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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 $0$.

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

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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.

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  • $\begingroup$ 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. $\endgroup$
    – Figaro
    Commented Dec 11, 2020 at 22:19

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