# A question about the degree of an extension field

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 $$\mathbb{Z_3}(a) \simeq \mathbb{Z_3}[x]/$$ 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.

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

## 2 Answers

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.

• 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 $\alpha$? – manifolded Apr 21 at 21:49
• Thanks @egreg, my arithmetic is suspect. – Ehsaan Apr 21 at 21:58
• @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. – 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.