Worksheet 8.2.4 Activity: Extending Fields to Factor
Goal: Build the smallest field possible in which \(p(x) = x^3 + 3x^2 - x + 2\) is NOT irreducible.
Note that \(p(x)\) is irreducible over \(\Q\) because it has no roots in \(\Q\) (why is this and why is that enough?). So let's invent a new number, call it \(\s\text{,}\) and insist that \(\s\) is a root of \(p(x)\text{.}\) Then consider the smallest field \(E\) larger than \(\Q\) that also contains \(\s\text{.}\)
1.
List five elements in \(E\) that are NOT already in \(\Q\text{.}\)
For example, \(\s\) (which is not in \(\Q\) because, by the rational roots theorem, there are no roots to \(p(x)\) in \(\Q\)), as well as \(\s+1\text{,}\) \(4\s\text{,}\) \(\s^2\text{,}\) and \(3+\frac{1}{2}\s - \s^2\text{.}\)
2.
The element \(\s^3\) is in \(E\text{,}\) but this can also be written using smaller powers of \(\s\text{.}\) How?
Since \(\s^3 + 3\s^2 - \s + 2 = 0\) we see that \(\s^3 = -3\s^2 + \s - 2\text{.}\)
3.
Describe \(E\) as a set using set builder notation. In other words, \(E\) is the set of all elements of the form …
\(E = \{a + b \s + c \s^2 \st a, b, c, \in \Q\}\text{.}\) Certainly everything of this form is in \(E\) since \(E\) must be closed under addition and multiplication. That nothing else is is a consequence of the previous question: since \(\s^3\) can already be expressed as a linear combination of \(1\text{,}\) \(\s\text{,}\) and \(\s^2\text{,}\) any power of \(\s\) greater or equal to 3 can be as well.
4.
Wait: why are we doing this? Our goal is for \(p(x)\) to factor. Does it? What would one of the factors be?
Yes, we get \(p(x) = (x-\s)(x^2 + (3+\s)x - 1 + 3\s + \s^2)\text{.}\) Do this by long division. The remainder might not look like zero right away, except that it is precisely \(p(\s)\text{.}\) This is not a surprise: the division algorithm says that \(p(x) = b(x)(x-\s) + r(x)\) where \(r(x)\) is either the zero polynomial or has degree 0 (less than the degree of \((x-\s)\)). Now plug in \(\s\) for \(x\text{.}\) Long division is just there to help us find \(b(x)\text{.}\)
5.
Wait again: we want \(E\) to be a field. Is it? What would we need to check?
This is a hard question that we will return to. Note that all we must check still is that every non-zero element in \(E\) has a multiplicative inverse. That is, given \(a+b\s + c\s^2\text{,}\) find an element \(a' + b'\s + c'\s^2\) such that
You could do this algebraically, although it is not an easy computation, and requires you to know that the polynomial \(p(x)\) is irreducible. We will see why later.
6.
List five elements in the quotient ring \(\Q[x]/\langle p(x)\rangle\) (using the same \(p(x)\) from the previous page). Remember, these will all be cosets.
For example, \(x + \langle p(x) \rangle\text{,}\) \(x+1 +\langle p(x) \rangle\text{,}\) \(4x + \langle p(x) \rangle\text{,}\) \(x^2 + \langle p(x) \rangle\text{,}\) and \(3+\frac{1}{2}x - x^2 + \langle p(x) \rangle\text{.}\)
7.
The element \(x^3+\langle p(x) \rangle\) is an element of \(\Q[x]/\langle p(x) \rangle\text{,}\) but it can also be written as a “simpler” coset. How?
Since \(x^3 + 3x^2 - x + 2 \in \langle p(x) \rangle\) we see that \(x^3 + \langle p(x) \rangle = -3x^2 + x - 2 + \langle p(x) \rangle\text{.}\) You can get here either by dividing \(x^3\) by \(p(x)\) and looking for the remainder, or by asking, what would I need to add to \(p(x)\) to get \(x^3\text{.}\) That is because the coset \(x^3 + \langle p(x) \rangle\) contains all the polynomials in \(\Q[x]\) that result from adding \(x^3\) to a multiple of \(p(x)\text{.}\) One of these elements is \(x^3\) (since the polynomial \(0\) is a multiple of \(p(x)\)). So if we can find another polynomial \(a(x)\) such that \(a(x)\) added to a multiple of \(p(x)\) gives \(x^3\text{,}\) then \(a(x) + \langle p(x)\rangle\) will be exactly the same coset as \(x^3 + \langle p(x) \rangle\text{.}\)
8.
Describe \(\Q[x]/\langle p(x) \rangle\) as a set using set builder notation. In other words, this quotient ring is the set of all cosets of the form …
\(\Q[x]/\langle p(x) \rangle = \{a + bx + cx^2 + \langle p(x) \rangle \st a, b, c\in \Q\}\text{.}\) While we can certainly add a polynomial of degree greater to 2 to get a coset, we can always divide it by \(p(x)\) and get a remainder of degree 2 or less.
This looks very much like the definition of \(E\text{,}\) and in fact it suggests that \(E \cong \Q[x]/\langle p(x) \rangle\text{.}\) This is correct, as we could prove using the fundamental homomorphism theorem. To do that, we would need to identify a surjective homomorphism from \(\Q[x]\) onto \(E\) for which \(p(x)\) was the kernel.
9.
Wait: if we want to show that \(E\) is a field, and \(E\) is basically the same as \(\Q[x]/\langle p(x) \rangle\text{,}\) then we could just show \(\Q[x]/\langle p(x) \rangle\) is a field. What would this mean? What do we need to verify?
We need to verify that for any coset \(a+bx + cx^2 + \langle p(x) \rangle\text{,}\) there is a coset \(a' + b'x + c'x^2 + \langle p(x) \rangle\) such that the product of the two cosets is the multiplicative identity coset, namely \(1 + \langle p(x) \rangle\text{.}\) We will see how to do this next.