $f(x)$ in some extension field $K$ of $F$, then, if $F(a)$ denotes the smallest subfield of $K$ containing $F$ and $a$, we have $$ F(a)\cong F[x]/\langle f(x)\rangle $$ and moreover $F[a]$, the smallest subring of $K$ containing $F$ and $a$ is the same as $F(a)$. Therefore we can see $F(a)=F[a]=\{g(a):g(x)\in F[x]\}$.

On the other hand, as $f(a)=0$, given $g(x)\in F[x]$, we can perform the division and write $g(x)=f(x)q(x)+r(x)$, where $r$ has degree less than the degree of $f$. Thus we also have $$ F(a)=F[a]=\{g(a):g(x)\in F[x],\deg g<\deg f\} \tag{*} $$ which is probably what you refer to by saying “any polynomial in $\mathbb{Z}_3[x]$ can have degree at most $2$” (which isn't a good way to express the fact).

Now, suppose $g(x)$ is a monic polynomial satisfying $g(a)=0$. Take $g$ of minimal degree. Since we can perform the division $f(x)=g(x)q(x)+r(x)$, the assumptions give us that $r(a)=0$; by minimality of $\deg g$, we infer that $r(x)=0$. Therefore $g$ divides $f$. Since $f$ is irreducible, we deduce that $g(x)=f(x)$ (they can differ up to a nonzero multiplicative constant, but being both monic, the constant is $1$).

Hence $f(x)$ is the minimal polynomial of $a$.

Now we can see that the set $\{1,a,a^2,\dots,a^{n-1}\}$ (where $n=\deg f$) is a basis of $F[a]$ as a vector space over $F$. The fact it is a spanning set follows from (*); it is linearly independent because $f$ is the minimal polynomial and a linear combination of those elements is the value of a polynomial of lesser degree than $f$, so it cannot vanish unless all the coefficients are zero.

Finally apply this to your particular case: $\mathbb{Z}_3[a]$ is a three-dimensional vector space over $\mathbb{Z}_3$, so it has $3^3=27$ elements.


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