Constructing a field so that a polynomial has a root.

Question

This was on my study guide given out by my professor. I could not find this anywhere in my book though. Any advice on how to do this. I understand the concept that a polynomial may not have a root over one field, but have a root over another. But what does he mean when he says "construct a field"? Is there some sort of homomorphism that must be constructed? Any advice would be appreciated. Thanks.

Answer 1

You define the algebraic extension $F/K$: $$F=K(\alpha)= K[x]/\langle p(x)\rangle$$

Where $\alpha = x+\langle p(x)\rangle$

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Note that: $$p(\alpha) = \sum_{i=1}^n a_i \alpha^i= \sum_{i=1}^n a_i(x+\langle p(x)\rangle)^i=\sum_{i=1}^n a_ix^i + \langle p(x)\rangle = p(x)+\langle p(x)\rangle = 0+\langle p(x)\rangle$$

Hence $p(\alpha)=0$

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When $p(x)$ is irreducible, $R[x]/\langle p(x)\rangle$ is a field, where $p(x)$ is maximal, which it will be since it is irreducible.

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Classical Example:

$f(x)=x^2+1$ is irreducible over $\Bbb R[x]$ since it has no real roots.

Let $L = \Bbb R[x]/\langle x^2+1\rangle$, then the elements of $L$ are cosets, $a+\langle x^2+1\rangle$. I.e. $3x^3+2x^2+2+\langle x^2+1 \rangle = 3x^3+2(x^2+1)+\langle x^2+1\rangle = 3x^2$.

Anyway, back to the main discussion:

The elements of $L$ are of the form $f(x)+\langle x^2+1\rangle$ where $f(x)\in \Bbb R[x]$ we have the homomorphism $i:\Bbb R\to L$, $i:a\mapsto a+\langle x^2+1\rangle$. So we have that $\Bbb R\cong i(\Bbb R)$ is embedded in $L$.

$$i(\Bbb R) = \{ a+\langle x^2+1\rangle: a\in \Bbb R\} \subseteq L$$

Now by Kroneckers theorem, $L$ is a $\Bbb R$-vectorspace, with scalar multiplication:

$$a\cdot (f(x)+\langle x^2+1\rangle) = i(a)(f(x)+\langle x^2+1\rangle)$$ $$=(a+\langle x^2+1\rangle)(f(x)+\langle x^2+1\rangle)= af(x) + \langle x^2+1\rangle$$

Let $f(x)\in \Bbb R[x]$. We use the division algorithm to write $f=q(x)(x^2+1) + (a+bx)$ for some $q(x)\in \Bbb R[x]$ and $a,b\in \Bbb R$ so we get the typical element of $L$ to look like $a+bx+\langle x^2+1\rangle$.

$$L=\{ (a+bx)+\langle x^2+1\rangle: a,b\in \Bbb R\}$$

Now we denote $1_L = 1+\langle x^2+1 \rangle$ and $\alpha = x+\langle x^2+1\rangle\in L$

$$\alpha^2 = (x+\langle x^2+1\rangle)(x+\langle x^2+1\rangle) = x^2+\langle x^2+1\rangle = (x^2+1)-1 + \langle x^2+1\rangle = -1_L$$

What have we shown? $L = \{ a\cdot 1_L + b\cdot \alpha: a,b\in \Bbb R, \alpha^2 = -1_L \} \cong \Bbb C$

Note, it might be very easy to see if you write $1_L=1$ and $\alpha = i$, then $$L = \{ a + bi: a,b\in \Bbb R, i^2 = -1 \} $$

Answer 2

You can always get a field from a commutative ring considering the quotient of the ring by a maximal ideal. In this case you can consider:

$$F=K[x]/I$$

Where $I$ is the ideal generated by $p$. You would have to prove that it is a maximal ideal because $p$ is irreducible.

$p$ has at least one root in this new field.