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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 \(\frac{1}{2} - \frac{9}{10}\s - \frac{3}{10}\s^2\)