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What are the irreducible elements in $\mathbb{R}[x]$ and $\mathbb{C}[x]$?

For $\mathbb{C}[x]$, I guess the irreducible elements are all polynomials $x-z$ where $z\in \mathbb{C}$.

For $\mathbb{R}[x]$, I guess the irreducible elements are all polynomials $x-a$ and some other polynomials?

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    For 1), you also have $aX+b$ but no other since $\Bbb C$ is algebraically closed. For 2), you also have some degree 2 polynomials, but that's all since $[\Bbb C:\Bbb R]=2$. – 2017-02-03

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For $\Bbb C[X]$, you also have the polynomials $aX+b$ ($a\neq 0$) but no other since $\Bbb C$ is algebraically closed.

For $\Bbb R[X]$, you also have the polynomials $aX+b$ ($a \neq 0$) and some degree 2 polynomials $aX^2+bX+c$ ($a \neq 0$), but that's all since $[\Bbb C:\Bbb R]=2$. Notice that for a polynomial of degree $\leq 3$, irreducibility is equivalent to "having no roots", so $aX^2+bX+c \in \Bbb R[X]$ is irreducible iff $b^2-4ac<0$.

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    Why $[\mathbb{C}:\mathbb{R}]$ implies that there is no degree$\geq 3$ irreducible polynomials in $\mathbb{R}[x]$? – 2017-02-03
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    @Kenneth.K : because a root $z$ of such an irreducible polynomial of degree $\geq 3$ belongs to $\Bbb C$, so it has degree at most $2$ over $\Bbb R$, meaning that its minimal polynomial over $\Bbb R$ has degree at most $2$. Therefore any irreducible polynomial $f \in \Bbb R[X]$ such that $f(z)=0$ has degree at most $2$, so it can't be $\geq 3$. – 2017-02-03
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    This is a general fact: if $F$ is a field and $f \in F[X]$ is irreducible, then the degree of $f$ divides $[K:F]$, where $K \subset \overline F$ is the splitting field of $F$ – 2017-02-03
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For $\mathbf C[X]$, you're (almost) right: the irreducible polynomials are linear polynomials $\;c_1X+c_0$, $c_0, c_1\in\mathbf C$. Such a polynomial is associated to a unique monic polynomial. This results from the D'Alembert-Gauß theorem.

As for $\mathbf R[X]$, as the complex roots of a polynomial with real coefficient are pairwise conjugate, there results irreducible polynomials are linear polynomials and quadratic polynomials with a negative discriminant.

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    With _D'Alembert-Gauß theorem_ do you mean what everyone else in the world calls fundamental theorem of algebra? :) – 2017-02-03
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    I'm not the only one… – 2017-02-03
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    Of course I am overexagerrating, but one has to agree that most people (at least in English-speaking communities) call it FTA. – 2017-02-03
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Let be $R[x]$ is a polynomial ring. If $f(x)\in R[x]$ is an irreducible polynomial, then $(f(x))$ is a prime ideal.

we know that $\mathbb R[x]$ and $\mathbb C[x]$ are $PID$ then every prime ideal is maximal.

In $\mathbb C[x]$ we know that maximal ideal is the form $(x-a)$ , then irreducible polynomial in the $\mathbb C[x]$ are the form $f(x)=x-a$.

In $\mathbb R[x]$, $f(x)$ is irreducibe polynomial if $(f(x))$ is a maximal ideal