Check the statements about irreducibility We have the ring $R$ and the polynomial $a=x^3+x+1$ in $R[x]$. I want to check the following statements:

*

*If $R=\mathbb{R}$ then $a$ is irreducible in $\mathbb{R}[x]$.
This statement is false, since according to Wolfram the polynomial has a real solution, right? But how can we calculate this root without Wolfram, by hand?


*If $R=\mathbb{Z}_5$ then $a$ is irreducible in $R[x]$.
The possible roots are $0,1,2,3,4$. We substitute these in $a$ and we get the following:
\begin{align*}&0^3+0+1=1\neq 0\pmod 5 \\ &1^3+1+1=3\neq 0\pmod 5 \\ &2^3+2+1=11\equiv 1\pmod 5\neq 0\pmod 5 \\ &3^3+3+1=31\equiv 1\pmod 5\neq 0\pmod 5 \\ &4^3+4+1=69\equiv 4\pmod 5\neq 0\pmod 5\end{align*}
So since none of these elements is a root of $a$, the statement is correct, right?


*If $R=\mathbb{C}$ then $a$ is irreducible in $R[x]$.
This statement is wrong, because from the first statement we have that $a$ is reducible in $\mathbb{R}[x]$, and so it is also in $\mathbb{C}[x]$. Is this correct?


*If $R=\mathbb{Q}$ then $a$ is not irreducible in $R[x]$.
Since this is a cubic polynomial, it is reducible if and only if it has roots.
By the rational root test, the only possible rational roots are $\pm 1$.
Since neither of these is a root, it follows that $a$ is irreducible over $\mathbb{Q}$, right?


*If $R=\mathbb{C}$ then $a$ has no root in $R$.
This statement is wrong, since from statemenet 3 we have that $a$ is reducible in $\mathbb{C}[x]$ and so it has roots in $\mathbb{C}$, or not?
 A: Let's let aside the five element field, for the moment.
Cardan's formula for the roots of the polynomial $x^3+px+q$ require to compute
$$
\Delta=\frac{p^3}{27}+\frac{q^2}{4}
$$
In your case
$$
\Delta=\frac{1}{27}+\frac{1}{4}>0
$$
so the polynomial has a single real root, precisely
$$
r=\sqrt[3]{-\frac{1}{2}+\sqrt{\Delta}}+\sqrt[3]{-\frac{1}{2}-\sqrt{\Delta}}
$$
This shows that $a$ is reducible over $\mathbb{R}$. Of course it is reducible over $\mathbb{C}$ and has three complex roots that you can, in principle, compute by factoring out $x-r$.
The polynomial is, however, irreducible over $\mathbb{Q}$, because the only possible rational roots are $1$ and $-1$, which aren't roots by direct substitution.

Let $F$ be a field.

Theorem. A polynomial $f(x)\in F[x]$ of degree $2$ or $3$ is irreducible if and only if it has no roots in $F$.

Proof. If $f(x)$ has a root $r$, then it is divisible by $x-r$, so it is reducible. If $f(x)$ is reducible, then an irreducible factor must have degree $1$ (just count the degrees). QED
This can be applied to the case $\mathbb{Z}_5$: no element is a root, so the polynomial is irreducible.
Important note. The above criterion does not extend to polynomials of degree $>3$.

Over the reals there is a simpler criterion, instead of considering Cardan's formula.
Theorem. A polynomial of odd degree in $\mathbb{R}[x]$ has at least a real root.
This follows from continuity of polynomials as functions and the fact that the limit of a monic polynomial of odd degree at $-\infty$ is $-\infty$ and the limit at $\infty$ is $\infty$. The intermediate value theorem allows us to conclude.
If you know that $\mathbb{C}$ is algebraically closed, you can also classify the irreducible polynomials over $\mathbb{R}$: a polynomial in $\mathbb{R}[x]$ is irreducible if and only if it has degree $1$ or has degree $2$ and negative discriminant.
A: *

*Notice that the polynomial is of odd degree, thus it must have a zero by the intermediate value theorem.

*Yes, you are correct.

*Follows from 1.

*Indeed you are right.

*Follows from 1.

