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I was trying like this: Let $I=\langle f(x),g(x) \rangle$. Then $4=f(x)q_{1}(x)+g(x)q_{2}(x)$. But $deg (4)=0$. So $deg(f(x))=0$ and $deg(g(x))=0$. So $f(x)=a$ and $g(x)=b$ where $a,b \in \mathbb{Z}$. Then I am stuck. Is this the correct way? Any help is appreciated.

Unknown
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  • Of course this is wrong : we are not multiplying $f,g$ together but rather adding some multiples of both together. This could very well result in a polynomial of lower degree, for example a constant. – Sarvesh Ravichandran Iyer Sep 18 '20 at 16:10
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    I'm the \langle \rangle fairy, here to let you know that $\langle, \rangle$ plays nicer with TeX than <, > does :) – Patrick Stevens Sep 18 '20 at 16:15
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    Another typographic convention is to put a space after the full stop at the end of a sentence. – Angina Seng Sep 18 '20 at 16:17
  • For the record, I originally gold-badge closed this with the above listed $4$ dupes, It was then gold-badge reopened by the answerer, then mod reclosed, after which I updated the dupe list again. – Bill Dubuque Sep 18 '20 at 18:23

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Suppose one has a commutative unital ring $R$ with an ideal $I$, and a maximal ideal $M$. Then $R/M$ is a field $k$, say. Also $I/MI$ is a module over $R/M$, a vector space over $k$. Suppose its dimension as a vector space is $d$. If $a_1,\ldots,a_n$ are generators of $I$, then their images in $I/MI$ span $I/MI$ as a $k$-vector space. Therefore $n\ge d$.

Here, $R=\Bbb Z[X]$, $I=\left<x^2,2x,4\right>$, $M=\left<x,2\right>$ and $d=3$.

Angina Seng
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  • Please strive not to post further dupe answers to dupes of FAQs.For site organization is is better to instead link to prior answers (which hopefully have been polished over time due to feedback). Further it helps to localize information on a topic. – Bill Dubuque Sep 18 '20 at 17:39
  • I have reclosed this question as a duplicate. Even to me, a non-expert, it appears that the linked question is identical to this one. Can you please explain why you chose to exercise your gold-badge power to reopen this question? – Xander Henderson Sep 18 '20 at 18:17
  • @XanderHenderson In fact is is an exact dupe of this question that I listed (and no expertise is needed to see that). It is a special case or abstract dupe of the others. – Bill Dubuque Sep 18 '20 at 18:18