Decomposition of a biquadratic polynomial

Question

Let $f(t):=T^4+aT^2+b$ be a polynomial in a field $K[T]$ such that it can be decomposed in two irreducible factors of degree 2.

My intuition then is screaming to me that then we neccessarily have that the decomposition is $f(T)= (T^2-u^2)(T^2-v^2)$, where $\pm u, \pm v$ are the roots of $f$.

My reasoning here is that since $u$ and $-u$ are related through an element in $K$ then they are bound tighter together than $u$ and $v$, so they must always be paired roots.

Is this true? Under which conditions? What is an easy proof of this fact?

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My half assed attempt:

Let's suppose this is not the case, so wlog $(T-u)(T-v)\in K[T]$.

Then the possible K-automorphisms of the splitting field $K_f=K(u,v)$ are those described by the images of $u$ and $v$. But since the roots of irreducible polynomials must be mapped to roots of irreducible polynomials, then $u$ goes to itself or to $v$, and $-u$ goes also to itself or $-v$ due to how we decomposed $f$.

But either choice of image for $u$ fixes the image of $-u$, so there are only two $K$-automorphisms and thus the degree of the expansion $K_f | K$ is equal to 2.

And this is a contradiction because the degree of the expansion is $4$... except that I can't actually prove this last part.

Any ideas?

Answer 1

Let $f(X)=X^4+aX^2+b$ be a polynomial of degree four with irreducible factorization $f(X)=(X^2-d_1)(X^2-d_2)$, then $a=-d_1-d_2$ and $b=d_1d_2$. Hence the necessary and sufficient condition, to have such a factorization, is that the quadratic polinomial $X^2+aX+b$ has a root in $K$.

Answer 2

Consider $T^4+1$ over $K=\mathbb Q(\sqrt 2)$. The roots in a splitting field are the fourth roots of $-1$, which in $\mathbb C$ are complex numbers and are not elements of $K$, which is a real field.

We have the factorisation in $K$ into irreducible quadratic factors $$T^4+1=(T^2+\sqrt 2T+1)(T^2-\sqrt 2T+1)$$ and this is not of the form you surmise.

So you perhaps need to clarify how you are thinking about this case. It is true that you can solve the quartic by solving two quadratics, but not that you have to pick roots in the pairs you have specified.