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Is the ring $\mathbb{Z}_5[x]$ isomorphic to the ring of polynomial functions from $\mathbb{Z}_5$ to $\mathbb{Z}_5$?

If not, what is a good counterexample?

If yes, how can we prove that there's a bijection between the two rings?

Thanks in advance.

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4  
What do you mean by $\mathbb Z_5$? The localisation of $\mathbb Z$ in $5$, that is $\mathbb Z[5^{-1}]$ or the quotient $\mathbb Z/(5)$? –  martini Oct 17 '13 at 9:55
    
Isn't it just the definition of $F[x]$ for any field $F$? –  studiosus Oct 17 '13 at 10:07
4  
@studiosus Not exactly. If the OP has $\mathbb Z_5 = \mathbb Z/(5)$, than note that $\mathbb Z/(5)$ is a finite set, and there are only finitely many functions $\mathbb Z/(5) \to \mathbb Z/(5)$. But $\mathbb Z/(5)[X]$ has infinitely many elements, for example $1$, $X$, $X^2$, $\ldots$. (Note that f. e. $X$ and $X^5$ represent the same function). –  martini Oct 17 '13 at 10:21
    
Ah, of course! Thank you, martini! –  studiosus Oct 17 '13 at 10:37
    
thanks guys! god bless. –  PandaMan Oct 21 '13 at 6:28

2 Answers 2

up vote 5 down vote accepted

No, it is not and the difference is a rather important, formal one.

For example: take the elements $\;f(x)=x\;,\;\;g(x)=x^5\;$ . Clearly, $\;f\neq g\;$ in $\;\Bbb F_5[x]\;$ , yet as functions $\;\Bbb F_5\to\Bbb F_5\;$ they are identical.

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thanks! god bless. –  PandaMan Oct 21 '13 at 6:28

How does this look?


Let $T$ be the ring of polynomial functions from $\mathbb{Z}_5$ to $\mathbb{Z}_5$

By Fermat's Little Theorem, $x^5 = x (mod 5)$.

Therefore, any polynomial $p(x)$ in $T$ may be rewritten in the form $ax^4 + bx^3 + cx^2 + dx + e (mod 5)$ for some $a,b,c,d,e \in \{0,1,2,3,4 \}$. Since there are $5$ possibilities for each coefficient, we see that $T$ has $5^5=3125$ elements, which is a finite ring.

$\mathbb{Z}_5[x]$ has infinitely many elements, since only the coefficients work $mod 5$ and not necessarily $x$.

Since $\mathbb{Z}_5[x]$ has infinitely many elements, while $T$ only has $3125$ elements, these two rings can't possibly be isomorphic.

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1  
Looks fine.$\phantom{}$ –  Baby Dragon Oct 23 '13 at 5:41
    
thanks baby dragon! –  PandaMan Oct 25 '13 at 20:57

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