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I am trying to show that

I have said:

f=0 is in I so I is non-empty.

Let $f,g$ be members of $I$

$(f+g)(\sqrt5)=f(\sqrt5)+g(\sqrt5)=0+0=0 ==> f+g ∈ I$ so I is closed under addition

Let $g∈Q[x], f ∈ I$

$fg(\sqrt5)=f(\sqrt5)g(\sqrt5)=0*g(\sqrt5)=0 ==> fg ∈ I $ so I closed under multiplication from R.

Is this sufficient to show that I is an ideal?

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3 Answers

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Let's rephrase it slightly to gain a little more insight. Let $\,\ell(f) = f(\sqrt{5})\,$ be the map that evaluates $\,f(x)\,$ at $\,x = \sqrt{5}.\,$ Note that it is a ring hom: $\ \ell(f\pm g) \,=\, \ell f \pm \ell g ,\, $ and $\ \ell(fg) = (\ell f)\,(\ell g).\ $

More generally let $\,\ell \,$ be any ring hom and let $I$ be its kernel, i.e. all $f$ such that $\,\ell f = 0.\,$

Then $\,f,g \in I\,\Rightarrow\, \ell f,\,\ell g=0\,\Rightarrow\ \ell(f\pm g) = \ell f \pm\ell g = 0\pm0 = 0\ \Rightarrow\, f\pm g\in I,\ $ and

further $\ \ell(fh) = \ell f\, (\ell h) = 0\cdot \ell h = 0,\ $ so $\,fh\in I.\,$ Thus the kernel of a ring hom is an ideal.

Likewise, many ideals arise naturally as kernels of ring homomorphisms.

Remark $\ $ This raises an important question: why is the evaluation map a ring hom? Have you proved that yet? Note that this is not true for any ring, since, if so, evaluating $\, ax = xa $ at $\,x = b\,$ shows $\,ab = ba\,$ so the coefficient ring is commutative. In fact, this necessary condition is also sufficient for evaluation to be a ring hom. You should prove this at some point, since it helps to understand essential properties of polynomial rings $\,R[x]\,$ (as universal $R$-algebras).

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You've shown it to be closed to multiplication from the ring and to additions. Can you show it's also an additive subgroup?

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The usual definition of an ideal is that it is an additive subgroup of $R$ stable under multiplication by elements of $R$.

Your first point does not in itself show the first part, as an additively closed subset might not be a subgroup.

It is however possible that you have either an unusual defintion or a lemma that allows for your proof. The point is that by the multiplicative closedness you get additve inverses "for free" as $-g = (-1)g$.

A common criterion to show that something is an ideal is to show for the additive part that $f-g \in I$ for $f,g \in I$; that is using a common criterion for a non-empty subset being a subgroup.

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