0

On a field $K$ with $char(K)$ not equal to 2, all irreducible polynomials of a quadratic extension are separable. The proof is straightforward: Assume the opposite, namely $P=X^2+aX+b = (X-\alpha)(X-\beta)$ with $P \in K[X]$, so $a = -\alpha -\beta$. Assuming $\alpha$ and $\beta$ are the same, we get: $a = -2\alpha$. We can solve for $\alpha$ and see that $\alpha \in K$.

With $char(K)=2$, we cannot do that. I figured out that $Q=X^2+X+1$ is irreducible in $\mathbb{F}_2$ and separable because $Q$ and $Q'$ are coprime. How do the roots look like in a splitting field of $Q$?

mdot
  • 947
  • Polynomials over finite fields are always separable. There are non-separable polynomials over fields of characteristic not 2, just not quadratics. – Gerry Myerson Jul 15 '15 at 07:32
  • A polynomial has multiple roots if and only if derivative of that polynomial is zero. – Ripan Saha Jul 15 '15 at 07:49
  • @Gerry Myerson Well, what about $F = a_0 + a_1t^p+... + a_rt^{rp}$ with $K$ field of $char(K) = p$, in our case taking $r = 1$, $p=2$. – mdot Jul 15 '15 at 07:52
  • What about it? If $K$ is a finite field of characteristic 2, then every element of $K$ is a square, and $a+bt^2=(c+dt)^2$, where $a=c^2$ and $b=d^2$, so $a+bt^2$ is not irreducible. – Gerry Myerson Jul 15 '15 at 12:52
  • @GerryMyerson but it gives a counter example that polynomials over finite fields are always separable. – mdot Jul 15 '15 at 13:08
  • I thought we were just talking about irreducible polynomials. – Gerry Myerson Jul 15 '15 at 13:10
  • @GerryMyerson probably my fault. Would you mind telling me how to explicity construct and write down the roots from $X^2+X+1$ in an extension field? How do they look like in a field extension of $\mathbb{F}_{2q}$ for a suiting integer $q$ (made an update on the question) – mdot Jul 15 '15 at 13:28
  • 1
    I'm not quite sure if this question is well-posed. The roots do not lie in $\mathbb{F}_2$, since it is irreducible. So first you'd have to choose an algebraic closure in which the roots lie in. If you're interested in how the roots look, then a Google-search in Artin-Schreier-Theory might help, where you consider polynomials of the form $X^2 - X - a$. $Q$ above is one such as $-1 = 1$ in $\mathbb{F}_2$. – Steven Jul 15 '15 at 14:32
  • 1
    I assume, max, that when you write ${\bf F}{2q}$, you mean, ${\bf F}{2^q}$. $K={\bf F}_2[x]/(x^2+x+1)$ is a field of 4 elements. The elements $x+(x^2+x+1)$ and $x+1+(x^2+x+1)$ of $K$ are the roots of $x^2+x+1$ in $K$. – Gerry Myerson Jul 15 '15 at 23:00
  • @GerryMyerson Alright. I can now see why I was that confused: I always tried to associate the extension field with $\frac{\mathbb{Z}}{(4)\mathbb{Z}}, which is no field. A finite field with 4 elements is {0, 1, x, 1+x}. Glad you made that clear! – mdot Jul 16 '15 at 06:36

1 Answers1

1

Let $F$ be a field, write $F[x]$ for the ring of polynomials with coefficients in $F$, and let $f$ be in $F[x]$ and irreducible over $F$. Then $(f(x))$ (which means the ideal generated by $f(x)$, which is just the set of all multiples of $f(x)$ in $F[x]$) is a maximal ideal in $F[x]$, so the quotient ring, $K=F[x]/(f(x))$, is a field (and an extension field of $F$ --- see note below). The elements of $K$ are cosets of the ideal $(f(x))$ in $F[x]$, so they are of the form $g(x)+(f(x))$ for various $g(x)$ in $F[x]$. The element $x+(f(x))$ is a zero of the polynomial $f(x)$ in $K$.

How this works in the particular case of $f(x)=x^2+x+1$ over the field of two elements is elaborated in the comments on the question.

Readers not familiar with maximal ideals, quotient rings, and other concepts used above are referred to any textbook that covers Field Theory.

Note: we identify the set of all elements $c+(f(x))$ of $K$ with $c$ in $F$ as the inclusion of $F$ in $K$, justifying the view of $K$ as an extension of $F$.

Gerry Myerson
  • 179,216