Fulton, Algebraic Curves, Exercise 1.37

Question

I am stuck trying to prove this:

Let $k$ be any field, $F\in k[X]$ a monic polynomial of degree $n>0$. Then the residues $\bar{1},\bar{X},...,\overline{X^{n-1}}$ form a basis for $k[X]/(F)$ over $k$.

I thought I had done it correctly but my proof isn't convincing me, so back to square one.

Any help will be greatly appreciated!

EDIT: The question does say "any field" but it is in the section that says $k$ is algebraically closed for the whole section.

Here's the attempt.

Let $G$ be any polynomial with residue $\bar{G}\in k[X]/(F)$, so $G-aF^m=0$ for some $a\in k$ and $m>0$. Since $k$ is algebraically closed, we may divide $G$ by the irreducible factors of $F$ so that $G$ does not vanish for any $P\in V(F)$. Moreover, since $F$ is monic, $\deg(G)=\deg(G-aF^m)+1$ for some $a\in k$ and $m>0$. It follows then that $\{\bar{1},\bar{X},...,\overline{X^{n-1}}\}$ is a basis for $k[X]/(F)$.

How about showing us that proof of yours? We could perhaps show you if you've done something wrong? — The Count, Jan 02 '17 at 20:55
It isn't net the case that any $G=aF^m$, instead you want to say that any polynomial is a sum of monomials and then use my answer to show that only the first $n$ monomials are required — Rene Schipperus, Jan 02 '17 at 21:25

score -2 · Answer 1 · answered Jan 02 '17 at 20:58

-2

The fact that $F$ is monic is useful it says that $$x^{n}=a_1x^{n-1}+\cdots +a_n.$$

answered Jan 02 '17 at 20:58

Rene Schipperus

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I don't mean to be rude, but this doesn't even sort of answer the question. – The Count Jan 02 '17 at 21:03
I don't think that's true :/ $F$ being monic and of degree $n$ means it is in one variable and with coefficient 1 for its $X^n$ term. – Purple Jan 02 '17 at 21:04
What the hell are you guys talking about ? $x^k$ $k\in \mathbb{N}$ generates $k[x]$ and the above relation which holds modulo $F$ show that only the first $n$ of these elements are required to generate the ring. – Rene Schipperus Jan 02 '17 at 21:16
@Rene Schipperus Apologies, the way you worded your answer I thought you giving me the definition of "monic". I'll try this exercise again. – Purple Jan 02 '17 at 21:43
Course because $k$ is a field any poly can be made to be monic. – Rene Schipperus Jan 02 '17 at 21:54
No need to get angry, Rene. But I stand by my statement. If you disagree, then I suppose we will all get on with our lives. – The Count Jan 03 '17 at 00:02
@TheCount Not angry, just that you guys are talking out of your hats. (seasonal joke.) – Rene Schipperus Jan 03 '17 at 01:20
@ReneSchipperus But I have 10 hats! I have to do something with them! – The Count Jan 03 '17 at 01:27
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@TheCount one ought be able to wear more than one hat at a time... they should do that. – Rene Schipperus Jan 03 '17 at 01:29

Fulton, Algebraic Curves, Exercise 1.37

1 Answers1