Radical of an ideal

In ring theory, a branch of mathematics, the radical of a ring ${displaystyle R}$ is the ideal on the left ${displaystyle N}$ which is the intersection of all ideals on the maximal left of ${displaystyle R}$. There are different types of radicals, like the nilradical or radical Jacobsonas well as a theory of radical general properties.

Definition of radical of an ideal

Sea ${displaystyle R}$ a commutative ring ${displaystyle I}$ an ideal ring. The whole ${displaystyle {sqrt {I}}={text{rad }}I:={rin RUBexists nin mathbb {n}r^{n}in I}}}$ is called radical ideal ${displaystyle I}$ (or simply radical ${displaystyle I}$).

Yeah. ${displaystyle ain {hbox{rad }}I}$ is that there is an integer ${displaystyle ngeq 0}$ such as ${displaystyle a^{n}in I}$. Yeah. ${displaystyle rin R}$ That's it. ${displaystyle(ar)^{n}=a^{n}r^{n}in I}$.

Yes. ${displaystyle bin {hbox{rad }}I}$ There will be another integer ${displaystyle mgeq 0}$ so that ${displaystyle b^{m}in I}$.

By the Binomial Theorem:

${displaystyle (a+b)^{n+m}=sum _{i=0}^{n+m}{n+m choose i}a^{i}b^{n+m-i}}}$

• Yeah. ${displaystyle i within}$ That's it. ${displaystyle n+m-n=m marginn+m-i}$Then the exponent of ${displaystyle b}$ is greater or equal than ${displaystyle m}$And so ${displaystyle a^{i}b^{n+m-i}=a^{i}(b^{m})b^{n+m-i-m}in I}$.

• Yeah. ${displaystyle igeq n}$ That's it. ${displaystyle a^{i}b^{n+m-i}=(a^{n})a^{i-n}b^{n+m-i}in I}$ since ${displaystyle a^{n}in I}$.

In any case, every sum of ${displaystyle (a+b)^{n+m}}$ It's in. ${displaystyle I}$, which is an ideal ${displaystyle R}$Then ${displaystyle (a+b)^{n+m}in I}$ and it will ${displaystyle a+bin {hbox{rad }}I}$.

So. ${displaystyle {hbox{rad }I}$ is an ideal ${displaystyle R}$.

An ideal ${displaystyle I}$ of a commutative and unitary ring ${displaystyle R}$ is said to be radical ideal if it matches his radical, that's, if ${displaystyle {hbox{rad}I=I}$. As is obvious, the radical of an ideal is always a radical ideal.

Every ideal cousin is radical: Indeed, yes ${displaystyle P}$ It's an ideal cousin, then ${displaystyle R/P}$ is an integral domain, that is, it has no divisors of zero, and in particular it cannot have nilpotents.

It's easy to check if we take ${displaystyle pi:Rlongrightarrow R/I}$ the canonical projection ${displaystyle R}$ on ${displaystyle I}$, then ${displaystyle {hbox{rad }I=pi ^{-1}(N(R/I)}}$ (in fact through this demonstration it is shown immediately that ${displaystyle {hbox{rad }I}$ is an ideal ${displaystyle R}$Here, ${displaystyle N(R)}$ It's him. nilradical ${displaystyle R}$defined below). To see this, notice in the first place that if ${displaystyle rin pi ^{-1}(R/I)}}$So for some ${displaystyle ngeq 0}$, ${displaystyle pi (r)^{n}=pi (r^{n})}$ It's zero. ${displaystyle R/I}$and therefore ${displaystyle r^{n}}$ It's in. ${displaystyle I}$. reciprocally, yes ${displaystyle r^{n}}$ It's in. ${displaystyle I}$ for some ${displaystyle ngeq 0}$ It will be ${displaystyle r^{n}in I}$, then ${displaystyle (r)}^{n}=pi (r^{n}}}$It's zero. ${displaystyle R/I}$and therefore ${displaystyle pi (r)}$ It's in. ${displaystyle N(R/I)}$.

By using the location, we can see that ${displaystyle {hbox{rad }I}$ is the intersection of all ideals cousins ${displaystyle R}$ containing ${displaystyle I}$: every prime ideal is radical, so the intersection of the prime ideals that contain ${displaystyle I}$ contain a ${displaystyle {hbox{rad }I}$. Yeah. ${displaystyle r}$ is an element of ${displaystyle R}$ He's not in. ${displaystyle {hbox{rad }I}$So be ${displaystyle S}$ the whole ${displaystyle {r^{n}:nin mathbb {Z}n rigid0}$. ${displaystyle S}$ It's multiplyingly closed, so we can form the location. ${displaystyle S^{-1}R}$.

The nilradical

Sea ${displaystyle R}$ A switching ring. First we will show that the nilpotent elements of ${displaystyle R}$ form an ideal ${displaystyle N}$. Sean. ${displaystyle a}$ and ${displaystyle b}$ nilpotent elements ${displaystyle R}$ with ${displaystyle a^{n}=0}$ and ${displaystyle b^{m}=0}$. Let's prove that ${displaystyle a+b}$ It's nilpotent. We can use the Binomial Theorem to expand (a+b)^(n+m):

${displaystyle (a+b)^{n+m}=sum _{i=0}^{n+m}{n+m choose i}a^{i}b^{n+m-i}}}$

For each ${displaystyle i}$is given one and only one of the following conditions:

• ${displaystyle igeq n}$
• ${displaystyle n+m-igeq m}$

This says in every expression ${displaystyle a^{i}cdot b^{n+m-i}}$or the exponent of ${displaystyle a}$ will be big enough to override the expression (if ${displaystyle i within}$ That's it. ${displaystyle n+m-i/2005n+m-n=m}$Then the exponent of ${displaystyle b}$ is greater or equal than ${displaystyle m}$And so ${displaystyle a^{i}b^{n+m-i}=a^{i}cdot (b^{m)^{n+m-i-m}=a^{i}0^{n-i}=0}$), or the exponent of ${displaystyle b}$ will be big enough to override the expression (if ${displaystyle igeq n}$ That's it. ${displaystyle a^{i}b^{n+m-i}=(a^{n})^{i-n}b^{n+m-i}=0^{i-n}b^{n+m-i}=0}$). So we have to ${displaystyle a+b}$ He's nilpotent, and so he's in. ${displaystyle N}$.

To finish checking that ${displaystyle N}$ is an ideal, we take an arbitrary element ${displaystyle rin R}$. ${displaystyle (rcdot a)^{n}=r^{n}cdot a^{n}=r^{n}cdot 0=0}$, so ${displaystyle rcdot a}$ He is nilpotent, and he is therefore in ${displaystyle N}$. With what ${displaystyle N}$ It's an ideal.

${displaystyle N}$ is then called nilradical ${displaystyle R}$or radical nilpotent ${displaystyle R}$and denotes by ${displaystyle N(R)}$. The ring ${displaystyle {frac {R}{N(R)}}}}$ is called small ring (associated ${displaystyle R}$), although this denomination is falling into disuse.

It's immediate to check ${displaystyle N(R/N(R))={0}}$.

It's easy to prove ${displaystyle N(R)={hbox{rad }}0}$, that is, that the nilradical of a ring is precisely the radical of the null ideal. For this, the nilradical of ${displaystyle R}$ is the intersection of all ideals cousins ${displaystyle R}$.