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Section 8.3 Extension Fields

Recall that we say a field \(E\) is an extension field of a field \(F\) if \(F\) is a subfield of \(E\text{.}\) The field \(F\) is called the base field. We write \(F \subset E\text{.}\) Or goal is to build extension fields so that particular polynomials that were previously irreducible can now be factored in the extension field.

One obvious way to ensure this happens is to include the root of that polynomial in the field. We previously defined \(E = \Q(\sqrt{2}) = \{a + b\sqrt{2} \st a, b \in \Q\}\) as an extension field of \(\Q\text{.}\) In this field, the polynomial \(x^2 - 2\) has a root, so can be factored.

In general, given a polynomial \(p(x) \in F[x]\) for some field \(F\text{,}\) and a root \(\alpha\text{,}\) we want to build an extension field containing \(\alpha\) so that \(p(x)\) factors. We could do the same thing as above, taking \(F(\alpha)\text{.}\) That is, add a new element \(\alpha\) to \(F\) and close the set under addition and multiplication. The question is, will this always be a field?

Our salvation comes from quotient rings. If it has been a while, you might review Section 5.3. Then try the following activity, which works through the general strategy in the simple example of \(\alpha = \sqrt{2}\text{.}\)

Worksheet 8.3.1 Activity: Quotient Rings that are Extension Fields
A4US

Let's explore the connection between extension fields and quotient rings. We will see that \(F[x]/\langle p(x) \rangle \cong F(\alpha)\text{,}\) where \(\alpha\) is a root of its minimal polynomial \(p(x)\) (i.e., \(p(x)\) is the unique monic irreducible polynomial that has \(\alpha\) as a root). The goal of this activity is to see how working in quotient rings help us realize \(E = F(\alpha)\) as a field.

We will start easy. For now, let \(E = \Q(\sqrt{2})\text{.}\)

1.

What quotient ring is \(E\) isomorphic to?

Solution.

An irreducible polynomial for \(\alpha = \sqrt{2}\) is \(p(x) = x^2 -2\text{,}\) so \(\Q(\sqrt{2}) \cong \Q[x]/\langle x^2 - 2 \rangle\text{.}\)

2.

One element in \(E\) is \(1+3\sqrt{2}\text{.}\) What element in the quotient ring does this correspond to?

Solution.

This will be the coset \(1+3x + \langle x^2 - 2\rangle\text{.}\)

3.

What will \(\gcd(3x+1, x^2 - 2)\) be? How do you know? Then verify you are correct using the Euclidean algorithm.

Solution.

The gcd will be 1, since \(x^2 - 2\) is irreducible. We have \(x^2 - 2 = (\frac{1}{3}x -\frac{1}{9}(3x+1)) - \frac{17}{9}\text{.}\) The next step will necessarily give us zero remainder, so the gcd is \(-\frac{17}{9}\text{.}\) But we can multiply by any constant, so we also have gcd equal to 1.

4.

Bezout's identity says that for any polynomials \(a(x)\) and \(b(x)\text{,}\) there are polynomials \(s(x)\) and \(t(x)\) such that

\begin{equation*} \gcd(a(x), b(x)) = s(x)a(x) + t(x)b(x)\text{.} \end{equation*}

Find \(s(x)\) and \(t(x)\) in our case, by working backwards from the Euclidean algorithm above.

Solution.

Go backwards to get \(1 = s(x)(x^2-2)+t(x)(3x+1)\text{.}\) In fact, we get \(s(x) = -\frac{9}{17}\) and \(t(x) = \frac{3}{17} x - \frac{1}{17}\text{.}\)

5.

What does Bezout's identity have to do with the expression

\begin{equation*} 1+\langle x^2 - 2\rangle = (3x+1 + \langle x^2 - 2\rangle)(t(x) + \langle x^2 - 2\rangle) \end{equation*}

and what does this have to do with finding inverses? In particular, what is \((1+3\sqrt{2})\inv\) in \(E\text{?}\)

Solution.

Notice that \(s(x)(x^2 - 2) \in \langle x^2 - 2\rangle\) and so what we have here shows us that \(1 \in \langle x^2 - 2\rangle + (\frac{3}{17} x - \frac{1}{17})(3x + 1) = (\langle x^2 - 2\rangle + \frac{3}{17} x - \frac{1}{17})(\langle x^2 - 2\rangle + 3x+1)\text{.}\) Thus \(\langle x^2 - 2\rangle + \frac{3}{17} x - \frac{1}{17}\) is the multiplicative inverse of \(\langle x^2 - 2\rangle 3x + 1\text{.}\) Going back to \(E\text{,}\) we see that the inverse of \(1+3\sqrt{2}\) must therefore be \(-\frac{1}{17} + \frac{3}{17}\sqrt{2}\text{.}\) This will always work, because for any coset \(\langle x^2 - 2 \rangle + r(x)\text{,}\) with \(r(x)\) not a multiple of \(x^2 - 2\) (so for any non-zero coset), the gcd of \(x^2 - 2\) and \(r(x)\) will be 1, so we can find \(s(x)\) and \(t(x)\) such that \(1 = s(x)(x^2 - 2) + t(x)r(x)\text{.}\) Then \(\langle x^2 - 2\rangle + t(x)\) will be the multiplicative inverse.

6.

Now let's try this again with a more complicated polynomial. As in the earlier activity, take \(p(x) = x^3 + 3x^2 - x + 2\) and let \(\s\) be a root. Use quotient rings to find the inverse of the element \(2 + 3\s^2\) in \(E = \Q(\s)\text{.}\)

Solution.

We work in \(\Q[x]/\langle p(x) \rangle\text{,}\) where \(p(x) = x^3 + 3x^2 - x + 2\text{.}\) We must find a polynomial \(t(x) \in \Q[x]\) such that

\begin{equation*} 1 + \langle p(x) \rangle = (2+3x^2)(t(x)) + \langle p(x) \rangle\text{,} \end{equation*}

which is equivalent to finding polynomials \(t(x)\) and \(s(x)\) such that

\begin{equation*} 1 = s(x)p(x) + t(x)(2+3x^2)\text{.} \end{equation*}

We know such \(s(x)\) and \(t(x)\) exist by Bezout's theorem and the fact that \(p(x)\) is irreducible, so \(\gcd(p(x), 2+3x^2) = 1\text{.}\)

The Euclidean Algorithm gives us:

\begin{align*} x^3 + 3x^2 - x + 2 \amp = \left(\frac{x}{3}+1\right)\left(3x^2 + 2\right) + \frac{-5x}{3} \\ 3x^2 + 2 \amp = \left(\frac{-9x}{5}\right)\left(\frac{-5x}{3}\right) + 2\text{.} \end{align*}

Then working backwards we find:

\begin{align*} 1 \amp = \frac{1}{2}(3x^2 + 2) + \left(\frac{9x}{10}\right)\left(\frac{-5x}{3}\right) \\ 1 \amp = \frac{1}{2}(3x^2+2) +\left(\frac{9x}{10}\right)\left((x^3 + 3x^2 - x + 2) - \left(\frac{x}{3}+1\right)\left(3x^2 + 2\right) \right) \\ 1 \amp = \left(\frac{1}{2} - \frac{9x}{10}\frac{x}{3} - \frac{9x}{10}\right)(3x^2 + 2) + \left(\frac{9x}{10}\right)(x^3 + 3x^2 - x + 2) \text{.} \end{align*}

So we have \(s(x) = \frac{9x}{10}\) and crucially \(t(x) = \frac{1}{2} - \frac{9}{10}x - \frac{3}{10}x^2\text{.}\)

Shifting back to \(E\text{,}\) we see that the inverse of \(2+ 3\s^2\) is \(\frac{1}{2} - \frac{9}{10}\s - \frac{3}{10}\s^2\)

Subsection 8.3.2 Extension Fields and Polynomials

To summarize our strategy: we want to build an extension field of \(F\) that contains an element \(\alpha\text{,}\) presumably not already in \(F\text{.}\) As long as \(\alpha\) is the root of a polynomial, we can take \(p(x)\) to be an irreducible polynomial (over \(F\)) that has \(\alpha\) as a root, and consider the quotient ring \(F[x]/\langle p(x) \rangle\text{.}\) Using the Fundamental Homomorphism Theorem, this quotient ring will be isomorphic to our desired extension field.

That is precisely the content of the statement and proof of the following theorem, due to Kronecker. The result is so important and so basic to our understanding of fields that it is often known as the Fundamental Theorem of Field Theory.

Proof.

We can assume that \(p(x)\) is an irreducible polynomial (otherwise, just factor it and work with an irreducible part that has \(\alpha\) as a root). We wish to find an extension field \(E\) of \(F\) containing an element \(\alpha\) such that \(p(\alpha) = 0\text{.}\) The ideal \(\langle p(x) \rangle\) generated by \(p(x)\) is a maximal ideal in \(F[x]\) by Theorem 6.29; hence, \(F[x]/\langle p(x) \rangle\) is a field. We claim that \(E = F[x]/\langle p(x) \rangle\) is the desired field.

We first show that \(E\) is a field extension of \(F\text{.}\) We can define a homomorphism of commutative rings by the map \(\psi:F \rightarrow F[x]/\langle p(x) \rangle\text{,}\) where \(\psi(a) = a + \langle p(x)\rangle\) for \(a \in F\text{.}\) It is easy to check that \(\psi\) is indeed a ring homomorphism. Observe that

\begin{equation*} \psi( a ) + \psi( b ) = (a + \langle p(x) \rangle) + (b + \langle p(x) \rangle) = (a + b) + \langle p(x) \rangle = \psi( a + b ) \end{equation*}

and

\begin{equation*} \psi( a ) \psi( b ) = (a + \langle p(x) \rangle) (b + \langle p(x) \rangle) = ab + \langle p(x) \rangle = \psi( ab )\text{.} \end{equation*}

To prove that \(\psi\) is one-to-one, assume that

\begin{equation*} a + \langle p(x) \rangle = \psi(a) = \psi(b) = b + \langle p(x) \rangle\text{.} \end{equation*}

Then \(a - b\) is a multiple of \(p(x)\text{,}\) since it lives in the ideal \(\langle p(x) \rangle\text{.}\) Since \(p(x)\) is a nonconstant polynomial, the only possibility is that \(a - b = 0\text{.}\) Consequently, \(a = b\) and \(\psi\) is injective. Since \(\psi\) is one-to-one, we can identify \(F\) with the subfield \(\{ a + \langle p(x) \rangle : a \in F \}\) of \(E\) and view \(E\) as an extension field of \(F\text{.}\)

It remains for us to prove that \(p(x)\) has a zero \(\alpha \in E\text{.}\) Set \(\alpha = x + \langle p(x) \rangle\text{.}\) Then \(\alpha\) is in \(E\text{.}\) If \(p(x) = a_0 + a_1 x + \cdots + a_n x^n\text{,}\) then

\begin{align*} p( \alpha ) & = a_0 + a_1( x + \langle p(x) \rangle) + \cdots + a_n ( x + \langle p(x) \rangle)^n\\ & = a_0 + ( a_1 x + \langle p(x) \rangle) + \cdots + (a_n x^n + \langle p(x) \rangle)\\ & = a_0 + a_1 x + \cdots + a_n x^n + \langle p(x) \rangle\\ & = 0 + \langle p(x) \rangle\text{.} \end{align*}

Therefore, we have found an element \(\alpha \in E = F[x]/\langle p(x) \rangle\) such that \(\alpha\) is a zero of \(p(x)\text{.}\)

We will often take \(F = \Q\text{,}\) but of course other fields can be used as well and our strategy of using quotient rings will still work. The following two examples show what can happen. They are not essential for understanding what we do next, but serve as a neat example of how you can apply what we are doing in a new context.

Example 8.11.

Let \(p(x) = x^2 + x + 1 \in {\mathbb Z}_2[x]\text{.}\) Since neither 0 nor 1 is a root of this polynomial, we know that \(p(x)\) is irreducible over \({\mathbb Z}_2\text{.}\) We will construct a field extension of \({\mathbb Z}_2\) containing an element \(\alpha\) such that \(p(\alpha) = 0\text{.}\) By Theorem 6.29 , the ideal \(\langle p(x) \rangle\) generated by \(p(x)\) is maximal; hence, \({\mathbb Z}_2[x] / \langle p(x) \rangle\) is a field. (That modding out by a maximal ideal produces a fields is a result that applies to rings in general, not just polynomial rings; we could have argued that \(\Z_2[x]/\langle p(x)\rangle\) is a field using Bezout's identity as well.)

Let \(f(x) + \langle p(x) \rangle\) be an arbitrary element of \({\mathbb Z}_2[x] / \langle p(x) \rangle\text{.}\) By the division algorithm,

\begin{equation*} f(x) = (x^2 + x + 1) q(x) + r(x)\text{,} \end{equation*}

where the degree of \(r(x)\) is less than the degree of \(x^2 + x + 1\text{.}\) Therefore,

\begin{equation*} f(x) + \langle x^2 + x + 1 \rangle = r(x) + \langle x^2 + x + 1 \rangle\text{.} \end{equation*}

The only possibilities for \(r(x)\) are then \(0\text{,}\) \(1\text{,}\) \(x\text{,}\) and \(1 + x\text{.}\) Consequently, \(E = {\mathbb Z}_2[x] / \langle x^2 + x + 1 \rangle\) is a field with four elements and must be a field extension of \({\mathbb Z}_2\text{,}\) containing a zero \(\alpha\) of \(p(x)\text{.}\) The field \({\mathbb Z}_2( \alpha)\) consists of elements

\begin{align*} 0 + 0 \alpha & = 0\\ 1 + 0 \alpha & = 1\\ 0 + 1 \alpha & = \alpha\\ 1 + 1 \alpha & = 1 + \alpha\text{.} \end{align*}

Notice that \({\alpha}^2 + {\alpha} + 1 = 0\text{;}\) hence, if we compute \((1 + \alpha)^2\text{,}\)

\begin{equation*} (1 + \alpha)(1 + \alpha)= 1 + \alpha + \alpha + (\alpha)^2 = \alpha\text{.} \end{equation*}

Other calculations are accomplished in a similar manner. We summarize these computations in the following tables, which tell us how to add and multiply elements in \(E\text{.}\)

Addition Table for \({\mathbb Z}_2(\alpha)\text{:}\)

\begin{equation*} \begin{array}{c|cccc} + & 0 & 1 & \alpha & 1 + \alpha \\ \hline 0 & 0 & 1 & \alpha & 1 + \alpha \\ 1 & 1 & 0 & 1 + \alpha & \alpha \\ \alpha & \alpha & 1 + \alpha & 0 & 1 \\ 1 + \alpha & 1 + \alpha & \alpha & 1 & 0 \end{array} \end{equation*}

Multiplication Table for \({\mathbb Z}_2(\alpha)\text{:}\)

\begin{equation*} \begin{array}{c|cccc} \cdot & 0 & 1 & \alpha & 1 + \alpha \\ \hline 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & \alpha & 1 + \alpha \\ \alpha & 0 & \alpha & 1 + \alpha & 1 \\ 1 + \alpha & 0 & 1 + \alpha & 1 & \alpha \end{array} \end{equation*}
Example 8.12.

Let \(p(x) = x^5 + x^4 + 1 \in {\mathbb Z}_2[x]\text{.}\) Then \(p(x)\) has irreducible factors \(x^2 + x + 1\) and \(x^3 + x + 1\text{.}\) For a field extension \(E\) of \({\mathbb Z}_2\) such that \(p(x)\) has a root in \(E\text{,}\) we can let \(E\) be either \({\mathbb Z}_2[x] / \langle x^2 + x + 1 \rangle\) or \({\mathbb Z}_2[x] / \langle x^3 + x + 1 \rangle\text{.}\) We will leave it as an exercise to show that \({\mathbb Z}_2[x] / \langle x^3 + x + 1 \rangle\) is a field with \(2^3 = 8\) elements.

Exercises 8.3.3 Collected Homework

1.

The number \(\phi = \frac{1+\sqrt{5}}{2}\) (which happens to be the Golden Ratio) is not an element of \(\Q\text{,}\) but of course is an element of \(\Q(\phi)\text{.}\) Let's explore that field a bit.

(a)

Find a polynomial with rational coefficients that has \(\phi\) as a root. Call this polynomial \(p(x)\text{.}\)

Hint.

Try setting \(\phi = x\) and do some algebra to get rid of any square roots.

(b)

What quotient ring is isomorphic to the field \(\Q(\phi)\text{?}\) List two non-zero elements of that quotient ring and say which elements of \(\Q(\phi)\) they correspond to.

(c)

Which coset does \(\phi\) itself correspond to? Take this coset and plug it into \(p(x)\) to verify that the coset is a root of \(p(x)\) (show your work).

(d)

The element \(\phi^3\) is an element of \(\Q(\phi)\) but can be expressed as \(a+b\phi\) for \(a, b \in \Q\text{.}\) Find \(a\) and \(b\text{.}\) Then find the corresponding element for both \(\phi^3\) and \(a+b\phi\) in the quotient ring and check that these two cosets are the same as each other.