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I always thought an ideal generated by a subset $S=\{a_1,a_2,\dots,a_n\}$ of a ring $R$ would be defined as $(S)=\{a_1r_1 + a_2r_2 +\cdots+ a_nr_n\}$ where the $a_i$'s come from the set $S$ and the $r_i$'s are elements of the Ring $R$.

If I go by this, then an ideal generated by $S=\{2,x\}$ in $\mathbb Z[x]$ will look like $$ (S)=(2,x)=\{f(x)\cdot 2 + g(x)\cdot x\} $$ such that $f(x)$ and $g(x)$ are polynomials from $\mathbb Z[x]$.

For a generator $d$ of such an ideal to exist such that $(2,x)=(d)$, then $d$ must be a polynomial of zero degree so that it divides $2$. I am not able to proceed further. And so also $x$ must divide $d$. This means d must be a polynomial of degree $> 0$. This is a contradiction I believe.

I dont know if my arguments are right.
Can someone kindly assist me?

mrtaurho
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Krishan
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    The expression is correct. Are you looking for a more detailed description of $(S)$? – Johnny El Curvas Jun 04 '21 at 10:46
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    Here it can be simplified, hint: $,f = x, g + f(0)\in (2,x)\iff f(0)\in (2,x)\iff 2\mid f(0).\ $ This question is a dupe, but a target may not be easy to locate. – Bill Dubuque Jun 04 '21 at 10:54
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    $2\in(d)$ implies $d\mid d$ and $x\in(d)$ implies $d\mid x$. This shows that $d=\pm1$. – user26857 Jun 04 '21 at 11:07
  • Please don't change your question after comments or answers are posted (here from "what does the ideal look like" to "is it principal") since it typically invalidates comments and answers and wastes the time of others (answering, searching for dupoes, etc). – Bill Dubuque Jun 04 '21 at 20:01
  • Got that Bill. Sorry – Krishan Jun 07 '21 at 09:28

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