4

1) Let $K/k$ an extension field. If $[K:k]=2$ show that there is $a\in K$ s.t. $K=k(\alpha )$ and $\alpha ^2=a$.

2) Does it work for $[K:k]=p$ when $p$ is prime ? Or for other number ? (like non-prime)

My work

For 1), if $[K:k]=2$ there is an $a\in K$ that is not in $k$. Let $X^2-\alpha $ its minimal polynomial. Then $$a^2-\alpha =0\implies (a-\sqrt \alpha )(a+\sqrt \alpha )=0\implies a=\pm\sqrt \alpha $$

It looks to be very confusing... any idea ?

Jyrki Lahtonen
  • 133,153
MSE
  • 3,153
  • 1
    Why is the minimum polynomial of $\alpha$ necessarily $X^2 - \alpha$? What about, for example, the extension $\mathbb{F}_4/\mathbb{F}_2$? – anomaly Dec 16 '15 at 21:45
  • $\mathbb F_4=\mathbb Z/4\mathbb Z$ is not a field... But anyway, by definition if $L/K$ is an extension of degree $n$ there is automatically an element of $L$ of degree $n$, and thus if $\beta $ is such an element, its minimal polinomial is of the form $X^n+a_{n-1}X^{n-1}+...+a_0\in K[X]$, isn't it ? – MSE Dec 16 '15 at 22:11
  • 1
    $\mathbb{F}_4$ is the field of order $4$, not $\mathbb{Z}/4\mathbb{Z}$. – anomaly Dec 16 '15 at 22:24
  • I actually see what you mean... in fact it could be $X^2+aX+b$ the minimal polynomial, right ? So, I don't know way it is $X^2-a$, my teacher told me but I don't understand why. – MSE Dec 16 '15 at 23:04
  • Anomaly (and Tsemo Aristide) got it right. This is true only if $k$ is not of characteristic two. The simplest counterexample is the extension $\Bbb{F}_4/\Bbb{F}_2$. The two elements of $\Bbb{F}_4$ that are not also elements of $\Bbb{F}_2$ are each others squares - settling the argument immediately. – Jyrki Lahtonen Dec 17 '15 at 06:03

2 Answers2

2

You have to be careful in the logical order of your arguments here.

It is correct to say that $[K:k] = 2$ implies there is an $a \in K \setminus k$. If its minimal polynomial is $X^2 - \alpha$ for $\alpha \in k$ then you certainly have that $a^2 - \alpha = 0$. This implies that $a^2 = \alpha$. This means that $a$ is a square root of $\alpha$ and that $-a$ is another (indeed, the only other) square root of $\alpha$ in $K$. At this point, however, the symbol $\sqrt{\alpha}$ is meaningless, believe it or not. You have not chosen any embedding of $K$ into a field where $\sqrt{\alpha}$ has already been defined (for example, some algebraic closure of $k$) and so it is not yet actually defined. You could define $\sqrt{\alpha} := a$ in $K$ and then you would of course have $-\sqrt{\alpha} = -a$ but there is little to be gained from this.

Instead of all of that, since it is clear that there is some element in $K\setminus k$ (namely $a$), your only mission is to show that in fact $a$ generates the extension - i.e. $K = k(a)$. This will be true because $K$, being a finite field extension of $k$ of degree 2, is necessarily a $k$-vector space of dimension $2$. Since $a \not\in k$, it is linearly independent from $1 \in k$ and so $\{1,a\}$ is a maximally linearly independent set in the vector space $K$ and is thus a basis. This means every element $b \in K$ can be written as $b = c_1\cdot 1 + c_2\cdot a$ for $c_1, c_2 \in k$. Hence $K = k(a)$.

As for your second question, I think the answer to be given here is that for 'nice' extensions (those being finite, separable ones), the statement remains true. That is, an extension $K/k$ is generated by a single element (called a 'primitive element'). This result is known as the Primitive Element Theorem. The caution is that for extension degree higher than 2, a primitive element need not be a $d^{th}$ root of an element of the base field (for $d$ the degree of the extension).

john
  • 2,376
1

Suppose the characteristic is not 2 and $[K:k]=2$. Let $x\in K$, $x$ is not in $k$, $1,x$ is a basis of the 2-dimensional space $K$, $x^2=ax+b, a,b\in K$, $x^2-ax-b=(x-a/2)^2-a^2/4-b=0$ thus $(x-a/2)^2=b+a^2/4$. We have $K=k(x-a/2)$ and $(x-a/2)^2\in k$.

Tsemo Aristide
  • 87,475