Degree is the answer to a recurring algebraic question: how much information is needed to build an object from a smaller one? A polynomial has a highest power, a [field extension](/page/Field%20Extension) has a dimension over its base field, and a graded algebra remembers which pieces sit in which layer. These are different constructions, but each uses degree to measure algebraic size in a way that survives change of presentation. The first warning is that visible complexity is not the same as degree. The polynomial $x^4 - 1$ has degree $4$, even though it factors over $\mathbb{Q}$ as $(x^2 - 1)(x^2 + 1)$ and over $\mathbb{C}$ as $(x - 1)(x + 1)(x - i)(x + i)$. The field $\mathbb{Q}(\sqrt[3]{2})$ is generated by one element, but it has degree $3$ over $\mathbb{Q}$. Degree is designed to ignore the accidental form of an expression and record the invariant algebraic size behind it. [example: Same Formula, Different Algebraic Size] Consider the same element $\sqrt{2}$ with three different base fields: $\mathbb{Q}$, $\mathbb{Q}(\sqrt{2})$, and $\mathbb{R}$. Over $\mathbb{Q}$, it satisfies the polynomial relation \begin{align*} (\sqrt{2})^2-2=2-2=0. \end{align*} Here $\mathbb{Q}[x]$ denotes the ring of polynomials in the variable $x$ with rational coefficients. Thus $x^2-2\in\mathbb{Q}[x]$ vanishes at $\sqrt{2}$. We check that no nonzero linear polynomial in $\mathbb{Q}[x]$ vanishes at $\sqrt{2}$. Let $ax+b\in\mathbb{Q}[x]$ and suppose \begin{align*} a\sqrt{2}+b=0. \end{align*} If $a=0$, then the equation becomes \begin{align*} b=0, \end{align*} so $ax+b$ is the zero polynomial. If $a\ne 0$, then subtracting $b$ and dividing by $a$ gives \begin{align*} \sqrt{2}=-\frac{b}{a}. \end{align*} Since $a,b\in\mathbb{Q}$ and $a\ne 0$, the quotient $-b/a$ lies in $\mathbb{Q}$. It remains to rule out $\sqrt{2}\in\mathbb{Q}$. Suppose $\sqrt{2}=p/q$ with $p,q\in\mathbb{Z}$, $q\ne 0$, and $\gcd(p,q)=1$. Squaring gives \begin{align*} 2=\frac{p^2}{q^2}. \end{align*} Multiplying by $q^2$ gives \begin{align*} p^2=2q^2. \end{align*} Hence $p^2$ is even, so $p$ is even. Write $p=2r$ with $r\in\mathbb{Z}$. Substituting gives \begin{align*} (2r)^2=2q^2. \end{align*} Thus \begin{align*} 4r^2=2q^2. \end{align*} Dividing by $2$ gives \begin{align*} 2r^2=q^2. \end{align*} So $q^2$ is even, hence $q$ is even. This contradicts $\gcd(p,q)=1$, since both $p$ and $q$ are divisible by $2$. Therefore $\sqrt{2}\notin\mathbb{Q}$, so the least possible degree of a nonzero polynomial in $\mathbb{Q}[x]$ vanishing at $\sqrt{2}$ is $2$. Over $\mathbb{Q}$, the element $\sqrt{2}$ therefore has algebraic degree $2$. Over $\mathbb{Q}(\sqrt{2})$, the element $\sqrt{2}$ is already in the base field, so $x-\sqrt{2}$ is a polynomial in $\mathbb{Q}(\sqrt{2})[x]$. Evaluating it at $\sqrt{2}$ gives \begin{align*} \sqrt{2}-\sqrt{2}=0. \end{align*} The polynomial $x-\sqrt{2}$ is monic and has degree $1$. No nonzero polynomial has degree less than $1$ and root $\sqrt{2}$, because a degree-$0$ polynomial is a nonzero constant and cannot evaluate to $0$. Hence the same element has algebraic degree $1$ over $\mathbb{Q}(\sqrt{2})$. The same calculation works over $\mathbb{R}$ because $\sqrt{2}\in\mathbb{R}$. The polynomial $x-\sqrt{2}\in\mathbb{R}[x]$ is monic, linear, and satisfies \begin{align*} \sqrt{2}-\sqrt{2}=0. \end{align*} The same element therefore has algebraic degree $2$ over $\mathbb{Q}$, but algebraic degree $1$ over any chosen base field that already contains $\sqrt{2}$. [/example] This example captures the central theme of the chapter. Degree is never just a property of an isolated symbol. It is a property of a polynomial relative to its coefficient ring, of an element relative to a base field, or of a homogeneous object relative to a grading. To use degree well, the ambient algebra must be part of the statement. ## Definition Before separating the polynomial, field-theoretic, and graded meanings, it is useful to name the common pattern. A degree is never just a bare number; it is an assignment made after a class of objects and a base structure have been fixed. [definition: Degree] Let $B$ be a fixed base structure, let $\mathcal{C}_B$ be a specified class of admissible algebraic objects over $B$, and let $\mathcal{D}$ be either $\{0\} \cup \mathbb{N}$ or a class of cardinal numbers. A degree on $\mathcal{C}_B$ is a map \begin{align*} \deg_B: \mathcal{C}_B &\to \mathcal{D} \end{align*} that assigns to each object $A \in \mathcal{C}_B$ a numerical invariant $\deg_B A$. [/definition] This is a meta-definition, not a formal universal construction. The precise definition comes from the particular degree function in use: polynomial degree records a highest exponent, extension degree records a vector-space dimension, and graded degree records membership in a homogeneous layer. ### Polynomial Degree The most elementary version of degree comes from polynomials. It measures the last nonzero coefficient, so it is sensitive to cancellation but insensitive to factorisation. The next definition is needed because later field-theoretic degrees are built from the same idea of measuring the first place where a nonzero leading term occurs. [definition: Degree of a Polynomial] Let $R$ be a commutative ring. The polynomial degree over $R$ is the map \begin{align*} \deg_R: R[x] \setminus \{0\} &\to \{0\} \cup \mathbb{N}. \end{align*} For a nonzero polynomial \begin{align*} f &= a_0 + a_1x + \cdots + a_nx^n \end{align*} with $a_n \ne 0_R$, \begin{align*} \deg_R f &= n. \end{align*} [/definition] The zero polynomial has no degree unless a convention is explicitly stated. The exclusion of the zero polynomial is not cosmetic. If $\deg 0$ were treated as an ordinary integer, the usual rules for multiplication would already fail when a product vanished through zero divisors. More broadly, over rings with zero divisors degree additivity can fail even for nonzero products, because the highest-degree term can cancel while lower terms survive. Many contexts adopt $\deg 0 = -\infty$ for bookkeeping, but the ordinary algebraic definition keeps the zero polynomial separate. The leading coefficient often controls whether degree behaves exactly as expected under division. Over a general ring, a nonunit leading coefficient can disappear after multiplication or fail to support division. The next distinction is not just terminology: it singles out the polynomials whose top term is fixed by the unit of the coefficient ring. This is the case in which the leading term is protected from many ring-theoretic pathologies and can later serve as the normal form for minimal polynomials and integrality. [definition: Monic Polynomial] Let $R$ be a commutative ring. A nonzero polynomial $f \in R[x]$ of degree $n$ is monic if the coefficient of $x^n$ is $1_R$. [/definition] Monic polynomials are important because their leading coefficient remains the image of $1_R$ under ordinary unital base change, so the leading term cannot disappear in the way a nonunit leading coefficient can. Minimal polynomials are also required to be monic. The monic condition is what makes integrality over rings parallel algebraicity over fields. ### Extension Degree The field-theoretic version of degree asks a different question: not how high a polynomial goes, but how many independent directions a larger field has over a smaller one. The next definition is needed because a field extension is also a [vector space](/page/Vector%20Space), so [linear algebra](/page/Cambridge%20IB%20Linear%20Algebra) supplies a numerical invariant. [definition: Degree of a Field Extension] Let $k$ be a field. The extension degree over $k$ is the map \begin{align*} \deg_k^{\mathrm{ext}}: \{K / k : K \text{ is a field extension of } k\} \to \{\text{cardinal numbers}\}. \end{align*} The notation $K / k$ means that $K$ is a field containing a specified copy of $k$; it sends such a field extension to $[K : k]$. The value $[K : k]$ is defined by \begin{align*} [K : k] &= \dim_k K. \end{align*} A field extension $K / k$ is finite if $[K : k]$ is finite. [/definition] The notation $[K : k]$ deliberately resembles the index of a subgroup. In both settings, the number measures how many independent copies of the smaller object are needed to account for the larger one. For fields, the copies are vector-space directions. A single element can generate a field extension, but not every generated element is governed by finitely many powers. The obstruction is the possible absence of any polynomial relation over the base field. To distinguish elements whose powers eventually become dependent from elements that behave like new variables, we name the relation-bearing case first. [definition: Algebraic Element] Let $K / k$ be a field extension. An element $\alpha \in K$ is algebraic over $k$ if there exists a nonzero polynomial $f \in k[x]$ such that $f(\alpha) = 0$. [/definition] Algebraicity is the point at which polynomial degree and extension degree begin to coincide. To make that coincidence precise, we need the canonical polynomial attached to an algebraic element, not just any polynomial that happens to vanish at it. The ideal theorem below explains why this monic polynomial exists and is unique; the definition records the object whose existence is supplied by that degree-descent argument. [definition: Minimal Polynomial] Let $K / k$ be a field extension, and let $\alpha \in K$ be algebraic over $k$. The [minimal polynomial](/page/Minimal%20Polynomial) of $\alpha$ over $k$ is the monic polynomial $m_{\alpha,k} \in k[x]$ of least degree such that \begin{align*} m_{\alpha,k}(\alpha) &= 0. \end{align*} [/definition] After the minimal polynomial is named, the degree of an algebraic element becomes a polynomial degree rather than a vector-space dimension. The next definition is needed because it gives a local degree attached to a single element before we discuss the whole field it generates. [definition: Degree of an Algebraic Element] Let $K / k$ be a field extension. The algebraic-element degree over $k$ is the map \begin{align*} \deg_k: \{\alpha \in K : \alpha \text{ is algebraic over } k\} \to \mathbb{N}. \end{align*} It sends an algebraic element $\alpha$ to the polynomial degree of its minimal polynomial: \begin{align*} \deg_k(\alpha) &= \deg m_{\alpha,k}. \end{align*} [/definition] This definition depends on the base field. Enlarging the base field can lower the degree because more coefficients are allowed in the polynomial relation. The guiding principle is that degree measures what remains unexplained by the chosen base. ## Polynomial Degree and Arithmetic ### Multiplication and Cancellation Polynomial degree behaves like a valuation of complexity: sums can cancel, but products usually add degrees. The word usually matters. Over a ring with zero divisors, the leading coefficients of two nonzero polynomials can multiply to zero, so the highest term of the product can disappear. [example: Cancellation in Polynomial Degree] In $(\mathbb{Z}/4\mathbb{Z})[x]$, let \begin{align*} f=2x+1 \end{align*} and \begin{align*} g=2x+1, \end{align*} where each coefficient is a residue class modulo $4$. In $\mathbb{Z}/4\mathbb{Z}$, the class $2$ is nonzero, so the highest nonzero term of both $f$ and $g$ is $2x$. Hence \begin{align*} \deg f=1 \end{align*} and \begin{align*} \deg g=1. \end{align*} Now multiply term by term in the [polynomial ring](/page/Polynomial%20Ring): \begin{align*} fg=(2x+1)(2x+1)=2x\cdot 2x+2x\cdot 1+1\cdot 2x+1\cdot 1. \end{align*} Each product is \begin{align*} 2x\cdot 2x=(2\cdot 2)x^2=4x^2, \end{align*} \begin{align*} 2x\cdot 1=2x, \end{align*} \begin{align*} 1\cdot 2x=2x, \end{align*} and \begin{align*} 1\cdot 1=1. \end{align*} Therefore \begin{align*} fg=4x^2+2x+2x+1. \end{align*} Combining the two $x$-terms gives \begin{align*} 2x+2x=(2+2)x=4x, \end{align*} so \begin{align*} fg=4x^2+4x+1. \end{align*} In $\mathbb{Z}/4\mathbb{Z}$, the class $4$ is equal to the class $0$, because $4\equiv 0 \pmod 4$. Thus \begin{align*} 4x^2+4x+1=0x^2+0x+1=1. \end{align*} So \begin{align*} \deg(fg)=\deg(1)=0, \end{align*} whereas \begin{align*} \deg f+\deg g=1+1=2. \end{align*} The leading term disappears because the leading coefficients satisfy $2\cdot 2=4=0$ in $\mathbb{Z}/4\mathbb{Z}$, so this product shows explicitly how zero divisors can break additivity of polynomial degree. [/example] The example shows why polynomial degree cannot be expected to multiply well over every coefficient ring. The obstruction is the possible disappearance of the leading term: if the product of the leading coefficients is zero, the top-degree term of the product is lost. Integral domains remove exactly this obstruction, so the degree of a product can again be read from the degrees of the factors. In the sum statement below, the zero polynomial is still treated as having no degree: the inequality for $\deg(f+g)$ is to be read with the explicit caveat that either $f+g=0$, or else $f+g\ne 0$ and $\deg(f+g)\leq \max(\deg f,\deg g)$. [quotetheorem:3234] The theorem tells us that over an integral domain the leading term of a product remembers the leading terms of the factors. Its hypotheses are doing real work: without an integral domain, the highest-degree term of a product can vanish, and without the zero-sum caveat, even addition can fall outside the degree convention adopted here. This is the algebraic mechanism behind unique factorisation arguments in polynomial rings. ### Division and Descent Division by an arbitrary polynomial over a ring is too much to expect, because leading coefficients may not be invertible. The obstruction is already visible in the first step of long division: to cancel the highest-degree term, one must divide by the divisor's leading coefficient. Over a field this division is always possible for a nonzero divisor, and the degree of the remainder then strictly descends. In the theorem below, the remainder condition is interpreted with the same zero-polynomial convention as before: either $r=0$, or $r\ne 0$ and $\deg(r)<\deg(g)$. [quotetheorem:1706] Polynomial division makes degree into a descent parameter. A nonzero remainder has smaller degree than the divisor, while a zero remainder means the division has ended exactly. This is why arguments about ideals, greatest common divisors, and irreducibility can be reduced step by step. The most important application is the ideal structure of $k[x]$. If a collection of polynomial relations is closed under addition and under multiplication by arbitrary polynomials, it forms an ideal, but it is not obvious a priori that all those relations should be consequences of one polynomial. Degree descent in $k[x]$ solves this by choosing a nonzero relation of least degree and reducing every other relation by it. We write $I \trianglelefteq k[x]$ to mean that $I$ is an ideal of the polynomial ring $k[x]$. For a polynomial $d \in k[x]$, the notation $(d)$ denotes the principal ideal generated by $d$, namely the set of all multiples $qd$ with $q \in k[x]$. The theorem below says that, over a field, every ideal of $k[x]$ has exactly this one-generator form. [quotetheorem:7994] This theorem is the hidden engine behind minimal polynomials. The set of polynomials vanishing at an algebraic element is an ideal, so a single monic generator controls all polynomial relations satisfied by that element. ## Algebraic Elements and Simple Extensions Suppose a field $K$ contains $k$ and a new element $\alpha$. To study the contribution of just this element, we must separate the field forced by $k$ and $\alpha$ from the rest of the ambient field $K$. That smallest forced field is the object whose degree measures exactly what adjoining $\alpha$ adds. [definition: Simple Field Extension] Let $K / k$ be a field extension, and let $\alpha \in K$. The simple field extension generated by $\alpha$ over $k$ is the smallest subfield of $K$ containing $k$ and $\alpha$, denoted $k(\alpha)$. [/definition] The notation $k(\alpha)$ can hide two different constructions. If $\alpha$ is transcendental, it behaves like a variable and produces a rational function field. If $\alpha$ is algebraic, powers of $\alpha$ eventually satisfy a linear relation over $k$. The key question is how many powers of $\alpha$ are genuinely independent before the first polynomial relation forces reduction. The minimal polynomial gives the first such relation, so its degree should determine both the spanning process and the number of basis elements in $k(\alpha)$ over $k$. The quotient notation records this reduction process formally. If $P_\alpha$ is the minimal polynomial, then $\langle P_\alpha \rangle$ denotes the ideal generated by $P_\alpha$, consisting of all polynomial multiples of $P_\alpha$. The quotient ring $K[t]/\langle P_\alpha \rangle$ means that two polynomials in $K[t]$ are identified when their difference is a multiple of $P_\alpha$. The evaluation map sends a polynomial in $t$ to its value at $\alpha$; after passing to the quotient, the relation $P_\alpha(\alpha)=0$ is built into the notation. What remains is to turn this notation into an actual description of the generated field. The obstruction is that $k(\alpha)$ is defined inside an ambient field, while the quotient is built purely from polynomials; comparing them shows that every element of $k(\alpha)$ has a unique reduced representative of degree below $\deg P_\alpha$. [quotetheorem:1251] This theorem is the point where the two definitions of degree meet. The polynomial relation of least degree gives the first power of $\alpha$ that can be rewritten in terms of lower powers, and all higher powers can then be reduced. [example: Computing the Cubic Extension Degree] Let $\alpha=\sqrt[3]{2}\in\mathbb{R}$, so $\alpha^3=2$. Therefore \begin{align*} \alpha^3-2=2-2=0. \end{align*} Thus $x^3-2\in\mathbb{Q}[x]$ vanishes at $\alpha$. We verify irreducibility over $\mathbb{Q}$ using *[Eisenstein's criterion](/theorems/859)* with the prime $2$. The leading coefficient of $x^3-2$ is $1$, and $2\nmid 1$. The coefficients of $x^2$ and $x$ are both $0$, and $2\mid 0$. The constant term is $-2$, so $2\mid -2$, but $4\nmid -2$. Hence $x^3-2$ is irreducible in $\mathbb{Q}[x]$. Since $x^3-2$ is monic, irreducible over $\mathbb{Q}$, and vanishes at $\alpha$, it is the minimal polynomial of $\alpha$ over $\mathbb{Q}$: \begin{align*} m_{\alpha,\mathbb{Q}}=x^3-2. \end{align*} Its degree is \begin{align*} \deg m_{\alpha,\mathbb{Q}}=\deg(x^3-2)=3. \end{align*} By *[Structure of Simple Algebraic Extensions](/theorems/1251)*, \begin{align*} [\mathbb{Q}(\alpha):\mathbb{Q}]=\deg m_{\alpha,\mathbb{Q}}=3. \end{align*} The same theorem also gives the $\mathbb{Q}$-basis \begin{align*} 1,\alpha,\alpha^2. \end{align*} Therefore every element of $\mathbb{Q}(\sqrt[3]{2})$ has a unique expression \begin{align*} a+b\alpha+c\alpha^2 \end{align*} with $a,b,c\in\mathbb{Q}$. Adjoining the single element $\sqrt[3]{2}$ therefore creates three independent $\mathbb{Q}$-directions, not just one new symbol. [/example] The example shows why adjoining one symbol need not produce a degree-one extension. The quotient part of *Structure of Simple Algebraic Extensions* encodes the same computation by an isomorphism $k[x]/(m_{\alpha,k})\cong k(\alpha)$, which lets us compare generated fields without choosing an ambient real or complex model. Algebraicity is what makes the evaluation map reach the whole field $k(\alpha)$: once a nonzero polynomial relation exists, every nonzero element of $k[\alpha]$ has an inverse represented by another polynomial in $\alpha$, so $k[\alpha]=k(\alpha)$. This quotient viewpoint is the bridge to [commutative algebra](/page/Cambridge%20III%20Commutative%20Algebra). Degree is no longer only a number attached to an expression; it controls the rank of a quotient algebra over its coefficient field. ## Towers and Multiplicativity Algebraic constructions rarely happen in one step. We may first adjoin $\sqrt{2}$ and then adjoin $\sqrt{3}$, or pass from $k$ to $L$ and then from $L$ to $K$. To compare the one-step degree with the degrees of the separate stages, the intermediate field must be part of the notation rather than hidden inside the larger extension. [definition: Tower of Field Extensions] A tower of field extensions is a chain of fields \begin{align*} k \subset L \subset K, \end{align*} where $L / k$ and $K / L$ are field extensions. [/definition] The question is whether the number of $k$-directions in $K$ can be found by first counting $k$-directions in $L$ and then counting $L$-directions in $K$. A basis of $L$ over $k$ and a basis of $K$ over $L$ should combine to describe $K$ over $k$, but this has to be justified because the scalars change between the two counts. [quotetheorem:1248] The [tower law](/theorems/1248) turns degree into a multiplicative invariant. It is the field-theoretic analogue of multiplying dimensions after a change of scalars. The hypotheses are doing real work here. The middle field $L$ has to be a field of scalars for $K$, and the inclusion $k\subset L\subset K$ has to be fixed; without that tower structure, the two separate degrees do not have a common setting in which they can be multiplied. In practice, this is what lets us compute a complicated degree by inserting a convenient intermediate field, and it is also what constrains possible intermediate fields because their degrees must divide the total degree in the finite case. [example: The Biquadratic Extension $\mathbb{Q}(\sqrt{2},\sqrt{3})$] Let $L=\mathbb{Q}(\sqrt{2})$ and $K=\mathbb{Q}(\sqrt{2},\sqrt{3})$. Since $(\sqrt{2})^2=2$, we have \begin{align*} (\sqrt{2})^2-2=2-2=0. \end{align*} Thus $x^2-2\in\mathbb{Q}[x]$ vanishes at $\sqrt{2}$. To see that it has no rational root, suppose $p/q\in\mathbb{Q}$ is in lowest terms, with $q\ne 0$, and \begin{align*} (p/q)^2=2. \end{align*} Multiplying by $q^2$ gives \begin{align*} p^2=2q^2. \end{align*} Hence $p^2$ is even, so $p$ is even. Write $p=2r$ with $r\in\mathbb{Z}$. Substituting gives \begin{align*} (2r)^2=2q^2. \end{align*} Therefore \begin{align*} 4r^2=2q^2. \end{align*} Dividing by $2$ gives \begin{align*} 2r^2=q^2. \end{align*} So $q^2$ is even, hence $q$ is even, contradicting $\gcd(p,q)=1$. Therefore $x^2-2$ has no root in $\mathbb{Q}$. A reducible quadratic over a field has a linear factor, hence a root in that field, so $x^2-2$ is irreducible over $\mathbb{Q}$. By *Structure of Simple Algebraic Extensions*, \begin{align*} [L:\mathbb{Q}]=2. \end{align*} We next show that $\sqrt{3}\notin L$. Since $1,\sqrt{2}$ is a $\mathbb{Q}$-basis of $L$ by *Structure of Simple Algebraic Extensions*, every element of $L$ has the form $a+b\sqrt{2}$ with $a,b\in\mathbb{Q}$. Suppose, for contradiction, that \begin{align*} \sqrt{3}=a+b\sqrt{2}. \end{align*} Squaring both sides gives \begin{align*} 3=(a+b\sqrt{2})^2=a^2+2ab\sqrt{2}+b^2(\sqrt{2})^2. \end{align*} Using $(\sqrt{2})^2=2$, this becomes \begin{align*} 3=a^2+2ab\sqrt{2}+2b^2. \end{align*} Subtracting $a^2+2b^2$ from both sides gives \begin{align*} 2ab\sqrt{2}=3-a^2-2b^2. \end{align*} The right-hand side lies in $\mathbb{Q}$. If $ab\ne 0$, then division by $2ab$ gives \begin{align*} \sqrt{2}=\frac{3-a^2-2b^2}{2ab}\in\mathbb{Q}, \end{align*} contradicting the irrationality of $\sqrt{2}$ proved above. Therefore $ab=0$. If $b=0$, then $\sqrt{3}=a$, so $a^2=3$. Writing $a=p/q$ in lowest terms gives \begin{align*} p^2=3q^2. \end{align*} Thus $3$ divides $p^2$, so $3$ divides $p$. Write $p=3r$. Substitution gives \begin{align*} 9r^2=3q^2. \end{align*} Dividing by $3$ gives \begin{align*} 3r^2=q^2. \end{align*} Thus $3$ divides $q$, contradicting $\gcd(p,q)=1$. Hence $b\ne 0$. Since $ab=0$ and $b\ne 0$, we must have $a=0$. Then $\sqrt{3}=b\sqrt{2}$, so squaring gives \begin{align*} 3=2b^2. \end{align*} Thus \begin{align*} b^2=\frac{3}{2}. \end{align*} Writing $b=p/q$ in lowest terms gives \begin{align*} \frac{p^2}{q^2}=\frac{3}{2}. \end{align*} Multiplying by $2q^2$ gives \begin{align*} 2p^2=3q^2. \end{align*} Thus $3$ divides $2p^2$. Since $3$ does not divide $2$, it follows that $3$ divides $p^2$, hence $3$ divides $p$. Write $p=3r$. Substitution gives \begin{align*} 2(3r)^2=3q^2. \end{align*} Therefore \begin{align*} 18r^2=3q^2. \end{align*} Dividing by $3$ gives \begin{align*} 6r^2=q^2. \end{align*} Thus $3$ divides $q^2$, hence $3$ divides $q$, contradicting $\gcd(p,q)=1$. Therefore $\sqrt{3}\notin L$. The polynomial $x^2-3\in L[x]$ has root $\sqrt{3}$ in $K$, since \begin{align*} (\sqrt{3})^2-3=3-3=0. \end{align*} If $-\sqrt{3}\in L$, then $\sqrt{3}=-(-\sqrt{3})\in L$, contradicting what we just proved. Hence neither root $\sqrt{3}$ nor $-\sqrt{3}$ lies in $L$. A reducible quadratic over a field has a root in that field, so $x^2-3$ is irreducible over $L$. Therefore, by *Structure of Simple Algebraic Extensions*, \begin{align*} [K:L]=2. \end{align*} Applying *Tower Law* to the tower $\mathbb{Q}\subset L\subset K$ gives \begin{align*} [K:\mathbb{Q}]=[K:L][L:\mathbb{Q}]. \end{align*} Substituting the two computed degrees gives \begin{align*} [K:\mathbb{Q}]=2\cdot 2=4. \end{align*} The basis also follows explicitly. By *Structure of Simple Algebraic Extensions*, $1,\sqrt{2}$ is a $\mathbb{Q}$-basis of $L$, and $1,\sqrt{3}$ is an $L$-basis of $K$. Thus every element of $K$ can be written as \begin{align*} u+v\sqrt{3} \end{align*} with $u,v\in L$. Writing \begin{align*} u=a+b\sqrt{2} \end{align*} and \begin{align*} v=c+d\sqrt{2} \end{align*} with $a,b,c,d\in\mathbb{Q}$ gives \begin{align*} u+v\sqrt{3}=(a+b\sqrt{2})+(c+d\sqrt{2})\sqrt{3}. \end{align*} Distributing the last product gives \begin{align*} u+v\sqrt{3}=a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6}. \end{align*} Since $[K:\mathbb{Q}]=4$, these four spanning elements form a $\mathbb{Q}$-basis: \begin{align*} 1,\sqrt{2},\sqrt{3},\sqrt{6}. \end{align*} So the two independent quadratic adjunctions produce a degree-$4$ extension, not merely two unrelated square roots. [/example] The calculation also shows how degree detects hidden containment. If $\sqrt{3}$ had already belonged to $\mathbb{Q}(\sqrt{2})$, the second degree would have been $1$, and the total degree would have remained $2$. A powerful consequence is that degrees constrain possible subfields and possible algebraic elements. If an intermediate field $L$ lies between $k$ and $K$, then the tower law forces the total degree $[K:k]$ to factor through $[L:k]$. Thus a proposed intermediate field can be impossible for purely numerical reasons: its degree over the base must divide the total degree. [quotetheorem:7995] This divisibility test is often the fastest way to rule out proposed intermediate fields. It is also the numerical shadow of [the Galois correspondence](/theorems/1898), where intermediate fields correspond to subgroups. ## Algebraic, Transcendental, and Finite Degree separates algebraic extensions from transcendental ones. For an algebraic element, one polynomial relation eventually rewrites high powers in terms of lower powers. The opposite situation is the failure of every possible nonzero polynomial relation over the base field, so the powers of the element cannot be controlled by a finite reduction rule. [definition: Transcendental Element] Let $K / k$ be a field extension. An element $\alpha \in K$ is transcendental over $k$ if there is no nonzero polynomial $f \in k[x]$ such that $f(\alpha)=0$. [/definition] The contrast with algebraic elements is sharp. Algebraic elements generate finite extensions; transcendental elements generate extensions that contain all rational functions in a variable. This raises the structural question of whether finite vector-space size automatically rules out transcendence. The issue is that algebraicity is a condition on each individual element, while finite degree is a condition on the whole extension. The bridge is that, in a finite-dimensional space, the infinite list $1,\alpha,\alpha^2,\ldots$ cannot remain independent forever. [quotetheorem:1303] This result explains why finite degree is a strong condition. It does not merely bound the size of the field as a vector space; it forces every element to satisfy a polynomial equation over the base. [example: Why $\mathbb{Q}(t)$ Has Infinite Degree over $\mathbb{Q}$] Let $t$ be transcendental over $\mathbb{Q}$. We prove that the powers \begin{align*} 1,t,t^2,t^3,\ldots \end{align*} are linearly independent over $\mathbb{Q}$. Fix $n\ge 0$, and suppose a finite $\mathbb{Q}$-linear relation among the first $n+1$ powers is zero: \begin{align*} a_0+a_1t+a_2t^2+\cdots+a_nt^n=0 \end{align*} with $a_0,a_1,\ldots,a_n\in\mathbb{Q}$. Define \begin{align*} f(x)=a_0+a_1x+a_2x^2+\cdots+a_nx^n\in\mathbb{Q}[x]. \end{align*} Evaluating at $t$ gives \begin{align*} f(t)=a_0+a_1t+a_2t^2+\cdots+a_nt^n. \end{align*} By the assumed relation, this is \begin{align*} f(t)=0. \end{align*} If some coefficient $a_i$ were nonzero, then $f$ would be a nonzero polynomial in $\mathbb{Q}[x]$ satisfying $f(t)=0$. This contradicts the definition of $t$ being transcendental over $\mathbb{Q}$, which says that no nonzero polynomial with rational coefficients vanishes at $t$. Hence no coefficient is nonzero, so \begin{align*} a_0=a_1=\cdots=a_n=0. \end{align*} Since this holds for every $n$, each finite set \begin{align*} \{1,t,\ldots,t^n\} \end{align*} is linearly independent over $\mathbb{Q}$. Now suppose, for contradiction, that $\mathbb{Q}(t)$ were finite-dimensional over $\mathbb{Q}$, say \begin{align*} [\mathbb{Q}(t):\mathbb{Q}]=d. \end{align*} A vector space of dimension $d$ cannot contain a linearly independent set with $d+1$ elements. But the set \begin{align*} \{1,t,\ldots,t^d\} \end{align*} has $d+1$ elements and is linearly independent over $\mathbb{Q}$, a contradiction. Therefore $\mathbb{Q}(t)$ is not finite-dimensional over $\mathbb{Q}$, so \begin{align*} [\mathbb{Q}(t):\mathbb{Q}]=\infty. \end{align*} A transcendental generator never creates a first polynomial relation, so the powers of $t$ never collapse to a finite basis. [/example] This failure mode matters because many formal expressions look algebraic until the base field is specified. Degree is finite only when polynomial relations exist over that base. When several algebraic elements are adjoined, the main difficulty is that the generators may interact: a later element might already lie in the field generated by the earlier ones, or it might contribute a new finite degree. The tower law turns this problem into a step-by-step test, multiplying only the finite contributions that remain after each enlargement of the base field. [quotetheorem:1309] This theorem justifies computing complicated degrees one generator at a time, but its algebraicity hypothesis is essential. If one generator is transcendental over the field already built, the process stops producing finite factors, as in $\mathbb{Q}(t)/\mathbb{Q}$. If a generator is already present, its contribution is instead degree $1$, so adjoining it changes notation but not size. For example, adjoining $\sqrt{2}$ and then another element asks two separate questions: first how large $\mathbb{Q}(\sqrt{2})$ is over $\mathbb{Q}$, and then whether the next element is algebraic over $\mathbb{Q}(\sqrt{2})$ and what degree it contributes there. This stepwise viewpoint is what makes later degree arguments workable: the tower law records the total, while algebraicity at each stage guarantees that every factor being multiplied is finite. ## Degree in Graded Algebra The same word degree appears in graded algebra, where it no longer counts the dimension of a field extension. Instead, it records the layer in which a homogeneous element lives. The next definition is needed because graded rings make the additive behaviour of degrees part of the algebraic structure itself. [definition: Graded Ring] A graded ring is a ring $R$ together with additive subgroups $(R_n)_{n \ge 0}$ such that \begin{align*} R &= \bigoplus_{n \ge 0} R_n \end{align*} as abelian groups, and \begin{align*} R_mR_n &\subset R_{m+n} \end{align*} for all $m,n \ge 0$. [/definition] The direct-sum decomposition means that every element may have several pieces living in different degrees, so it is not always meaningful to assign one degree to the whole element. To make degree usable inside a graded ring, we need a name for the elements that occupy exactly one graded piece. These are the elements for which multiplication has a predictable degree, and they are the pieces from which arbitrary elements are assembled. [definition: Homogeneous Element] Let $R = \bigoplus_{n \ge 0} R_n$ be a graded ring. A nonzero element $r \in R$ is homogeneous of degree $n$ if $r \in R_n$. [/definition] Homogeneous elements are the atoms of graded calculations. General elements are sums of homogeneous pieces, and many arguments proceed by comparing the highest or lowest degree piece. [example: Polynomial Rings as Graded Rings] Let $k$ be a field and set $R=k[x_1,\ldots,x_m]$. For each $n\ge 0$, let $R_n$ be the $k$-vector space spanned by all monomials $x_1^{a_1}\cdots x_m^{a_m}$ with exponents $a_i\ge 0$ satisfying \begin{align*} a_1+\cdots+a_m=n. \end{align*} Every polynomial in $R$ is a finite $k$-linear combination of monomials. Each monomial has a unique exponent tuple $(a_1,\ldots,a_m)$, so it belongs to exactly one piece $R_n$, namely the piece indexed by its total exponent sum $a_1+\cdots+a_m$. Hence every polynomial decomposes uniquely as a finite sum of its total-degree pieces, and therefore \begin{align*} R=\bigoplus_{n\ge 0}R_n. \end{align*} We check that multiplication respects these pieces. Suppose \begin{align*} x_1^{a_1}\cdots x_m^{a_m}\in R_n \end{align*} and \begin{align*} x_1^{b_1}\cdots x_m^{b_m}\in R_p. \end{align*} By membership in $R_n$ and $R_p$, the exponent sums are \begin{align*} a_1+\cdots+a_m=n \end{align*} and \begin{align*} b_1+\cdots+b_m=p. \end{align*} Multiplying the two monomials combines equal variables by adding exponents: \begin{align*} (x_1^{a_1}\cdots x_m^{a_m})(x_1^{b_1}\cdots x_m^{b_m})=x_1^{a_1+b_1}\cdots x_m^{a_m+b_m}. \end{align*} The total exponent sum of this product is \begin{align*} (a_1+b_1)+\cdots+(a_m+b_m)=(a_1+\cdots+a_m)+(b_1+\cdots+b_m). \end{align*} Substituting the two displayed sums gives \begin{align*} (a_1+b_1)+\cdots+(a_m+b_m)=n+p. \end{align*} Thus the product of a monomial in $R_n$ and a monomial in $R_p$ lies in $R_{n+p}$. Since polynomial multiplication is bilinear, any product of a finite $k$-linear combination of monomials from $R_n$ with a finite $k$-linear combination of monomials from $R_p$ is a finite $k$-linear combination of monomials in $R_{n+p}$. Therefore \begin{align*} R_nR_p\subset R_{n+p}. \end{align*} So $R=k[x_1,\ldots,x_m]$ is graded by total degree. For instance, when $m\ge 3$, \begin{align*} x_1^2x_3=x_1^2x_2^0x_3^1x_4^0\cdots x_m^0. \end{align*} Its total degree is \begin{align*} 2+0+1+0+\cdots+0=3, \end{align*} so $x_1^2x_3\in R_3$ and this monomial is homogeneous of degree $3$. By contrast, $x_1^2+x_2+1$ is not homogeneous: $x_1^2\in R_2$, $x_2\in R_1$, and $1\in R_0$. Since these three nonzero terms lie in different graded pieces, the sum has distinct homogeneous components rather than one homogeneous degree. [/example] This graded meaning of degree is compatible with polynomial degree for homogeneous polynomials, but it contains more structure. In commutative algebra, the objects acted on by a graded ring often carry their own decompositions, and the obstruction is that scalar multiplication must not mix degrees unpredictably. A module is graded only when its pieces are organized so that multiplying by a degree-$m$ ring element moves degree-$n$ module elements into degree $m+n$. [definition: Graded Module] Let $R = \bigoplus_{n \ge 0} R_n$ be a graded ring. A graded $R$-module is an $R$-module $M$ together with additive subgroups $(M_n)_{n \in \mathbb{Z}}$ such that \begin{align*} M &= \bigoplus_{n \in \mathbb{Z}} M_n \end{align*} and \begin{align*} R_mM_n &\subset M_{m+n} \end{align*} for all $m \ge 0$ and $n \in \mathbb{Z}$. [/definition] Allowing module degrees to range over $\mathbb{Z}$ makes it possible to compare modules whose algebraic elements are the same but whose generators are meant to sit in different internal degrees. To make that comparison precise, one needs a way to change only the bookkeeping of degrees while leaving all module elements and operations intact. The degree shift is this controlled reindexing of the grading: it moves every homogeneous piece by a fixed integer, so degree placement becomes explicit data that can be tracked across constructions. [definition: Degree Shift] Let $M = \bigoplus_{n \in \mathbb{Z}} M_n$ be a graded module, and let $a \in \mathbb{Z}$. The degree shift $M(a)$ is the graded module whose degree-$n$ part is \begin{align*} M(a)_n &= M_{n+a}. \end{align*} [/definition] Degree shifts record where generators live. They are indispensable in minimal free resolutions, where the shape of a resolution includes both its homological length and the internal degrees of its generators. ## Degrees as Invariants The many uses of degree share a common pattern: a complicated object is expressed through simpler pieces, and degree measures how far the construction has gone. Polynomial degree measures the highest power. Extension degree measures vector-space size. Graded degree measures layer. The danger is to transfer a theorem from one setting to another without checking which meaning is in use. [remark: The Base Object Is Part of the Data] The phrase "the degree of $\alpha$" is incomplete unless the base field is specified. The phrase "the degree of a homogeneous element" is incomplete unless the grading is specified. The phrase "the degree of a polynomial" is incomplete if coefficient cancellation or the zero polynomial is relevant. [/remark] This warning becomes concrete when changing base fields. A polynomial may factor, an element may drop in degree, and a graded algebra may acquire a different presentation. [example: Degree Drops after Base Change] Let $\alpha=\sqrt[4]{2}$ and let $F=\mathbb{Q}(\alpha)$. Since $\alpha^4=2$, evaluating $x^4-2$ at $\alpha$ gives \begin{align*} \alpha^4-2=2-2=0. \end{align*} Thus $x^4-2\in\mathbb{Q}[x]$ vanishes at $\alpha$. By *Eisenstein's criterion* with the prime $2$, this polynomial is irreducible over $\mathbb{Q}$: the leading coefficient $1$ is not divisible by $2$, the lower coefficients $0,0,0,-2$ are all divisible by $2$, and the constant term $-2$ is not divisible by $4$. Since $x^4-2$ is monic, irreducible over $\mathbb{Q}$, and vanishes at $\alpha$, it is the minimal polynomial: \begin{align*} m_{\alpha,\mathbb{Q}}=x^4-2. \end{align*} Therefore \begin{align*} \deg_{\mathbb{Q}}(\alpha)=\deg(x^4-2)=4. \end{align*} By *Structure of Simple Algebraic Extensions*, \begin{align*} [F:\mathbb{Q}]=\deg m_{\alpha,\mathbb{Q}}=4. \end{align*} Now set $L=\mathbb{Q}(\sqrt{2})$. Since $\alpha^2=\sqrt{2}$, the field $F=\mathbb{Q}(\alpha)$ contains $\sqrt{2}$, so $L\subset F$. Over $L$, evaluating $x^2-\sqrt{2}$ at $\alpha$ gives \begin{align*} \alpha^2-\sqrt{2}=\sqrt{2}-\sqrt{2}=0. \end{align*} Thus $x^2-\sqrt{2}\in L[x]$ is a polynomial relation for $\alpha$ over $L$. We next show that this relation cannot be replaced by a linear relation over $L$. First, $x^2-2$ is irreducible over $\mathbb{Q}$: if $\sqrt{2}=p/q$ with $p,q\in\mathbb{Z}$, $q\ne 0$, and $\gcd(p,q)=1$, then squaring gives \begin{align*} 2=\frac{p^2}{q^2}. \end{align*} Multiplying by $q^2$ gives \begin{align*} p^2=2q^2. \end{align*} Hence $p^2$ is even, so $p$ is even. Write $p=2r$ with $r\in\mathbb{Z}$. Substituting gives \begin{align*} (2r)^2=2q^2. \end{align*} Expanding the left-hand side gives \begin{align*} 4r^2=2q^2. \end{align*} Dividing by $2$ gives \begin{align*} 2r^2=q^2. \end{align*} Thus $q^2$ is even, so $q$ is even, contradicting $\gcd(p,q)=1$. Therefore $\sqrt{2}\notin\mathbb{Q}$, so $x^2-2$ has no rational root and is irreducible over $\mathbb{Q}$. By *Structure of Simple Algebraic Extensions*, every element of $L=\mathbb{Q}(\sqrt{2})$ has the form $a+b\sqrt{2}$ with $a,b\in\mathbb{Q}$. Suppose, for contradiction, that $\alpha\in L$. Then \begin{align*} \alpha=a+b\sqrt{2} \end{align*} for some $a,b\in\mathbb{Q}$. Squaring both sides gives \begin{align*} \alpha^2=(a+b\sqrt{2})^2. \end{align*} Since $\alpha^2=\sqrt{2}$, this becomes \begin{align*} \sqrt{2}=a^2+ab\sqrt{2}+ab\sqrt{2}+b^2(\sqrt{2})^2. \end{align*} Using $(\sqrt{2})^2=2$ and combining the two middle terms gives \begin{align*} \sqrt{2}=a^2+2ab\sqrt{2}+2b^2. \end{align*} Subtracting $\sqrt{2}$ from both sides gives \begin{align*} 0=a^2+2b^2+(2ab-1)\sqrt{2}. \end{align*} If $2ab-1\ne 0$, then \begin{align*} \sqrt{2}=-\frac{a^2+2b^2}{2ab-1}. \end{align*} The right-hand side lies in $\mathbb{Q}$, contradicting $\sqrt{2}\notin\mathbb{Q}$. Hence \begin{align*} 2ab-1=0. \end{align*} Then the displayed equation $0=a^2+2b^2+(2ab-1)\sqrt{2}$ reduces to \begin{align*} 0=a^2+2b^2. \end{align*} Since $a,b\in\mathbb{Q}$, both $a^2$ and $2b^2$ are nonnegative [real numbers](/page/Real%20Numbers), so the equality forces \begin{align*} a=0 \end{align*} and \begin{align*} b=0. \end{align*} Substituting into $2ab-1=0$ gives \begin{align*} 2\cdot 0\cdot 0-1=-1\ne 0, \end{align*} a contradiction. Therefore $\alpha\notin L$. Also $-\alpha\notin L$, because if $-\alpha\in L$, then closure of the field $L$ under additive inverses would give $\alpha\in L$. The polynomial $x^2-\sqrt{2}$ has roots $\alpha$ and $-\alpha$, because \begin{align*} \alpha^2-\sqrt{2}=0 \end{align*} and \begin{align*} (-\alpha)^2-\sqrt{2}=\alpha^2-\sqrt{2}=0. \end{align*} Neither root lies in $L$. A reducible quadratic over a field has a linear factor, hence a root in that field, so $x^2-\sqrt{2}$ is irreducible over $L$. Since it is monic and vanishes at $\alpha$, it is the minimal polynomial over $L$: \begin{align*} m_{\alpha,L}=x^2-\sqrt{2}. \end{align*} Therefore \begin{align*} \deg_L(\alpha)=\deg(x^2-\sqrt{2})=2. \end{align*} By *Structure of Simple Algebraic Extensions*, \begin{align*} [F:L]=\deg m_{\alpha,L}=2. \end{align*} We have already shown that $x^2-2$ is irreducible over $\mathbb{Q}$, so *Structure of Simple Algebraic Extensions* gives \begin{align*} [L:\mathbb{Q}]=2. \end{align*} Applying *Tower Law* to the tower $\mathbb{Q}\subset L\subset F$ gives \begin{align*} [F:\mathbb{Q}]=[F:L][L:\mathbb{Q}]. \end{align*} Substituting the computed degrees gives \begin{align*} [F:\mathbb{Q}]=2\cdot 2. \end{align*} Thus \begin{align*} [F:\mathbb{Q}]=4. \end{align*} The same generator $\sqrt[4]{2}$ has degree $4$ over $\mathbb{Q}$ but degree $2$ over $\mathbb{Q}(\sqrt{2})$; enlarging the base field has already accounted for the intermediate square root $\sqrt{2}$. [/example] The lesson is that degree is stable only after the surrounding algebra is fixed. Once the base and structure are declared, degree becomes one of the most effective invariants in algebra. A final refinement asks what extra data is needed once the degree has been computed. Finite degree alone cannot determine a generated field: two simple extensions may have the same vector-space dimension while satisfying different polynomial relations. The quotient description in *Structure of Simple Algebraic Extensions* shows that the missing information is the actual minimal polynomial, because it records both the degree and the multiplication rule imposed on the generator. This does not say that two ambient fields are the same. It says that the portion generated by a root is determined, over $k$, by the minimal polynomial. Degree is therefore a partial classifier: it gives the size, while the polynomial gives the precise simple extension. ## Beyond and Connected Topics Degree is the entry point to Galois theory because finite extension degree constrains intermediate fields and because separability asks how many distinct roots a polynomial has after passing to a [splitting field](/page/Splitting%20Field). The tower law becomes the numerical backbone of the [Galois correspondence](/page/Galois%20Correspondence). In [commutative algebra](/page/Cambridge%20III%20Commutative%20Algebra), degree reappears in graded rings, Hilbert functions, Hilbert polynomials, and multiplicity. There the central question shifts from field dimension to the growth rate of graded pieces. In [homological algebra](/page/Homological%20Algebra%20I%3A%20Complexes%20and%20Resolutions), degree has another role: chain complexes are indexed by homological degree, while graded modules may carry internal degree at the same time. Keeping these degrees separate is part of the language of resolutions and derived functors. In [Lie algebras](/page/Lie%20Algebras%20I%3A%20Foundations), graded Lie algebras use degree to control brackets by the rule $[\mathfrak{g}_m,\mathfrak{g}_n] \subset \mathfrak{g}_{m+n}$. This is a structural version of the same additive behaviour seen in graded rings. ## References Androma, [Cambridge IB Linear Algebra](/page/Cambridge%20IB%20Linear%20Algebra). Androma, [Cambridge III Commutative Algebra](/page/Cambridge%20III%20Commutative%20Algebra). Androma, [Homological Algebra I: Complexes and Resolutions](/page/Homological%20Algebra%20I%3A%20Complexes%20and%20Resolutions). Androma, [Lie Algebras I: Foundations](/page/Lie%20Algebras%20I%3A%20Foundations). Lang, *Algebra* (2002). Atiyah and Macdonald, *Introduction to Commutative Algebra* (1969). Dummit and Foote, *Abstract Algebra* (2004).