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For the given field T and $g \in T[x]$ - polynomial of positive degree, prove that every nonzero and non-invertible quotient ring $T[x]/(g)$ element is indeed zero divisor.

This task was explained us during our zoom class, however I haven't really got it, but it seems to be fundamental one, so I need to realize its solution fully. Can you think of the most simple solution that would be really easy to undestand? Would be grateful for any help.

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  • $\begingroup$ Can you prove the analogous result in $\mathbf Z/m\mathbf Z$? It is the same idea. (The notation $T$ for a field is strange. Why not $F$? Or $K$?) $\endgroup$ – KCd May 15 at 14:29
  • $\begingroup$ I use F notation a lot in my solutions, so it is uncomfortable for me to use it specifically for that purpose, however haven't heard of K-notation, will use it then next :) $\endgroup$ – 9cloudalpha May 15 at 14:32
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    $\begingroup$ Writing about polynomials $f(x)$ in $F[x]$ looks okay. The letter $K$ is widely used for fields since it is the first letter of the German word for field (in algebra). $\endgroup$ – KCd May 15 at 14:38
  • $\begingroup$ Ok, got it, thanks for the info! $\endgroup$ – 9cloudalpha May 15 at 14:39
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Not sure what is the easiest solution, but here's one I like.

Since $g$ is of positive degree, $R=T[x]/(g)$ is a finite-dimensional vector space over $T$. Let $a_0\in R$ be an element which is not a zero divisor. Then the map $R\to R$ given by the formula $a\mapsto a_0a$ is injective and $T$-linear, and thus it is surjective (since $R$ is a finite-dimensional vector space over $T$). In particular, there is some $a$ such that $a_0a=1$.

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  • $\begingroup$ Sounds like great solution, however not sure whether "is a finite-dimensional vector space over" can use this transfer in terms of our college algebra course.. $\endgroup$ – 9cloudalpha May 15 at 14:36
  • $\begingroup$ @9cloudalpha: Why not? Did you not have linear algebra? If you know what a vector space is, this is pretty easy to show. The dimension is the degree of $g$ (provided it is positive). $\endgroup$ – tomasz May 15 at 14:37
  • $\begingroup$ Yeah, I know what vector space is, but that fact seems sound non-obvious to me, we have never observed $T[x]/(g)$ as vector space over T, actually $\endgroup$ – 9cloudalpha May 15 at 14:40
  • $\begingroup$ @9cloudalpha: it's quite straightforward. Any ring containing $T$ is a vector space over $T$. $\endgroup$ – tomasz May 15 at 15:50

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