2
$\begingroup$

I am having a litte trouble to understand what my professor told the last algebra class. Let me explain what I am talking about.

We know that if $R'$ is a ring and $R$ is a subring of it, we define $R[S]$ as being the smallest subring of $R`$ that contains $R$ and $S$. Explicity we can say that $$R[S]=\{f(s_1,...,s_n); f\in R[X_1,...,X_n], s_i \in S, n\in \mathbb{N}\}$$

Similarly if $K/F$ is a field extension and $S\subseteq K$ then $F(S)$ is the smallest subfield of $K$ contaning $S$ and $F$. Explicity $F(S)=Frac F[S]$.

Then, his WARNING: whenever you write $R[a,b,...]$ or $F(a,b,...)$ it is important that you work inside some fixed, specified ring $R`$ for field $F`$. For example, do not write $(\mathbb{Z}/2\mathbb{Z})[\sqrt{2}]$. But $\mathbb{Q}[\sqrt{2}]$ is okay.

I think I understood the warning. I just did not understand why the first example is wrong and the second one is correct. Can someone make it clear for me?

Thank you everybody! Have a good weekend!

2 Answers 2

1

In the second example, when you write $\mathbb Q[\sqrt 2]$ it's implicit that $\sqrt 2$ lies in $\mathbb C$. This is okay because $\mathbb Q$ is a subset of $\mathbb C$, and so $\mathbb Q[\sqrt 2]$ is the smallest subring of $\mathbb C$ containing $\mathbb Q$ and $\sqrt 2$.

However the meaning of the first example $\mathbb F_2[\sqrt 2]$ (where $\mathbb F_2=\mathbb Z/2\mathbb Z$) is unclear, because $\mathbb F_2$ is not a subset of $\mathbb C$, and cannot even be embedded in $\mathbb C$.

  • 0
    In the first example, we are looking for this as a subset of $\mathbb{C}$?? I thought it was of $\mathbb{R}$. – 2017-01-21
  • 0
    On the other hand, it's not uncommon to find something like $\mathbb{F}_3[\sqrt{2}]$, where $\sqrt{2}$ means an element of some extension field such that $(\sqrt{2})^2=2\in\mathbb{F}_3$, that is, a root of $x^2-2\in\mathbb{F}_3[x]$. It's just a question of being aware of what we're writing about. – 2017-01-21
  • 0
    @bttmbrcelo It doesn't matter whether you consider $\mathbb Q$ and $\sqrt 2$ inside $\mathbb C$ or $\mathbb R$, as $\mathbb R\subset\mathbb C$. (Any field containing $\mathbb R$ is fine.) – 2017-01-21
0

$\sqrt{2}$ makes no sense in an extension field of characteristic $2$ (which any extension field of $\Bbb Z/2\Bbb Z$ would necessarily be), since in such a field $2 = 0$.

For the second field, we might take $\sqrt{2}$ to be in $\Bbb C$, or $\Bbb R$, or the field of real algebraic numbers, all of which are legitimate field extensions of $\Bbb Q$.

  • 0
    Well, technically it makes sense, it's just that $\sqrt 2=0$. So if $\mathbb F_2[\sqrt 2]$ is interpreted as the splitting field of $x^2-2$ over $\mathbb F_2$, then $\mathbb F_2[\sqrt 2]=\mathbb F_2$. – 2017-01-21
  • 0
    Good point. It's usually not what we mean by $\sqrt{2}$, and its a trivial extension, but I suppose it technically counts. – 2017-01-22