## Jacobson Radical, Jacobson Ring

### Jacobson Ring

a ring R is jacobson if the jacobson radical
equals the nil radical in any homomorphic image of R.
This includes the identity map, hence jac(R) = nil(R).
A jacobson ring is also called a hilbert ring.

If R is not commutative nil(R) is actually the
lower nil radical.
In either case, it is the intersection of all prime ideals.

Note that the quotient of a jacobson ring is another jacobson ring.
Any homomorphic image of the quotient ring is a homomorphic image of the original ring,
thus each homomorphic image of the quotient ring
has its jacobson radical equal to its nil radical.

If R is jacobson, mod out by any prime ideal P,
and the nil radical is 0, whence the jacobson radical is 0.
Pull back to show P is the intersection of the maximal ideals containing P,
or the maximal left ideals if R is not commutative.

Now assume R is commutative, and each prime P is the intersection of maximal ideals.
Let P be a prime that is not maximal.
It is the intersection of maximal ideals containing P,
which is the intersection of all prime ideals that properly contain P.

Now we need to complete the circle and show this condition implies R is jacobson.
Suppose there is an element f in R/H which is in the jacobson radical,
but is not nilpotent.
Localize about the powers of f, and let M be a maximal ideal in (R/H)/f.
Pull this back to a prime Q in R/H, which pulls back to a prime P in R.
Note that Q does not contain f, and neither does P.
Thus P is not maximal,
and it becomes the intersection of every prime ideal properly containing P.
Such a prime becomes a prime containing Q in R/H, which includes f.
Therefore P and Q contain f, which is a contradiction.
The ring R is jacobson.

### Finitely Generated Ring

Let R be a finitely generated integral domain,
the image of **Z** adjoin finitely many indeterminants.
Let f be a nonzero element in R, and adjoin 1/f.
This gives the fraction ring R/f.
Let M be a maximal ideal in R/f.
Pull back to R and find a prime ideal P missing f.
Thus P injected into R/f creates M.

The quotient ring (R/f)/M is a finitely generated field.
This is a finite field.

Verify that the cosets of M in R/f correspond to the cosets of P in R.
Therefore R/P is a finite integral domain.
A finite integral domain is a finite field,
and P becomes a maximal ideal that misses f.
In other words, R is jacobson semisimple.

Now let R be any commutative finitely generated ring.
If P is a prime ideal,
the quotient R/P is a finitely generated integral domain, hence jacobson semisimple.
Each P is the intersection of maximal ideals, and R is a jacobson ring.