# Clarification needed: Smallest subfield of a field

Q: Given a field does there exist a smallest subfield ?

A: Yes, intersection of all subfields gives us the smallest subfield. Suppose a field has $p^n$ elements (where $p$ is prime), then the smallest subfield will be isomorphic to $\Bbb Z_p$. In terms of polynomials find an irreducible polynomial of degree $n$.

Confusions:

1. How are we being able to guarantee that there there will exist a smallest subfield having exactly $p$ elements? If we use Lagrange's theorem for the additive group formed by the field then we could say that the smallest subgroup of it would be of some order $p^m$ ($m\leq n$). Similarly if we use Lagrange's theorem for the multiplicative group we could only say that the smallest subgroup of it would be of some order $p^k$ ($k\leq n$). Thus the order of any subfield would probably be $\text{LCM}(p^k,p^m)$ (I'm just guessing this part). Anyhow, I have no idea how they claimed that the smallest subfield will/must be isomorphic to $\Bbb Z_p$

2. No idea what they mean by "in terms of polynomials find an irreducible polynomial of degree $n$" in this context.

Could someone please clarify?

• You have to have the image of the integers under $n\mapsto1+1+\cdots+1$, ($n$ appearances of the $1$). You prove that this is a morphism of rings, and that the kernel is either trivial (characteristic zero) or a proper prime ideal, thus generated by $p$ for some prime $p$. Since $\Bbb Z/(p)$ is a field, there you are, you’ve found a minimal subfield of the field you started with. – Lubin May 24 '18 at 3:52
• @Lubin Sorry, can't understand what you're saying. What do you mean by $n\mapsto1+1+\cdots+1$ ? – Sky May 24 '18 at 4:01
• Let's try parsing what's there: "You have the image of the integers under ..." So the thing in the ellipsis is a function defined on the integers. And the description tells you which function: it sends a (nonnegative) integer $n$ to the result of adding 1 to itself $n$ times in the field. And then Lubin suggests you show this is a ring homomorphism, and figure out its kernel. I know reading closely can be tough, but it's (mostly) easier than math, and it's a good skill to develop regardless. – John Hughes May 24 '18 at 4:41

Once you prove that the intersection of all subfields of a field is itself a field, you see that every field has a single smallest subfield. The function that Lubin suggests gives you a way to understand the structure of that smallest field. Lubin is defining the function $\varphi: \mathbb{Z} \rightarrow F$ (where $F$ is your original field) by $$\varphi(n) = \underbrace{1 + 1 + \dots + 1}_{\text{n times}}.$$

Equivalently, you can also define this function as the group homomorphism $\varphi: (\mathbb{Z}, +) \rightarrow (F, +)$ with the property $\varphi(1) = 1_F$. (Since $1$ generates $\mathbb{Z}$ as a group, knowing where $\varphi$ sends $1$ is enough to define the whole function.)

At this point, you have to prove that $\varphi$ is also a ring homomorphism (which I'll leave to you).

Consider the kernel of $\varphi$. Since $\mathbb{Z}$ is a principal ideal domain, we have that $\ker \varphi = p\mathbb{Z}$ for some nonnegative integer $p$. This $p$ will, in fact, be the characteristic of the field $F$, meaning $p = 0$ or is a prime.

Now, suppose $p = 0$. Then consider the subset $$S = \left\{\varphi(a) \varphi(b)^{-1}: a, b \in \mathbb{Z}, b \not = 0\right\}$$ of $F$. You can show that $S$ is isomorphic to $\mathbb{Q}$. Since $\mathbb{Q}$ has no proper subfields, $S$ must be the smallest subfield of $F$.

On the other hand, suppose $p$ is a prime. Then $\varphi(\mathbb{Z}) \cong \mathbb{Z}/p\mathbb{Z}$. Since $\mathbb{Z}/p\mathbb{Z}$ has no proper subfields, $\varphi(\mathbb{Z})$ must be the smallest subfield of $F$.

Now, how do we know that a field $F$ with $p^n$ elements (where $p$ is prime) has characteristic $p$ (and hence a smallest subfield isomorphic to $\mathbb{Z}/p\mathbb{Z}$)? Since $F$ is a finite field, it must have some non-zero prime characteristic, $q$. Thus, $F$ has a subfield isomorphic to $\mathbb{Z}/q\mathbb{Z}$, and hence $F$ is a vector space over $\mathbb{Z}/q\mathbb{Z}$. Let $m = \dim F$ as a $\mathbb{Z}/q\mathbb{Z}$-vector space. Because $F$ is finite, $m$ must be finite as well. That means $p^n = q^m$. Hence $p = q$ and $m = n$.

• Thank you for explaining my gnomic language. – Lubin May 24 '18 at 14:25