Algebraic Extensions of Z_7 with polynomials?

Question

I am studying fields and rings and I came across this statement in a textbook that I am having trouble visualising;

$\mathbb{Z_7}/\left<x^2-3\right>$ is an algebraic extension of $\mathbb{Z_7}$

The book mentions finding polynomials in the field that have its roots in the extension. For example, I can see why $\mathbb{Q}(\sqrt{2})$ is an algebraic extension of $\mathbb{Q}$ since the polynomial $h(x)=x^2-2$ is a non-zero polynomial in $\mathbb{Q[x]}$ with $h(\sqrt{2})=0$ but trying to find a similar non-zero polynomial in $\mathbb{Z_7}$ to show the above is proving more difficult and I cant think of one at all. Any help much appreciated, thank you

Answer 1

Assuming that you do mean $\mathbb{Z}_7[x]/\langle x^2 - 3 \rangle$, I'll try to explain what this field is.

Edit: tl;dr is that $x + \langle x^2 - 3 \rangle$ is a root of the polynomial $t^2 - 3 \in \mathbb{Z}[t]$, but I've tried to give some additional explanation because these ideas can be quite hard to grasp.

First of all, note that $x^2 - 3$ is an irreducible polynomial in $\mathbb{Z}_7[x]$, since it has no roots in $\mathbb{Z}_7$. Since $\mathbb{Z}_7$ is a field, $\mathbb{Z}_7[x]$ is principal ideal domain, which means that $\langle x^3 - 3\rangle$ is a maximal ideal (since it is generated by an irreducible element of the pid). It is a standard result that the quotient of a ring by a maximal ideal is always a field, so indeed $\mathbb{Z}_7[x]/\langle x^2 - 3 \rangle$ is a field.

But what actually is it? This is a bit harder to answer than you example of $\mathbb{Q}(\sqrt{2})$, because in that case $\sqrt{2}$ actually exists as one of the real numbers, so we are naturally comfortable adjoining it to $\mathbb{Q}$ and working with it. However, the key observation is that the properties of $\sqrt{2}$ all come merely from the fact that it is some element whose square is $2$.

Let E = $\mathbb{Z}_7[x]/\langle x^2 - 3 \rangle$, and define $\alpha \in E$ to be the element $x + \langle x^2 - 3 \rangle$. We then have by definition of quotient rings that $$ \alpha ^2 - 3 = (x + \langle x^2 - 3\rangle)^2 - (3 + \langle x^2 - 3\rangle) = x^3 - 3 + \langle x^2 - 3 \rangle = 0 + \langle x^2 - 3 \rangle $$

So $\alpha$ is actually a root of the polynomial $t^2 - 3 \in \mathbb{Z}_7[x]$. The natural inclusion of $\mathbb{Z}_7$ into $E$ allows us to view $\mathbb{Z}_7$ as a subfield of $E$, and thus think of $E$ as the field we obtain by adjoining this mysterious $\alpha$ to $K$. However, the key point is that $\alpha^2 = 3$, so $\alpha$ has all the same properties of a regular square root of $3$, since the only property we care about in a square root of $3$ is that its square is $3$. So in fact the field extension is basically equivalent to adjoining a square root of $3$ to $\mathbb{Z}_7$.

Where is this square root of $3$, you may ask. This is exactly the same as asking where $i$ is in relation to the real numbers. It doesn't matter where it is. It "just exists". So that's all we're doing here: we are "imagining" some element extending $\mathbb{Z}_7$ with $\alpha^2 = 3$, just as we might "imagine" some element $i$ extending $\mathbb{R}$ with $i^2 = -1$.

Note: This "just exists" argument is most certainly not a substitute for rigorous definitions. It is merely an aide for thinking about definitions that are initially very confusing.

Answer 2

Observe that the polynomial $p(x) = x^2 - 3$ is irreducible over $\mathbb Z_7 \stackrel{\text{def}}{=}\mathbb Z / 7 \mathbb Z$ since it is a quadratic polynomial with no roots in $\mathbb Z_7.$ (Indeed, one can check that $p(n) \neq 0$ for any integer $0 \leq n \leq 6.$) Consequently, we have that $\langle p(x) \rangle = \langle x^2 - 3 \rangle$ is a prime (and hence maximal) ideal in the PID (principal ideal domain) $\mathbb Z_7[x],$ from which it follows that $$R = \frac{\mathbb Z_7[x]}{\langle p(x) \rangle} = \frac{\mathbb Z_7[x]}{\langle x^2 - 3 \rangle}$$ is a field extension of $\mathbb Z_7.$ Observe that (by definition) every element of $R$ can be written as the class of some polynomial $ax + b$ modulo $p(x),$ i.e., $\overline{ax + b} = a \bar x + b \bar 1$ for some scalars $a, b \in \mathbb Z_7,$ hence we have that $\{\bar 1, \bar x\}$ forms a basis for $R$ as a $\mathbb Z_7$-vector space. We conclude therefore that $R$ is a 2-dimensional $\mathbb Z_7$-vector space. Every finite field extension is an algebraic field extension.

Answer 3

$x^2-3$ is irreducible over $\mathbb{Z}_7$. Hence, $ \frac{\mathbb{Z}_7}{<x^2-3>}$ is a field. The quotient map $\pi: \mathbb{Z}_7 \to \frac{\mathbb{Z}_7}{<x^2-3>}$ is a field homomorphism. Hence, $\frac{\mathbb{Z}_7}{<x^2-3>}$ is a finite extension of degree 2, therefore, algebraic.$