This lecture is part of an online course on Galois theory.We review some basic results about field extensions and algebraic numbers.We define the degree of a...The extension field degree (or relative degree, or index) of an extension field , denoted , is the dimension of as a vector space over , i.e., (1) Given a field , there are a couple of ways to define an extension field. If is contained in a larger field, .Some properties. All transcendental extensions are of infinite degree.This in turn implies that all finite extensions are algebraic. The converse is not true however: there are infinite extensions which are algebraic. For instance, the field of all algebraic numbers is an infinite algebraic extension of the rational numbers.. Let E be an extension field of K, and a ∈ E.$\begingroup$ Thanks a lot, very good ref. I almost reach the notion of linearly disjoint extensions. I just remark that, in the last result (Corollary 8) of your linked notes, it's enough to assume only L/K to be finite Galois, in fact in J. Milne's "Fields and Galois Theory" (version 4.40) Corollary 3.19, the author gives a more general formula. $\endgroup$A field extension of degree 2 is a Normal Extension. Let L be a field and K be an extension of L such that [ K: L] = 2 . Prove that K is a normal extension. What I have tried : Let f ( x) be any irreducible polynomial in L [ x] having a root α in K and let β be another root. Then I have to show β ∈ K.This cardinality is the transcendence degree of the extension. Then L is algebraic over the subfield generated by a transcendence basis. Briefly any field ...We know Q[(] is a cyclic Galois extension of degree p-1. Therefore, there is a tower of field extensions Q = K0 ( K1 ( ((( ( Km = Q[(], with each successive extension cyclic of order some prime q dividing p-1. Now, we would like these extensions to be qth root extensions, but we need to make sure we have qth roots of unity first.Pursuing a Master’s degree in CA (Chartered Accountancy) can be a wise decision for those who want to advance their careers and gain expertise in accounting, auditing, taxation, and other related fields.Here's a primitive example of a field extension: $\mathbb{Q}(\sqrt 2) = \{a + b\sqrt 2 \;|\; a,b \in \mathbb{Q}\}$. It's easy to show that it is a commutative additive group with identity $0$. ... (cannot be written as a product of nonconstant polynomials of strictly smaller degree); this polynomial is called "the monic irreducible (polynomial ...1 Answer. Sorted by: 1. Each element of L L that isn't in K K has a minimal polynomial of degree 3 3. At most three of them can share the same minimal polynomial. You may wish to count more accurately: e.g. you're only counting x3 x 3 as one sixth of a polynomial.2 Answers. Sorted by: 18. There are two kinds of quadratic extensions in characteristic 2 2. The first are the same as in other characteristics: namely, if α ∈ F ∖F2 α ∈ F ∖ F 2, then F( α−−√) F ( α) is a quadratic extension. It need not be the case that every element is a square in characteristic 2 2. This occurs iff the ...10.158 Formal smoothness of fields. 10.158. Formal smoothness of fields. In this section we show that field extensions are formally smooth if and only if they are separable. However, we first prove finitely generated field extensions are separable algebraic if and only if they are formally unramified. Lemma 10.158.1. Field extensions Jan Snellman1 1Matematiska Institutionen Link opings Universitet Link oping, fall 2019 ... [C : R] = 2, so R C is a nite dimensional extension of degree 2. [R : Q] = 1, so this extension is in nite dimensional. It is a theorem (as long as you accept the axiom of choice) that any vectorSTEM Designated Degree Program List Effective May 10, 2016 The STEM Designated Degree Program list is a complete list of fields of study that DHS considers to be science, technology, engineering or mathematics (STEM) fields of study for purposes of the 24-month STEM optional practical training extension described at 8 CFR 214.2(f).The dimension of F considered as an E -vector space is called the degree of the extension and is denoted [F: E]. If [F: E] < ∞ then F is said to be a finite extension of E. Example 9.7.2. The field C is a two dimensional vector space over R with basis 1, i. Thus C is a finite extension of R of degree 2. Lemma 9.7.3. Theorem 1: Multiplicativity Formula for Degrees. Let E be an field extension of K and F be a field extension of E. Then, [ F: K] = [ F: E] [ E: K] The real interesting part of this for me (and why I’m writing this in the first place) is the fact that the proof uses basic concepts from linear algebra to prove this. Proof.Subject classifications. For a Galois extension field K of a field F, the fundamental theorem of Galois theory states that the subgroups of the Galois group G=Gal (K/F) correspond with the subfields of K containing F. If the subfield L corresponds to the subgroup H, then the extension field degree of K over L is the group order of H, |K:L| = |H ...1 Answer. Suppose every odd degree equation has a solution. Let L / K be a finite extension. Go to a Galois closure M / K with group G. It has a Sylow 2-subgroup H. Consider the fixed field M H. This has odd degree over K, so M H = K and H = G. Thus | G | is a power of 2 and | M: K | and | L: K | are powers of 2.Field Extensions 1 Section V.1. Field Extensions Note. In this section, we define extension fields, algebraic extensions, and tran- ... ∼= K[x]/(f) where f ∈ K[x] is an irreduciblemonic polynomial of degree n ≥ 1 uniquely determined by the conditions that f(u) = 0 and g(u) = 0 (where g ∈ K[x]) if and only if f divides g;The dimension of F considered as an E -vector space is called the degree of the extension and is denoted [F: E]. If [F: E] < ∞ then F is said to be a finite extension of E. Example 9.7.2. The field C is a two dimensional vector space over R with basis 1, i. Thus C is a finite extension of R of degree 2. Lemma 9.7.3.A faster way to show that $\mathbb{C}$ is an infinite extension of $\mathbb{Q}$ is to observe that $\mathbb{C}$ is uncountable, while any finite extension of $\mathbb{Q}$ is countable. A more interesting question is showing that $\overline{\mathbb{Q}}$ is an infinite extension of $\mathbb{Q}$, which your argument in fact shows.I want to show that each extension of degree 2 2 is normal. Let K/F K / F the field extension with [F: K] = 2 [ F: K] = 2. Let a ∈ K ∖ F a ∈ K ∖ F. Then we have that F ≤ F(a) ≤ K F ≤ F ( a) ≤ K. We have that [K: F] = 2 ⇒ [K: F(a)][F(a): F] = 2 [ K: F] = 2 ⇒ [ K: F ( a)] [ F ( a): F] = 2. m ( a, F) = 2.De nition 12.3. The transcendence degree of a eld extension L=Kis the cardinality of any (hence every) transcendence basis for L=k. Unlike extension degrees, which multiply in towers, transcendence degrees add in towers: for any elds k L M, the transcendence degree of M=kis the sum (as cardinals) of the transcendence degrees of M=Land L=k.Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this siteIn wikipedia, there is a definition of field trace. Let L/K L / K be a finite field extension. For α ∈ L α ∈ L, let σ1(α),...,σn(α) σ 1 ( α),..., σ n ( α) be the roots of the minimal polynomial of α α over K K (in some extension field of K K ). Then. TrL/K(α) = [L: K(α)]∑j=1n σj(α) Tr L / K ( α) = [ L: K ( α)] ∑ j = 1 ...Our students in the Sustainability Master’s Degree Program are established professionals looking to deepen their expertise and advance their careers. Half (50%) have professional experience in the field and all work across a variety of industries—including non-profit management, consumer goods, communications, pharmaceuticals, and utilities.When ll algebraic extensions arechar²-³~ - or when is a finite field, a separable, but such is not the case with more unusual fields. As mentioned earlier, an extension of is ,-normal if it is the splitting field of a family of polynomials. An extension that is both separable and normal is called a Galois extension. Distinguished ExtensionsDetermine the degree of a field extension Ask Question Asked 10 years, 11 months ago Modified 9 years ago Viewed 8k times 6 I have to determine the degree of Q( 2-√, 3-√) Q ( 2, 3) over Q Q and show that 2-√ + 3-√ 2 + 3 is a primitive element ? Could someone please give me any hints on how to do that ? abstract-algebra extension-field Share CiteQuestion: 2. Find a basis for each of the following field extensions. What is the degree of each extension? (a) Q (√3, √6) over Q (b) Q (2, 3) over Q (c) Q (√2, i) over Q (d) Q (√3, √5, √7) over Q (e) Q (√2, 2) over Q (f) Q (√8) over Q (√2) (g) Q (i. √2+i, √3+ i) over Q (h) Q (√2+ √5) over Q (√5) (i) Q (√2, √6 ...A basic datum of a field extension is its degree [F : E], i.e., the dimension of F as an E-vector space. It satisfies the formula [G : E] = [G : F] [F : E]. Extensions whose degree is finite are referred to as finite extensions. The extensions C / R and F 4 / F 2 are of degree 2, whereas R / Q is an infinite extension. Algebraic extensions The degree (or relative degree, or index) of an extension field, denoted , is the dimension of as a vector space over , i.e., If is finite, then the extension is said to be finite; otherwise, it is said to be infinite.Thus $\mathbb{Q}(\sqrt[3]{2},a)$ is an extension of degree $6$ over $\mathbb{Q}$ with basis $\{1,2^{1/3},2^{2/3},a,a 2^{1/3},a 2^{2/3}\}$. The question at hand. I have to find a basis for the field extension $\mathbb{Q}(\sqrt{2}+\sqrt[3]{4})$. A hint is given: This is similar to the case for $\mathbb{Q}(\sqrt{1+\sqrt[3]{2}})$.The degree of E/F E / F, denoted [E: F] [ E: F], is the dimension of E/F E / F when E E is viewed as a vector space over F F .A lot of the other answers have espoused that your answer is ultimately ok, but you should be cautious with polynomials of higher degree. I can't say I fully agree with the first point - saying that the roots aren't in $\mathbb{Q}(i)$ feels to me like you are begging the question, because that is precisely what you are trying to prove.Field extensions 1 3. Algebraic extensions 4 4. Splitting fields 6 5. Normality 7 6. Separability 7 7. Galois extensions 8 8. Linear independence of characters 10 ... The degree [K: F] of a finite extension K/Fis the dimension of Kas a vector space over F. 1and the occasional definition or two. Not to mention the theorems, lemmas and so ...Mar 21, 2015 ... Definition 31.2. If an extension field E of field F is of finite dimension n as a vector space over F, then E is a finite extension of degree ...Definition 9.15.1. Let E/F be an algebraic field extension. We say E is normal over F if for all \alpha \in E the minimal polynomial P of \alpha over F splits completely into linear factors over E. As in the case of separable extensions, it takes a bit of work to establish the basic properties of this notion.Jul 1, 2016 · Galois extension definition. Let L, K L, K be fields with L/K L / K a field extension. We say L/K L / K is a Galois extension if L/K L / K is normal and separable. 1) L L has to be the splitting field for some polynomial in K[x] K [ x] and that polynomial must not have any repeated roots, or is it saying that. What things we have to take care of while finding the degree of field extension, splitting fields for some polynomial? 5. Finding the degree of an algebraic field extension. 0. There are infinitely many non-isomorphic cubic Galois extension of $\mathbb Q$ 1.Are you looking for a comprehensive and accessible introduction to the theory of field extensions? If yes, then you should check out this pdf document from Maharshi Dayanand University, which covers the basic concepts, examples, and applications of this important branch of abstract algebra. This pdf is also part of the study material for the Master of Science (Mathematics) course offered by ...The key element in proving that all these extensions are solvable over the base field is then to define a solvable extension as an extension which normal closure has solvable Galois group (equivalently such that there exist an extension which Galois group is solvable) (def (a)), this makes "being a solvable extension" transitive (it is ...The STEM OPT extension is a 24-month extension of OPT available to F-1 nonimmigrant students who have completed 12 months of OPT and received a degree in an approved STEM field of study as designated by the STEM list. ... (CIP code 40). If a degree is not within the four core fields, DHS considers whether the degree is in a STEM-related field ...AN INTRODUCTION TO THE THEORY OF FIELD EXTENSIONS 3 map ˇ: r7!r+ Iis a group homomorphism with kernel I(natural projection for groups). It remains to check that ˇis a …Oct 18, 2015 ... Let's consider K/k a finite field extension of degree n. The following theorem holds. Theorem: the following conditions are equivalent:.To Choose a Field of Study: Complete two courses at Harvard in a chosen field with grades of B or higher. Submit a field of study proposal form to the Office of ALB Advising and Program Administration. Maintain a B grade average in 32 Harvard credits in the field, with all B– grades or higher. Fields of study and minors appear on your ...The degree of the field extension is 2: $[\mathbb{C}:\mathbb{R}] = 2$ because that is the dimension of a basis of $\mathbb{C}$ over $\mathbb{R}$. As additive groups, $\mathbb{R}$ is normal in $\mathbb{C}$, so we get that $\mathbb{C} / \mathbb{R}$ is a group. The cardinality of this group is uncountably infinite (we have an answer for this here ...If K is an extension eld of F, thedegree [K : F] (also called the relative degree or very occasionally the \index") is the dimension dim F(K) of K as an F-vector space. The extension K=F is nite if it has nite degree; otherwise, the extension isin nite. In fact, de ning the degree of a eld extension was the entireDefine Field extension. Field extension synonyms, Field extension pronunciation, Field extension translation, English dictionary definition of Field extension. n. 1. A …Is every field extension of degree $2018$ primitve? 1. Calculate the degree of a composite field extension. 0. suppose K is an extension field of finite degree, and L,H are middle fields such that L(H)＝K.Prove that [K:L]≤[H:F] Hot Network QuestionsIn mathematics, a Galois extension is an algebraic field extension E/F that is normal and separable; or equivalently, E/F is algebraic, and the field fixed by the automorphism group Aut(E/F) is precisely the base field F.The significance of being a Galois extension is that the extension has a Galois group and obeys the fundamental theorem of Galois theory.. …The key element in proving that all these extensions are solvable over the base field is then to define a solvable extension as an extension which normal closure has solvable Galois group (equivalently such that there exist an extension which Galois group is solvable) (def (a)), this makes "being a solvable extension" transitive (it is ...STEM Designated Degree Program List Effective May 10, 2016 The STEM Designated Degree Program list is a complete list of fields of study that DHS considers to be science, technology, engineering or mathematics (STEM) fields of study for purposes of the 24-month STEM optional practical training extension described at 8 CFR 214.2(f).Such an extension is unique up to a K-isomorphism, and is called the splitting field of f(X) over K. If degf(X) = n, then the degree of the splitting field of f(X) over Kis at most n!. Thus if f(X) is a nonconstant polynomial in K[X] having distinct roots, and Lis its splitting field over K, then L/Kis an example of a Galois extension.In particular, all transcendence bases of a field extension have the same cardinality, called the transcendence degree of the extension. Thus, a field extension is a transcendental extension if and only if its transcendence degree is positive. Transcendental extensions are widely used in algebraic geometry.If K K is an extension field of Q Q such that [K: Q] = 2 [ K: Q] = 2, prove that K =Q( d−−√) K = Q ( d) for some square-free integer d d. Now, I understand that since the extension is finite-dimensional, so it has to be algebraic. So in particular if I take any element u ∈ K u ∈ K not in Q Q then it must be algebraic.21. Any finite extension of a finite field Fq F q is cyclic. For such an extension K K first recall that the Frobenius map x ↦ xq x ↦ x q is an Fq F q -linear endomorphism. If xq =yq x q = y q then (x − y)q = 0 ( x − y) q = 0, hence x = y x = y, so the Frobenius map is injective. Since it is an injective linear map from a finite ...We know Q[(] is a cyclic Galois extension of degree p-1. Therefore, there is a tower of field extensions Q = K0 ( K1 ( ((( ( Km = Q[(], with each successive extension cyclic of order some prime q dividing p-1. Now, we would like these extensions to be qth root extensions, but we need to make sure we have qth roots of unity first.De nition 12.3. The transcendence degree of a eld extension L=Kis the cardinality of any (hence every) transcendence basis for L=k. Unlike extension degrees, which multiply in towers, transcendence degrees add in towers: for any elds k L M, the transcendence degree of M=kis the sum (as cardinals) of the transcendence degrees of M=Land L=k.what is the degree of field extension over base field? 0. Degree of a field extension over $\mathbb{Q}$ 0. Find the degree of a field extension and proving polynomial irreducible. 0. Field theory questions about polynomials and extension. 1.In wikipedia, there is a definition of field trace. Let L/K L / K be a finite field extension. For α ∈ L α ∈ L, let σ1(α),...,σn(α) σ 1 ( α),..., σ n ( α) be the roots of the minimal polynomial of α α over K K (in some extension field of K K ). Then. TrL/K(α) = [L: K(α)]∑j=1n σj(α) Tr L / K ( α) = [ L: K ( α)] ∑ j = 1 ...An extension field of a field F that is not algebraic over F, i.e., an extension field that has at least one element that is transcendental over F. For example, the field of rational functions F(x) in the variable x is a transcendental extension of F since x is transcendental over F. The field R of real numbers is a transcendental extension of the field Q of rational numbers, since pi is ...The Division of Continuing Education (DCE) at Harvard University is dedicated to bringing rigorous academics and innovative teaching capabilities to those seeking to improve their lives through education. We make Harvard education accessible to lifelong learners from high school to retirement. Study part time at Harvard, in evening or online ...To Choose a Field of Study: Complete two courses at Harvard in a chosen field with grades of B or higher. Submit a field of study proposal form to the Office of ALB Advising and Program Administration. Maintain a B grade average in 32 Harvard credits in the field, with all B– grades or higher. Fields of study and minors appear on your ...Some field extensions with coprime degrees. 3. Showing that a certain field extension is Galois. 0. Divisibility between the degree of two extension fields. 0. Extension Degree of Fields Composite. Hot Network Questions How to take good photos of stars out of a cockpit window using the Samsung 21 ultra?Help clear the air and confusion by attending the Eco Markets and Carbon Dynamics Field Day with University of Illinois Urbana-Champaign , Illinois Extension, …FIELD EXTENSIONS 0. Three preliminary remarks. Every non-zero homomorphism between fields is injective; so we talk about field extensions F⊂ K. ... It is called the degree of the extension. 1. Algebraic and transcendental elements. Given K⊃ F, an element α∈ Kis called algebraic over F, if it is a root of a polynomial1. No, K will typically not have all the roots of p ( x). If the roots of p ( x) are α 1, …, α k (note k = n in the case that p ( x) is separable), then the field F ( α 1, …, α k) is called the splitting field of p ( x) over F, and is the smallest extension of F that contains all roots of p ( x). For a concrete example, take F = Q and p ...E. Short Questions Relating to Degrees of Extensions. Let F be a field. Prove parts 1−3: 1 The degree of a over F is the same as the degree of 1/a over F. It is also the same as the degrees of a + c and ac over F, for any c ∈ F. 2 a is of degree 1 over F iff a ∈ F. Find the degree $[K:F]$ of the following field extensions: (a) $K=\mathbb{Q}(\sqrt{7})$, $F=\mathbb{Q}$ (b) $K=\mathbb{C}(\sqrt{7})$, $F=\mathbb{C}$ (c) $K=\mathbb{Q}(\sqrt{5},\sqrt{7},\sqrt{... Stack Exchange Network Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this siteField of study courses must be completed with a B- or higher without letting your overall field of study dip below 3.0. The same is required for minor courses. ... Harvard Extension School. Harvard degrees, certificates and courses—online, in the evenings, and at your own pace.1. In Michael Artin states in his Algebra book chapter 13, paragraph 6, the following. Let L be a finite field. Then L contains a prime field F p. Now let us denote F p by K. If the degree of the field extension [ L: K] = r, then L as a vector space over K is isomorphic to K r. My three questions are:1Definition and notation 2The multiplicativity formula for degrees Toggle The multiplicativity formula for degrees subsection 2.1Proof of the multiplicativity formula in the finite caseOct 12, 2023 · The degree (or relative degree, or index) of an extension field K/F, denoted [K:F], is the dimension of K as a vector space over F, i.e., [K:F]=dim_FK. If [K:F] is finite, then the extension is said to be finite; otherwise, it is said to be infinite. In mathematics, more specifically field theory, the degree of a field extension is a rough measure of the "size" of the field extension. The concept plays an important role in many parts of mathematics, including algebra and number theory — indeed in any area where fields appear prominently. Oops something went wrong: 404 Enjoying Wikiwand?9.12 Separable extensions. 9.12. Separable extensions. In characteristic p something funny happens with irreducible polynomials over fields. We explain this in the following lemma. Lemma 9.12.1. Let F be a field. Let P ∈ F[x] be an irreducible polynomial over F. Let P′ = dP/dx be the derivative of P with respect to x.Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site2 Fields and Field Extensions Our goal in this chapter is to study the structure of elds, a subclass of rings in which every nonzero element has a multiplicative inverse, and eld extensions. An algebraic extension is a purely inseparable extension if and only if for every , the minimal polynomial of over F is not a separable polynomial. [1] If F is any field, the trivial extension is purely inseparable; for the field F to possess a non-trivial purely inseparable extension, it must be imperfect as outlined in the above section.The extension field $\mathbf{F}_{125}$ contains prime field as a subfield, and as a vector space over it has $\{1,\bar X, \bar X^2\}$ as its basis. This shows that $\bar X$ is not in the prime field. When the degree of a field extension is a prime number any element that is not in the base field will be a primitive element, as you have guessed ...Explore Programs Available at Harvard. Browse the graduate and undergraduate degrees and majors offered by Harvard's 13 Schools and learn more about admissions requirements, scholarship, and financial aid opportunities. We also offer executive education, certificate programs, and online courses for professional and lifelong learners.A B.A. degree is a Bachelor of Arts degree in a particular field. According to California Polytechnic State University, a Bachelor of Arts degree primarily encompasses areas of study such as history, language, literature and other humanitie...The dimension of F considered as an E -vector space is called the degree of the extension and is denoted [F: E]. If [F: E] < ∞ then F is said to be a finite extension of E. Example 9.7.2. The field C is a two dimensional vector space over R with basis 1, i. Thus C is a finite extension of R of degree 2. Lemma 9.7.3. In field theory, a branch of algebra, an algebraic field extension / is called a separable extension if for every , the minimal polynomial of over F is a separable polynomial (i.e., its formal derivative is not the zero polynomial, or equivalently it has no repeated roots in any extension field). There is also a more general definition that applies when E is not necessarily algebraic over F.A transcendence basis of K/k is a collection of elements {xi}i∈I which are algebraically independent over k and such that the extension K/k(xi; i ∈ I) is algebraic. Example 9.26.2. The field Q(π) is purely transcendental because π isn't the root of a nonzero polynomial with rational coefficients. In particular, Q(π) ≅ Q(x).the smallest degree such that m(x) = 0 is called the minimal polynomial of u over F. If u is not algebraic over F, it is called transcendental over F. K is called an algebraic extension of F if every element of K is algebraic over F; otherwise, K is called transcendental over F. Example. √ 2 + 3 √ 3 ∈R is algebraic over Q with minimal ...$\begingroup$ Moreover, note that an extension is Galois $\iff$ the number of automorphisms is equal to the degree of the extension. If it's not Galois, then the number of automorphisms divides the degree of the extension, which means there are either $1$ or $2$ automorphisms for this scenario, which should give you some reassurance that your ultimate list is complete.A polynomial f of degree n greater than one, which is irreducible over F q, defines a field extension of degree n which is isomorphic to the field with q n elements: the elements of this extension are the polynomials of degree lower than n; addition, subtraction and multiplication by an element of F q are those of the polynomials; the product ... If degree is nonzero, then name must be a string (or None, if this is a pseudo-Conway extension), and will be the variable name of the returned field. If degree is zero, the dictionary should have keys the divisors of the degree of this field, with the desired variable name for the field of that degree as an entry.finite field extensions of coprime aegrees is again a field. PROPOSITION 2.1. Let k be any field and Elk, F/k finit, Tour Start here for a quick overview of the site Help Cent, Some field extensions with coprime degrees. 3. Showing that a certain field extension is , 2. Complete Degree Courses for Admission. At Harvard , Are you fascinated by the idea of extending your lifespan and living a healthier, more vibrant life? Look no fu, Inseparable field extension of degree 2. I have searched for an example of a degree , The degree (or relative degree, or index) of an extension field, denoted , t. e. In mathematics, an algebraic number field (or si, Our results imply that over a large field extension , 2. Complete Degree Courses for Admission. At Harvard Exten, October 18, 2023 3:14 PM. Blog Post. An updated Corn , Every nite extension of F p is a Galois extension whose Galois g, A certificate is ideal for people attempting to move up in their cu, Mar 23, 2019 · The degree of the field extension , Tour Start here for a quick overview of the site Help Center Detail, Eligibility for 24-Month STEM OPT Extension You must: Be maint, A basic datum of a field extension is its degree [F :, The dimension of F considered as an E -vector space i.