For a polynomial ring $A[x]$, is $p(x) = x$ always irreducible?

Question

I am a little confused by the definition of irreducible elements. In particular, for a polynomial ring $A[x]$, is $p(x) = x$ always irreducible?

Or does it depend on the properties of $A$?

Answer 1

The short answer is "yes: divisibility properties like this can depend ont he coefficient ring."

Before I say much more, do you realize that irreducible elements are most often defined in the context of domains? I'm not saying this makes your question less interesting: I just wanted to make sure you were aware.

There is an easy answer to a very similar question: is $X$ always prime in $R[X]$? The answer is easily "no," and the easy way to see why is that $X$ will be prime iff $(X)$ is a prime ideal, iff $R[X]/(X)\cong R$ is a domain. Thus we see that (when we restrict our attention to domains) $X$ is indeed always prime and therefore irreducible.

For every nondomain $R$ then, $X$ isn't prime in $R[X]$, but that doesn't necessarily preclude $X$ from being irreducible.

If you break down what the condition says ("$X=ab$ implies $a$ or $b$ is a unit") it tells you that $(X)$ is a maximal among proper principal ideals of $R$. So the challenge is to find a nonzero principal ideal $(d)$ of $R[X]$ where $d$ isn't a unit, but $dp(X)=X$ for some polynomial $p(X)$.

Update - added example where $X$ is not irreducible

It turns out that $(3X+4)(4X+3)=X$ in $\Bbb Z_{12}[X]$. Neither of the two factors can be units, though. To see this, notice that if $3X+4$ were a unit in $\Bbb Z_{12}[X]$, then its projection into $\Bbb Z_{12}$ would also be a unit, but that's not possible since $4$ is a zero divisor. The same reason applies to $4X+3$.

Answer 2

In the polynomial ring $A[x]$, $x$ is transcendental over $A$ and $x$ is irreducible as long as $A$ is an integral domain.

Proof: Let $x = f \cdot g$, where $f, g \in A[x]$. It suffices to show that either $f$ or $g$ is a unit. Now the degree of $f$ and $g$ is at most 1, and they both can't be 1. So without loss of generality, let $f = ax+b$ have degree 1 where $a,b \in A$ and $g = c$ have degree 0. Now $$(ax+b) \cdot c = x$$ implies that $ac = 1$ and $bc = 0$. Since certainly $c \neq 0$, we have $b = 0$ and $g = c$ is a unit.

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However, one must be careful because something that looks like $A[x]$ is not always a polynomial ring. For example, $\mathbb{Q}[\sqrt{2}]$ and $(\mathbb{Q}[x^2])[x]$ are both not polynomial rings and the above statement won't be true in general.

Answer 3

Recall that an element $a$ of a commutative ring $A$ is called irreducible if $a \neq 0$, $a \notin A^*$ (these conditions are often forgotten) and for all decompositions $a=bc$ we have $b \in A^*$ or $c \in A^*$. The zero ring has no irreducible elements. Hence, $x \in 0[x]$ is not irreducible. But this is not the only exception. Consider $A=\mathbb{Z}/6$ and $b=2+3x$ and $c=3+2x$. Then $bc=2 \cdot 3 + (2 \cdot 2 + 3 \cdot 3)x + (3 \cdot 2)x = x$.