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Let $K$ be a field, $0 \neq I \subsetneq K[x]$ be an ideal. Show that $K[x] \ / \ I$ integral domain $\Rightarrow$ $K[x] \ / \ I$ field applies.

My idea was to show that $K[x] \ / \ I$ is a finite integral domain, which would imply it is a field, as every element $\neq 0$ is invertible. So I need to show two things:

  • $K[x] \ / \ I$ is an integral domain
  • $K[x] \ / \ I$ is finite

In order to show it is an integral domain, I need to show that there are no zero divisors in $K[x] \ / \ I$. I'm stuck here, can you please help me to show that there are no zero divisors?

We know that $I$ is principle ($K[x]$ is a PID), and therefore generated by a single element. Can I use this to show $K[x] \ / \ I$ is finite?

quid
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muffel
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2 Answers2

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First, you certainly cannot show that $K[X]/I$ is finite since this is plainly not true when $K$ is infinite (as pointed out in a comment).

Second, you do not have to show that $K[X]/I$ is an integral domain as the result is "if integral then field", again it is plainly not true that $K[X]/I$ is generally an integral domain.

On the positive said: yes, you could show that $K[X]/I$ is finite (assuming that $K$ is finite) in the way your propose by showing that the elements of $K[X]/(f)$ essentially can be considered as polynomials of degree less than $\deg f$, via considering euclidean division.

Going back to your problem here is a proposal how to proceed (as you said you can restrict to principal ideal as every ideal is principal, let us denote $I = (f)$ in particular $f \neq 0$):

  • Show that $K[X]/(f)$ is an integral domain if an only if $f$ is an irreducible polynomial.

  • Conclude that $(f)$ is a maximal ideal, which implies $K[X]/(f)$ is a field.

If the second point is not clear as you do not know the notion and or the result, use Bézout's identity to show that every polynomial $g$ that is not a multiple of $f$, for irreducible $f$, is invertible modulo $(f)$ as you can solve $P g + Q f = \gcd (f,g) = 1 $ in polynomials.

quid
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  • What if $f=0$? The quotient is an integral domain… – Bernard Jan 21 '15 at 16:21
  • Yes I know. I glossed over this as it is pointed out in a comment that the condition that $I$, and hence the $f$ that generates it, is nonzero must be added in OP. – quid Jan 21 '15 at 16:30
  • @quid thank you for your answer! For the first point (integral domain $\Leftrightarrow$ f irr. polynomial): Doesn't this follow from the definition of integral domain? If f is reducible, there is a zero divisor $a \in K[X]/I$, and some $b \in K[X]/I$ for which $ab = 0, a \neq 0, b \neq 0$, and the other way round, am I right? For the second point: Can you please give me a hint, who do I show this? – muffel Jan 21 '15 at 18:40
  • @muffel yes this is the argumentn basically. You might be a bit more careful with the converse implication as $ab= 0$ in $K[X]/I$ does not meant the product is equal to $f$ but equal to a multiple. Still this should work. For the second ppoint: suppose the ideal were not maximal, then you have a bigger one that is thus generated by a divisor of $f$. That a ring modulo a maximal ideal is a field is something you might know. If not rather go with the argument I sketched at the end. – quid Jan 21 '15 at 19:21
  • @Bernard I made it explcit now. – quid Jan 21 '15 at 19:31
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A direct proof: if $I\neq 0$, $I$ is generated by a monic polynomial of positive degree $d$ since $I$ is a proper ideal. As a consequence, the ring $A=K[x]/(f)$ is a $K$-vector space of dimension $d$; if we denote by $\xi$ the class $x+(f)$, a basis is the set $\{1, \xi, \dots, \xi^{d-1}\}$.

Now let's consider a non-zero element $u$ of $A$. If $A$ is an integral domain, multiplication by $u$ is an injective endomorphism of $A$. For an endomorphism of a finite dimensional vector space, injectivity is equivalent to surjectivity, hence $1$ is attained, i.e. there exists an element $v$ such that $uv=1$ – in other words every non-zero element has an inverse.

Bernard
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