Let $x \in \mathbb{R}$. Are the following statements true or false?

a. If $x^{7}$ and $x^{12}$ are rational, then $x$ is rational.

b. If $x^{9}$ and $x^{12}$ are rational, then $x$ is rational.

Corollary 3.55. If $P, Q \in \mathbb{Q}[X]$ are coprime, then, viewed as elements of $\mathbb{R}[X]$, they are coprime.

Proof. According to Bézout's theorem, there exist $\mathrm{U}, \mathrm{V} \in \mathbb{Q}[\mathrm{X}]$ such that $\mathrm{PU}+\mathrm{QV}=1$. A fortiori, $\mathrm{U}, \mathrm{V} \in \mathbb{R}[\mathrm{X}]$, so, according to the converse of Bézout's theorem, $\mathrm{P}$ and $\mathrm{Q}$ are coprime when viewed as elements of $\mathbb{R}[X]$.

From Bézout's theorem, we deduce Gauss's theorem.

Theorem 3.56. If $P, Q, R \in \mathbb{K}[X]$ are such that $P$ is coprime with $Q$ and $P$ divides $Q R$, then $P$ divides $R$.

Irreducible Polynomials in $\mathbb{K}[X]$ Irreducible polynomials play the role of prime numbers: they are the building blocks when we wish to factorize polynomials.

Definition 3.57. A polynomial $P \in \mathbb{K}[X]$ is said to be irreducible in $\mathbb{K}[X]$ if $P$ is not constant and if its only divisors in $\mathbb{K}[X]$ are the constants and the polynomials proportional to $P$ that are non-zero, or, equivalently, if there do not exist $Q, R \in \mathbb{K}[X]$ such that $P=Q R$ with $\operatorname{deg} Q \geqslant 1$ and $\operatorname{deg} R \geqslant 1$.

From this, we deduce the equivalent of the prime factorization theorem.

Theorem 3.58. Every polynomial in $\mathbb{K}[X]$ can be uniquely decomposed, up to the order of the factors, in the form:

$$
P=c P_{1}^{k_{1}} P_{2}^{k_{2}} \ldots P_{k}^{\alpha_{k}}
$$

where $c \in \mathbb{K}^{*}, \mathrm{k}_{i} \in \mathbb{N}^{*}$ and the $\mathrm{P}_{i}$ are distinct unitary and irreducible polynomials in $\mathbb{K}[X]$.

As an exercise, we leave the proof of this theorem (very close to its equivalent for integers).

The previous theorem invites us to seek the irreducible polynomials of $\mathbb{C}[X], \mathbb{R}[X]$ and $\mathbb{Q}[X]$. We start with a general proposition.

Proposition 3.59. A polynomial $P \in \mathbb{K}[X]$ of degree 2 or 3 is irreducible if, and only if, it has no root.

## Proof. Exercise.

Proposition 3.60. The irreducible polynomials in $\mathbb{C}[X]$ are the polynomials of the first degree.

Proof. It is clear that these polynomials are irreducible. Conversely, if $\mathrm{P} \in \mathbb{C}[X]$ of degree at least 2 is irreducible, according to the fundamental theorem of algebra, it can be written as $P(x)=(x-\alpha) Q(x)$, which contradicts its irreducibility.

We now move on to the study of polynomials with real coefficients.

Proposition 3.61. Every polynomial $P \in \mathbb{R}[X]$ can be decomposed in the form:

$$
P(x)=c \prod_{i=1}^{r}\left(x-\alpha_{i}\right) \prod_{i=1}^{s}\left(x^{2}+a_{i} x+b_{i}\right)
$$

Consequently, the irreducible polynomials in $\mathbb{R}[X]$ are the polynomials of the first degree and those of the second degree with a negative discriminant.

Proof. According to Corollary 3.49, we can write $P(x)=c\left(x-\alpha_{1}\right) \cdots\left(x-\alpha_{n}\right)$, where the $\alpha_{i}$ are complex. Using Proposition 3.51, we see that if $\alpha$ is a root of $P$, then $\bar{\alpha}$ is also a root of $P$. Indeed, if $P(x)=a_{0}+a_{1} x+\cdots+a_{n} x^{n}$ with the $a_{i}$ real, we have $0=\overline{P(\alpha)}=\overline{a_{0}+a_{1} \alpha+\cdots+a_{n} \alpha^{n}}=\overline{a_{0}}+\overline{a_{1} \alpha}+\cdots+\overline{a_{n} \alpha^{n}}=a_{0}+a_{1} \bar{\alpha}+\cdots+a_{n} \bar{\alpha}^{n}$. In the factorized form of $P$, we then group the complex (non-real) roots with their conjugates. Noting that for a complex number $z, (x-z)(x-\bar{z})=x^{2}+a x+b$, with $a, b \in \mathbb{R}$ such that $a^{2}-4 b \leqslant 0$, we conclude.

The previous reasoning shows that an irreducible polynomial in $\mathbb{R}[X]$ is a polynomial of the first degree or of the second degree with a negative discriminant. Conversely, such polynomials are irreducible according to Proposition 3.59.
The first assertion is true (use Bézout's theorem for integers with 7 and 12). The second assertion is false (take $x=2^{1 / 3}$).