# Fields from Elements & Elements from Fields

In the last post, We’ve seen how to extend a field using an irreducible polynomial. Here, I want to show another way to do so, however, it’s going to be from a different prespective. However, I would like to give a few reminders first from ring theory

## Prime Ideals and PID’s

Suppose that $R$ is a commutative ring, a prime ideal $I\vartriangleleft R$ is an ideal such that if $ab\in I$ then $a\in I$ or $b \in I$, and an element $a\in R$ is said to be prime if and only if the ideal $\langle a\rangle$ is prime.

Now, if you can recall, it turns out that if $R/I$ is an integral domain if and only if $I$ is a prime ideal. Let’s prove it for those who don’t remember:

Suppose that $R/I$ is an integral domain, and there are elements $a,b\in R$ such that $ab\in I$ but in the same time $a\notin I$ and $b \notin I$. Thus, in the quotient ring , $a+I,b+I$ both are not zero. On the other hand though: $ab+I$ is zero. We get that $a+I,b+I$ are zero-divisors which is a contradiction to $R/I$ being an integral domain. Thus, $I$ is a prime ideal.

The other direction is very similar – You start with a prime ideal $I$ and assume that $R / I$ is not an integral domain, thus, it has zero divisors – from that you can get a contradiction for $I$ being prime (Think about it!).

That’s a nice criteria that will help later in this post. One more thing that I would like to show, it that in a PID, a maximal ideal and a prime ideal are the same (it is actually also true in unique factorization domain (UFD) but I am not intereseted in them right now). Let’s prove it:

Suppose that $I\vartriangleleft R$ is a maximal ideal, wev’e seen in the last post that $R/I$ is a field. A field is in particular an integral domain, thus $R/I$ in an integral domain, therefore $I$ is prime (notice how I didn’t use the fact that $R$ is a PID here)

On the other hand, suppose that $I$ is prime, then since $R$ is a PID, then $I=\langle p \rangle$ where $p\in R$. By defintions, $p\in R$ is prime. Now suppose that $p=ab$ for some $a,b\in R$. Thus, $ab\in \langle p \rangle$. Since $\langle p\rangle$ is prime, we can assume WLOG that $a\in \langle p\rangle$, thus, $a=px$ for some $x\in R$. and we get

pxb=ab=p \Rightarrow xb=1

(I can ‘cancel out’ $p$ since $R$ is an integral domain) We now have got that $b$ is invertible, thus $p$ is irreducible, and we proved in the previous post that this means that $\langle p\rangle$ is maximal.

That’s all I wanted to show about rings, I am going to use what I have done here in this post, so I thought that adding it is necessary.

## Back to fields

Suppose that we already have a field extention $F \subseteq K$. and let’s pick some element $\alpha \in K$, now we can define a function $\Phi : F[x]\to K$ defined as:

\Phi(f(x))=f(\alpha)

What this function does is basically substitutes $x$ with $\alpha$. Let’s try to understand this function, first, notice that:

\Phi(\beta \cdot f(x))=\beta f(\alpha)=\beta \cdot \Phi(f(x))

Ok, we can ‘slide’ the scalar out of the argument – $\Phi$ is homogeneous, that’s nice, let’s see what else:

\Phi(f(x)+g(x))=f(\alpha)+g(\alpha)=\Phi(f(x))+\Phi(g(x))

great, so $\Phi$ is additive as well. It is actually also multiplicative:

\Phi(f(x)\cdot g(x))=f(\alpha)\cdot g(\alpha)=\Phi(f(x))\cdot\Phi(g(x))

So, in conclusion, this is a ring homomorphism (it’s not hard to see that the unit is being sent to the unit), while also being a linear map between two vector spaces over $F$. From the first isomorphism theorem we get:

F[x]/\ker \Phi\cong\text{Im}\Phi

and we know that $\text{Im}\Phi$ is a sub-ring of $K$ (it is a ring since it is isomorphic to a quotient ring). The only thing we can do now is to investigate the kernel of this map, which we denote as $J=\ker\Phi$. There are 2 cases now:

$J=0$. This is a boring, it means that the only polynomial $\alpha$ satisfies (i.e $f(\alpha)=0$) is the zero polynomial: $f(x)=0$. This means that $\alpha$ is not a root of any polynomial with coefficients from $F$. Such an elemet is said to be transcendental, and we get:

F[\alpha]=\text{Im}\Phi\cong F[x]/(0)\cong  F[x]

and there nothing new – $\alpha$ is exactly the same as $x$ and ‘nothing changed’.

The second option, is when $J\neq 0$, here something actually happens: since $F[x]$ is a PID, there exists a polynomial $h$ such that $J=\langle h \rangle$ (I’ll assume that $h$ is not a constant, since if it is, then $F=\{0\}$ (why?) which is a contradiction for it being a field). Then we get:

F[x]/\langle h\rangle =F[x]/\ker\Phi\cong \text{Im}\Phi=F[\alpha]\subseteq K

The interesting part here is: $F[x]/\langle h\rangle\cong F[\alpha]$, and since $F[\alpha]$ is a sub-ring of a field – it is an integral domain. Thus, $F[x] / \langle h \rangle$ is an integral domain as well. Now for the punch line:

F[x] / \langle h \rangle\text{ is an i.d} \iff \langle h \rangle\text{ is prime} \iff  \langle h \rangle\text{ is maximal} \iff F[x] / \langle h \rangle\text{ is a field}

We got something really special here – $F[\alpha]$ is not just a sub-ring, it is a sub-field (as it is isomorphic to a field) which contains $F$ and $\alpha$. What we have done here without even noticing, is adjoining an element from the ‘bigger’ field $K$ to the ‘smaller’ field $F$ to get a new sub-field (of course it is possible that $K=F[\alpha]$)

\begin{array}{c}
K\\
\vert\\
F[\alpha]\\
\vert\\
F
\end{array}

## The minimal polynomial

So we now know how to adjoin an element to a field to create a bigger field, however, we also know that the new field, $F[\alpha]$ is a vector space over $F$, but what do we know about it? what is it’s dimention? From out construction, we’ve seen that $F[\alpha]\cong F[x]/\langle h\rangle$ where $h$ is a polynomial that generates the kernel, thus, it satisfies $h(\alpha)=0$. That’s actually helpful, since we know what is the dimention of the vector space $F[x]/\langle h\rangle$ – it is the degree of the polynomial $h$!

So now we know that every time we are adjoining an element to a field, we also need to find this specific polynomial $h$ in order to know the dimention of our new vector space. So let’s find out what properties $h$ has.

First of all, I want $h$ to be a monic polynomial – i.e. it’s leading coefficient is 1 (I can assume that – indeed, if $h$ is not monic, we can just divide it by it’s leading coefficient to get a monic polynomial). We also know that since $F[x] / \langle h \rangle$ is a field, then $\langle h \rangle$ is maximal, thus $h$ is irreducible. Okay, now let’s suppose that $\alpha$ satisfies $f\in F[x]$, thus $f\in\ker\Phi=\langle h\rangle$. Therefore, $h|f$, and we can now conclude that $\deg h\leq \deg f$. So $h$ is the polynomial that satisfies $\alpha$ with the lowest degree. Let’s see what we have found out so far:

• $h$ is monic.
• $h$ is irreducible.
• $h$ is the polynomial with the lowest degree that satisfies $\alpha$.

Since $h$ is monic, then it is unique, and the other 2 properties, kind of make $h$ look small / minimal. This leads us to a definition. We say that the polynomial $h$ is the minimal polynomial of $\alpha$, and I’ll denote it by $m_\alpha(x)$, as we’ve seen, this is a pertty important polynomial. Let’s see a concrete example.

Consider the fields $K=\mathbb{R}$ and $F=\mathbb{Q}$, and let’s pick $\alpha = \sqrt[3]{5}$.

From now on I’ll denote the dimention of the field $F[\alpha]$ over $F$ as $[F[\alpha]:F]$ and I’ll call it the degree of the extenstion.

So I want to find the degree of the extention. In order to do so, we need to find the minimal polynomial $m_{\sqrt[3]{5}}$. Let’s try to find it:

\alpha=\sqrt[3]{5}\Rightarrow \alpha^3=5\Rightarrow\alpha^3-5=0

So it seems like $f(x)=x^3-5$ is a good candidate to be the minimal polynomial. This polynomial is monic, $f(\alpha)=0$ and irreducible (This is an eisenstien polynomial with $p=5$ – If you have no idea what I am talking about, don’t worry, I’ll write about it in the future when I’ll start writing posts about ring theory). Thus, it is indeed the minimal polynomial! Since it is of degree 3, we conclude that $[\mathbb{Q}(\alpha):\mathbb{Q}]=3$.

## Creating roots out of nowhere

So far we’ve discussed a way to create field using an existing element from a ‘bigger’ field. Unfortunately, we don’t always have such a field to take elements from, so what can we do? We actually talked about it before, we can create field by dividing the ring of polynomials by an irreducible polynomial. However, it turns out that this process actually generates the element we want:

Suppose that $F$ is a field and $h\in F[x]$ is irreducible. Let $K=F[x]/\langle h\rangle$ be the field quotient ring, which is actually a field. now I will denote $\alpha = x+\langle h\rangle \in K$, and let’s check what is the value of $h(\alpha)$, we will write first $h(x)=\sum_{i=0}^na_ix^i$, then:

h(\alpha)=\sum_{i=0}^na_i\alpha^i=\sum_{i=0}^na_i(x+\langle h\rangle)^i=\sum_{i=0}^na_i(x^i+\langle h\rangle)
=\sum_{i=0}^na_ix^i+\langle h\rangle=h+\langle h\rangle=0 +\langle h\rangle

We got that $\alpha = x +\langle h \rangle$ is a root of $h$ in the new field $K$. Therefore, for any irreducible polynomial $f$ in a field $F$ we can find a field $K$ that extends $F$ such that $f$ has a root in it!

At first sight it seems like cheating, is seems like I decided that I want the polynomial to have a root so I just made one. However, go over it again and observe carefully how Iv’e gotten to it and make sure you understand every step here. After you do so, you will understand that this is a real consturction that is valid and nothing here is cheating.

\begin{array}{c}
f\in F[x]\text{ is irreducible}\\
\downarrow\\
K=F[x]/\langle f(x)\rangle\text{ is a field}\\
\downarrow\\
f(x+\langle f(x)\rangle)=0_{K}\\
\downarrow\\
\alpha=x+\langle f(x)\rangle\text{ is a root of } f \text{ in } K
\end{array}

In my opinion, this is a really elegant way to ‘generate’ roots to a polynomial, and I would like to show one famous example for it:

Consider the field $F=\mathbb{R}$. As you know, there is not real root to the polynomial $f(x)=x^2+1\in \mathbb{R}[x]$. So let’s take look at the quotient ring:

K=\mathbb{R}[x]/\langle x^2+1 \rangle

we know that $\alpha = x+ \langle f \rangle$ is a root for $f$ in $K$. Thus, $f(\alpha)=\alpha ^2 + 1 = 0$ or $\alpha ^ 2 = -1$. Looks familiar? We just constructed the field of complex numbers $\mathbb{C}$. This is one of the many ways to construct this field, I definitly like it, although, it is not my favourite (my favourite is done by more geometric way – I hope to present it in the future when I’ll write about complex analysis), but it is valid, and it works, and it solves our problem (finding a root for an irrecducible polynomial), and it gives us information on the extesntion itself: Since the minimal polynomial is $x^2+1$ (Why?) then we know that $[\mathbb{C}:\mathbb{R}]=2$.

I think that’s enough to process for one post, in the next post I will start by showing how to calculate inverse elements in the field $F[x]/\langle h \rangle$, and I’ll start talking about splitting fields.