Factorize $f(X)=X^4+1$ on $L$.

Question

Let $L= \mathbb{Q}[Y]/(Y^4+1)$. Factorize $f(X)=X^4+1$ on $L$.

My attempt: We know that $\alpha = \overline{X}=X+(X^4+1)$ is one of the root of $f(X)$ in $L$. Now using the relation $\alpha^4+1=0$ and using simple division method, we can factorize $X^4+1 = (X-\alpha)(X^3+\alpha X^2+\alpha^2 X+\alpha^3)$. Is this the final factorization? How do we know that there are not more roots in $L$?

Edit: I have changed the variables in the field $L$ from $X$ to $Y$.

Answer 1

To avoid confusion I will give a new name to the indeterminate:

Factor $f(t)=t^4+1$ over $L$.

You already found the root $\alpha$ and hence the factor $t-\alpha$ of $f$ in $L$. But there are more! Notice that $f$ contains only even powers of $t$, so $-\alpha$ will also be a root. Now by division we get $f(t)=(t-\alpha)(t+\alpha)(t^2+\alpha^2)$. So we are reduced to factor $t^2+\alpha^2$. Notice that if there was an element $i\in L$ such that $i^2=-1$ we could write $t^2+\alpha^2=(t-i\alpha)(t+i\alpha)$. Can you find such an element?

Answer:

Take $i=\alpha^2$.

Answer 2

Hint: consider the $\mathbb Q$-algebra map $f \colon \mathbb Q[X] \to \mathbb Q(\xi_8)$, $X \mapsto \xi_8$, where $\xi_8$ is an $8$-th primitive root of unity. Since $\xi_8^8 = 1$, we have $(\xi_8^4)^2 = 1$. But $\xi_8^4\neq 1$, as $\xi_8$ is primitive, so we necessarily obtain $\xi_8^4 = -1$ and thus $\xi_8$ vanishes at $x^4+1$.

On the other hand, a standard fact of cyclotomic polynomials is that

$$ \deg m(\xi_8,\mathbb Q) = \varphi(8) = 4, $$

so $m(\xi_8,\mathbb Q) = X^4+1$. By construction, this is (a generator of) the kernel of $f$, hence

$$ \mathbb Q[X]/(X^4+1) \simeq \mathbb Q(\xi_8). $$ Maybe the latter presentation can help you.

The cyclotomic polynomial has as roots primitive roots of unity. These are characterized as powers of $\xi_8$ with exponent coprime with $8$; hence $T^4+1 = (T-\xi_8)(T-\xi_8^7)(T-\xi_8^3)(T-\xi_8^5)$.