# Generator for $\mathbb{Z}[x]$ ring?

Is the polynomial ring $$\mathbb{Z}[x]$$ generated exclusively by $$\langle 1,x\rangle$$? Or has to be generated by more or less elements? My intuition is that you can generate any of the "$$x$$'s" polynomials with just multiplying and adding $$x$$ to the equation but you need the "$$1$$" to generate the coefficient with degree $$0$$.

• As a ring with unity, it is generated by $x$ alone. As a ring (not necessarily with unity) it is generated by $1,x$.
– user700480
Dec 11, 2020 at 22:19

Your intuition would be correct. The only way to generate $$\mathbb{Z}\subset \mathbb{Z}[x]$$ is by having 1 in your ideal. Similarly, you must have x in your ideal to generate all the polynomials.
• Thank you very much! I've just read a powerful tool to prove that this specifically ring cannot be Principal; from the Dummit Abstract Algebra: Let $a,b \neq 0 \in R$. if $d$ is the unique generator for $R$ then the greatest common divisor is a power of $d$. However, when you use $[a = x \land b = x^2]$ or $[a = x \land b = 1]$ you see there're 2 generators 1 and x. Dec 11, 2020 at 22:19