# What does it mean by that $E$ is a finite extension of $F$, when $F\subseteq E$ is not clear?

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.
• 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. – zxcv Dec 31 '18 at 2:51
• @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. – zxcv Dec 31 '18 at 8:26
• Yes, the assumption I wanted was that $F$ is a field. – Qiaochu Yuan Dec 31 '18 at 11:06