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 $F\subseteq E$ 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) $\psi: F\to E$. 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 $\psi:F\to E$ 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 $r\geq 1$, 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 $F\subseteq K\subseteq E$, 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 $\beta_i\in F(\alpha_1, \alpha_2, \ldots, \alpha_r)$.
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 $f_i, g_i\in F[x_1, x_2, \ldots, x_r]$. 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 $f/g\in K$, 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 $A\subseteq B\subseteq C$ 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.

  • $\begingroup$ In 2, I see that $\varphi$ is a homomorphism. But why is $\varphi$ injective when $E$ is a field? I know that what I should show is that $(f_1 - f_2)\cdot 1_E=0$ implies $f_1 - f_2 = 0$. But I don't know how. $\endgroup$ – zxcv Dec 31 '18 at 2:51
  • $\begingroup$ @zxcv Actually I figured out myself the answer to the above comment. I proved that if $F$ is a division ring and if $1_E \neq 0_E$, then $\varphi$ is injective. I'll present the proof here so that other people can see. Assume that $f_1 - f_2 \neq 0_F$ and scalar multiply $(f_1 - f_2)^{-1}$ to both sides of $(f_1 - f_2) \cdot 1_E = 0_E$, then a contradiction arises. $\endgroup$ – zxcv Dec 31 '18 at 8:26
  • $\begingroup$ Yes, the assumption I wanted was that $F$ is a field. $\endgroup$ – Qiaochu Yuan Dec 31 '18 at 11:06

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