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**gotmejerry** Let p1(t), p2(t) be two polynomials in the F[t] polynomial ring, where F is a field. Let ß1 be a congruence relation for: r(t) ß1 s(t) <==> p1(t)|r(t) - s(t), and ß2 the same for r(t) ß2 s(t) <==> p2(t)|r(t) - s(t).

When will be F[t]/ß1 and F[t]/ß2 isomorph vector spaces. Give necessary and sufficient condition.

Thank you!

So, here's what you're saying. You have the polynomial ring $\displaystyle k[t]$$ for some field $\displaystyle k$ and you're thinking of it as a vector space over $\displaystyle k$ in the natural way. You're asking then when given $\displaystyle p_1(x),p_2(x)$ how can you tell if $\displaystyle k[t]/(p_1)\cong k[t]/(p_2)$ as vector spaces (where $\displaystyle (\cdot)$ denotes the ideal generated--not that this really is a subspace since $\displaystyle k$ sits nicely inside $\displaystyle k[t]$). Well, we know that vector spaces are exceedingly simple objects, in that their only invariant, the only thing that distinguishes them is their dimension. So, rephrasing your question we need only figure out when $\displaystyle \dim_k k[t]/(p_1)=\dim_k k[t]/(p_2)$. So, let's figure out in general, given any polynomial $\displaystyle p(x)=a_0+\cdots+a_nt^n\in k[t]$ what $\displaystyle \dim_k k[t]/(p)$ is. Note that since $\displaystyle \mathscr{B}=\{1,t,t^2,t^3,\cdots\}$ generates $\displaystyle k[t]$ that $\displaystyle \{1+(p),t+(p),t^2+(p),t^3+(p),\cdots,\}$ generates $\displaystyle k[t]/(p)$. What I claim though is that if $\displaystyle N\geqslant n$ then $\displaystyle t^N+(p)$ is a linear combination of the $\displaystyle \{1+(p),\cdots,t^{n-1}+(p)\}$. We can do this by induction. For example, $\displaystyle t^n-\frac{a_{n-1}}{-a_n}t^{n-1}+\cdots+\frac{a_0}{-a_n}=\frac{1}{a_n}\left(a_n t^n+\cdots+a_0\right)$ and so

$\displaystyle t^n+(p)=\frac{a_{n-1}}{a_n}t^{n-1}+\cdots+\frac{a_0}{a_{n}}+(p)=(\frac{a_{n-1}}{-a_n}t^{n-1}+(p))+\cdots+(\frac{a_0}{-a_n}+(p))$

Now, if we assume that $\displaystyle t^{n+j}+(p)$ is a linear combination of $\displaystyle \{1+(p),\cdots,+t^{n-1}+(p)\}$ for all $\displaystyle j<k$ then noting that

$\displaystyle t^{n+k}-\frac{t^k}{-a_n}\left(a_{n-1}t^{n-1}+\cdots+a_0\right)=\frac{t^k}{a_n}\left(a_nt^{n-1}+a_{n-1}t^{n-1}+\cdots+a_0\right)$

Allows us to conclude that $\displaystyle t^{n+k}$ is a linear combination of the $\displaystyle \{1+(p),\cdots,t^{n+k-1}+(p)\}$ and since each of these are a linear combination of the $\displaystyle \{1+(p),\cdots,t^{n-1}+(p)\}$ the induction follows.

From the fact that every element of a generating set of $\displaystyle k[t]/(p)$ is a linear combination of $\displaystyle \{1+(p),\cdots,t^{n-1}+(p)\}$ allows us to conclude that $\displaystyle \{1+(p),\cdots,t^{n-1}+(p)\}$ is a generating set for $\displaystyle k[t]/(p)$. What we now claim though is that $\displaystyle \{1+(p),\cdots,t^{n-1}+(p)\}$ is a linearly indepdent set. Indeed, the basic idea is that if we could write

$\displaystyle \displaystyle t^{j}+(p)=\sum_{m\ne j}a_mt^m+(p)$[/center]

then this tells us that $\displaystyle \displaystyle p\mid t^j-\sum_{m\ne j}a_m t^m$ but this is ridiculous since $\displaystyle \displaystyle \deg\left(t^j-\sum_{m\ne j}a_m t^m\right)<n$.

From all of this we may conclude that $\displaystyle \{1+(p),\cdots,t^{n-1}+(p)\}$ is a basis for $\displaystyle k[t]/(p)$ and so we may conclude that $\displaystyle \dim_k k[t]/(p)=n=\deg(p)$.

Thus, $\displaystyle k[t]/(p_1)\cong k[t]/(p_2)$ if and only if $\displaystyle \dim_k k[t]/(p_1)=\dim_k k[t]/(p_2)$ which by the above is true if and only if $\displaystyle \deg(p_1)=\deg(p_2)$.

*Remark:* Why you should have thought this to be true was to look at $\displaystyle \mathbb{C}[x]$ and look at polynomials $\displaystyle p(x)=(x-z_1)\cdots (x-z_n)$ with $\displaystyle z_i\ne z_j$ and then use the CRT to conclude that $\displaystyle \mathbb{C}[x]/(p)\cong \mathbb{C}[x]/(x-z_1)\times\cdots\times\mathbb{C}[x]/(x-z_n)\cong \mathbb{C}^n$.