Describe the idempotents in a polynomial ring $R[x]$.

In general, given that $R$ is a commutative ring, how to identify the idempotent elements in $R[x]$?

I tried by solving equalities of polynomials but nothing works...

So how can I see that in which form can we get some idempotents in $R[x]$?

• Compare the degrees of $p(x)$ and $p(x)^2$. – Xam Nov 13 '16 at 5:12
• @Charter I have tried that, but nothing works. – PropositionX Nov 13 '16 at 5:16
• What exactly did you try and why it didn't work? – Mariano Suárez-Álvarez Nov 13 '16 at 6:20
• @Y.X. Hint: the idempotents of $R$ are the only idempotents in $R[X]$. – user26857 Nov 13 '16 at 7:45

$f^2 = f\,\Rightarrow f(0)^2\! = f(0) := a.\,$ If $\,f\!-\!a\ne 0\,$ it has order $\,k\ge 1,\,$ say $\,f = a + bx^k+\cdots,\,b\ne 0.\,$ Then $\,f^2\! = f\,\Rightarrow 2ab = b,\,$ times $\,a\,\Rightarrow\,2ab = ab\,\Rightarrow\,ab=0\,\Rightarrow\,b = 2ab = 0,\,$ contradiction. Therefore $\,f = a = f(0)\,$ is constant.
• Remark $\$ Note that the proof also works also for power series. – Bill Dubuque Nov 14 '16 at 17:58
CASE I:$R$ is an integral domain .Then $p(x)^2=p(x)\implies \deg p(x)=0\implies p(x) \text{is a constant}$
So the idempotents in $R[x]$ are only the idempotents in $R$.
CASE II:$R$ is not an integral domain .Here though the idempotents in $R$ will be idempotents in $R[x]$ it can't be said whether they will be the only idempotents or not.