Ytukyg

I'm studying Lovett. "Abstract Algebra." Chapter 12. Multivariable Polynomial Rings. Section 3. The Nullstellensatz.

I don't understand the exact meaning of the following proposition.

Proposition 12.3.2 $;$ Let $F$ be a field and let $E$ be a finitely generated $F$-algebra. If $E$ is a field then it is a finite extension of $F$.

I don't understand three parts.

1) What does it mean by "$E$ is a field"? In the algebra structure $(F, +, times, E, +, cdot, [,])$, where the first $+$ and $times$ are addition and multiplication in $F$, the second $+$ is an addition in $E$, $cdot$ is a scalar multiplication, and $[,]$ is an $F$-bilinear map, is the statement saying that $(E, +, [,])$ is a field?

2) What does it mean by "$E$ is an extension of $F$"? I don't see why $Fsubseteq E$ should hold. So what I'm assuming is that the statement is saying that both $F$ and $E$ are fields, and there is an embedding(injective homomorphism) $psi: Fto E$. Am I right?

3) What does it mean by "$E$ is a finite extension of $F$"? Does it mean that the degreee $[E:F]$, or more precisely, $[E:psi(F)]$, where $psi:Fto E$ is an embedding, is finite?

I'll also write the proof in the book. (I couldn't follow the proof to find out the exact meaning of the proposition becuase I don't understand the proof either.)

Proof of Proposition 12.3.2 $;$ Suppose that the field $E$ is given by $E=F[alpha_1, alpha_2, ldots, alpha_n]$. Assume that $E$ is not algebraic over $F$. Then at least one of the generators is transcendental over $F$. In fact, we can renumber the generators of $E$ so that for some $rgeq 1$, the generators $alpha_1,ldots,alpha_r$ are algebraically independent over $F$ and that $alpha_{r+1},ldots,alpha_n$ are algebraic over $K=F(alpha_1,alpha_2,ldots,alpha_r)$. Then $E$ is a finite extension of $K$, that is a finite-dimensional vector space over $K$, and also finitely generated as a $K$-module. Since $Fsubseteq Ksubseteq E$, by Proposition 12.3.1, $K$ is finitely generated as a $F$-algebra, so in fact $K=F[beta_1, beta_2, ldots, beta_s]$, where $beta_iin F(alpha_1, alpha_2, ldots, alpha_r)$.

Now each $beta_i$ is of the form
$$beta_i = frac{f_i(alpha_1, alpha_2, ldots, alpha_r)}{g_i(alpha_1, alpha_2,ldots, alpha_r)}$$
for polynomials $f_i, g_iin F[x_1, x_2, ldots, x_r]$. Recall that $F[x_1, x_2, ldots, x_r]=F[alpha_1, alpha_2, ldots, alpha_r]$ is a UFD. There are a variety of ways to see that $F[x_1, x_2, ldots, x_r]$ has an infinite number of prime (irreducible) elements. Consequently, there exists an irreducible polynomial $h$ that does not divide any of the $g_i$. By properties of addition and multiplication of fractions, every rational expression $f/gin K$, when in reduced form, has a denominator that is divisible by some divisors of $g_1, g_2,ldots, g_r$. However, the rational expression $1/h$, which is in $F(alpha_1, alpha_2, ldots, alpha_r)$, does not satisfy this property. This is a contradiction. Therefore, $E$ is algebraic over $F$. Since $E$ is algebraic and generated over $F$ by a finite number of elements, then $E$ is a finite extension of $F$.

The proof uses proposition 12.3.1 in the book, so I'll write that down too.

Proposition 12.3.1 $;$ Let $A$ be a Noetherian ring. Let $Asubseteq Bsubseteq C$ be rings such that $C$ is finitely generated as an $A$-algebra and such that $C$ is finitely generated as a $B$-module. Then $B$ is finitely generated as an $A$-algebra.

1. Yes, it means that every nonzero element of $E$ is invertible, same as it usually does. That is, $E$ is in particular a ring, and it means $E$ is a field as a ring.




2. Every $F$-algebra $E$ comes with a canonical ring homomorphism $varphi : F to E$ given by scalar multiplying elements of $F$ with the multiplicative identity of $E$; that is, $varphi(f) = f cdot 1_E$. If $F$ is a field and $E$ is nonzero then this map is automatically injective. In general, if $E$ is a commutative ring, then equipping it with an $F$-algebra structure is in fact equivalent to equipping it with a ring homomorphism $F to E$ (exercise).




3. Yes, but the embedding is already fixed (it's $varphi$ above); the claim is not that an embedding exists, it's already determined by the $F$-algebra structure.