Let be an algebra. We say that an element is integral over if there exists some monic polynomial such that in . This is a generalization of the concept of algebraic elements in the theory of fields. The structure morphism is said to be an integral morphism if every element in is integral over ; if is an injection, it is said to be an integral extension.
Proposition. Let be an algebra. If there exists elements such that , and if induces integral morphisms for every , then is integral.
Proof. To simplify matters, we start by supposing that is an inclusion and that is a subring of . Pick any ; we want to show there exists a monic polynomial with coefficients in whose roots include . By hypothesis, for any , there exists a polynomial with such that . We can multiply each of these equations by the appropriate number of ’s until we have “cleared the denominators”, which produces new equations in , so that By setting , we have We can also set and multiply each by enough copies of to get it to be of the form , each multiplication by increasing the “degree” of the terms in its expression as a “polynomial in ” by one, so we finally find Since , it is true that generate the whole of as well (hint: what is the form of a general element of any power of ?) Hence there exists elements such that . Then so we find . This proves is the root of a polynomial with coefficients in , i.e. is integral over . Since the element was arbitrary, we have shown is integral in the case it is the inclusion of a subring into . In the general case, we have that is integral if and only if the induced inclusion is integral. Using this, together with the fact that the morphism factorizes through , one can reduce the general case to the one we just proved.
At the end of the previous proof, we used a special case of the following proposition in order to show that the general case held by reducing it to the special case of subring inclusion.
Proposition. Consider the following commutative diagram in the category of commutative rings with unity : If the top arrow is integral, then so is the bottom arrow . Moreover, if is surjective, then the converse of the previous statement is true as well. Note that injectivity of a morphism verifies the same kind of statement, so if the top arrow is an integral extension, then so is the bottom arrow .
Because the proof is really straightforward, I decided to omit it. But the gist of it is: given any , there’s a polynomial that kills it, and its coefficients look like for some ; therefore commutativity gives coefficients that look like , and reciprocally if is surjective.
Some immediate consequences of the previous proposition: let be a ring homomorphism; then
- (quotient of ) for any ideal of contained in , the induced morphism is integral if and only if is;
- (localization of ) for any multiplicative subset of , the induced morphism is integral when is.
Things are also well-behaved when taking quotients of . More precisely, let be any ideal of , and suppose is an integral morphism. Then is also integral. Indeed, pick any element ; because is integral, there exists an integer and elements such that we have Hence simply passing this equation to the quotient gives us an expression for zero as a sum of powers of , with coefficients . This shows the composite ring homomorphism is integral, as we wanted.
Things are less well-behaved when localizing on . More precisely, there exists a multiplicative subset and an integral morphism such that the composite morphism is not integral. Here’s an example: choose the polynomials with coefficients in a field , and take the identity as the (obviously integral) morphism . Now I will show that is not integral over . To do this, I need to exhibit some element that is not integral. After testing things out, I found that is probably the simplest element which is not integral over . Here’s a proof by contradiction: suppose it is integral, so there exists an integer and polynomials such that Putting everything over a common denominator and rewriting, we obtain Because is an integral domain, this equality in implies the following equality in : Evaluating the left-hand side at yields , which is absurd because is not the trivial ring. Therefore the rational expression cannot be an integral element of over , whence the ring homomorphism is not integral.
It is still possible to obtain an integral morphism by localization at and at the same time, in some cases: if is an integral morphism and if is a multiplicative subset of , then the induced morphism is also an integral morphism. The proof is as follows. Pick any in the localized ring . As usual, the fact is integral means there exists an integer and elements such that Denote by the induced morphism . Passing the previous equation to the localization and dividing out by gives an equation This shows is an integral morphism, as we wanted.
Proposition. Let be an algebra. An element is integral if and only if it is contained in a finite subalgebra of . Consequently, if itself is a finite algebra, then is an integral morphism.
Proof. If is integral, then it satisfies a monic polynomial of degree with coefficients in . Hence is a -submodule of finite type which is closed under multiplication (i.e. a finite subalgebra) and which contains . On the other hand, suppose is contained in a finite subalgebra generated by elements . For each , write as the linear combination , so we have the matrix equation We write for the square matrix, and for the vector. Then the previous equation may be written equivalently as . We would like to conclude that the determinant is zero, but we can’t do that in general: a matrix is invertible (in a general ring) if and only if its determinant is invertible. Thus we know the determinant is not invertible, but we can’t go further without the following trick. Recall that any square matrix has an adjugate matrix such that . In our case, multiplying our equation on both sides by yields The determinant is an element of that kills every generator of . Therefore, it kills every element in ; in particular, it kills so we get . Expanding out the determinant yields an integral expression for with coefficients in .
The previous result is quite practical. For instance, as a sort of complement of the earlier proposition with the commutative diagram, we use it to show integral morphisms behave well under composition:
Proposition. Let and be two integral morphisms. Then their composition is also an integral morphism.
Proof. Suppose first that and are actually inclusions of rings, so that . Pick any element . Because is integral over , there exists an integer and elements such that Likewise, since each is integral over , one can choose a large enough integer such that, for all , the power can be expressed as a -linear combination of strictly smaller powers of . Now consider the -submodule of defined as Now I claim is actually a -subalgebra, i.e. it is closed under multiplication. This fact can be verified by showing that the product of any two generators is again an element of . It is clear that the product of any two generators has the form since we work in commutative rings, but the exponents could be “too large”. However, this is not a problem because of the following algorithm. If is larger than , we can rewrite as a -linear combination of powers of that are small enough, and each coefficient of that -linear combination is a product of the ’s. After distributing everything, we get a new -linear combination, where each monomial has power of not larger than . If, after this operation, a monomial has a power of larger than , we apply the same idea to write it as a -linear combinations of smaller powers of . This roughly shows that is a -subalgebra, and it is clearly finite because it’s defined using a finite number of generators. Since it contains , the previous proposition ensures is integral over . Because was an arbitrary element of , the claim is proven in the case and are ring inclusions. The general case uses the same idea, so I won’t bother writing it down.