A multiple-choice quiz
by tralfaz.
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Quiz Answer Key and Fun Facts

Answer:
**Commutative multiplication**

Rings include all of the common properties for addition (closure, commutative, associative, identity, and inverses) along with the distributive law, closure under multiplication, and associative multiplication. Closure means that you cannot move out of the set when performing the operation. For example, integers are not closed under division because 3 divided by 19 is NOT an integer. Special rings can include combinations of commutative multiplicative (commutative rings), identity (rings with identity), and inverse (division rings). If a ring has ALL of these properties it is called a field.

These rules form a natural progression when learning arithmetic. We learn about adding and multiplying natural numbers but natural numbers are very limited (you can't have a problem like 3-8). We then proceed to integers which is the ring formed by the natural numbers, zero, and their inverses. This new system lets us do more but we still have a problem with division. What is -17 divided by 2 using long division? Is it -9r+1 or -8r-1? We need to have multiplicative inverses and so we take the integers and create the field of rational numbers commonly known as fractions. Now you know why math after 5th grade is so dependent on fractions.

Rings include all of the common properties for addition (closure, commutative, associative, identity, and inverses) along with the distributive law, closure under multiplication, and associative multiplication. Closure means that you cannot move out of the set when performing the operation. For example, integers are not closed under division because 3 divided by 19 is NOT an integer. Special rings can include combinations of commutative multiplicative (commutative rings), identity (rings with identity), and inverse (division rings). If a ring has ALL of these properties it is called a field.

These rules form a natural progression when learning arithmetic. We learn about adding and multiplying natural numbers but natural numbers are very limited (you can't have a problem like 3-8). We then proceed to integers which is the ring formed by the natural numbers, zero, and their inverses. This new system lets us do more but we still have a problem with division. What is -17 divided by 2 using long division? Is it -9r+1 or -8r-1? We need to have multiplicative inverses and so we take the integers and create the field of rational numbers commonly known as fractions. Now you know why math after 5th grade is so dependent on fractions.

Answer:
**125**

A group is a set of elements (one of which is the identity) with an operation where the operation is closed and associative and each element has an inverse in the set. An Abelian group is one where the operation is commutative, so every ring is an Abelian group under addition. Matrices form a group under multiplication but NOT an Abelian group since matrix multiplication is not commutative.

"Order" is the fancy math way to say the number of elements. A group whose order is a prime number is always Abelian (in fact, it is a cyclic group). If the order is the square of a prime, it is also Abelian - but this pattern does not extend to a cube of a prime. A very old (and bad) joke is: What is purple and commutes? An Abelian grape. I warned you that it was bad!

A group is a set of elements (one of which is the identity) with an operation where the operation is closed and associative and each element has an inverse in the set. An Abelian group is one where the operation is commutative, so every ring is an Abelian group under addition. Matrices form a group under multiplication but NOT an Abelian group since matrix multiplication is not commutative.

"Order" is the fancy math way to say the number of elements. A group whose order is a prime number is always Abelian (in fact, it is a cyclic group). If the order is the square of a prime, it is also Abelian - but this pattern does not extend to a cube of a prime. A very old (and bad) joke is: What is purple and commutes? An Abelian grape. I warned you that it was bad!

Answer:
**S3**

S3 is the symmetric group (every combination) of 3 elements. D6 is the dihedral group (rotating and flipping) of a hexagon. C3 x C3 are the ordered pair combining 0's, 1's, and 2's. A6 is the alternate 6 group which is formed by starting with (123456) and creating elements of S6 by applying an even number of transpositions (switching two adjacent numbers).

Symmetric groups are very important since an important theorem in algebra states that EVERY group is a subgroup of some symmetric group. They're also important in establishing an important result from Galois theory, but I'll save that for question 10.

S3 is the symmetric group (every combination) of 3 elements. D6 is the dihedral group (rotating and flipping) of a hexagon. C3 x C3 are the ordered pair combining 0's, 1's, and 2's. A6 is the alternate 6 group which is formed by starting with (123456) and creating elements of S6 by applying an even number of transpositions (switching two adjacent numbers).

Symmetric groups are very important since an important theorem in algebra states that EVERY group is a subgroup of some symmetric group. They're also important in establishing an important result from Galois theory, but I'll save that for question 10.

Answer:
**8**

By Lagrange's theorem, the order of a subgroup must divide the order of the group. Since 8 does not divide into 150, there's no subgroup of order 8 in a group of order 150. Every group has a subgroup of order 1 (the identity element).

An important result in abstract algebra is that if p^n (p prime) divides the order of a group, there is guaranteed to be at least one subgroup of order p^n. So a group of order 500 (2^2 x 5^3) will have at least one subgroup of each order: 2, 4, 5, 25, and 125.

By Lagrange's theorem, the order of a subgroup must divide the order of the group. Since 8 does not divide into 150, there's no subgroup of order 8 in a group of order 150. Every group has a subgroup of order 1 (the identity element).

An important result in abstract algebra is that if p^n (p prime) divides the order of a group, there is guaranteed to be at least one subgroup of order p^n. So a group of order 500 (2^2 x 5^3) will have at least one subgroup of each order: 2, 4, 5, 25, and 125.

Answer:
**The complex numbers**

An extension field is made by taking a field adding one or more elements and creating a new field using the properties discussed in Question One. The complex numbers start out with the field of real numbers and the square root of -1 (i). Since the field has to be closed under multiplication, we have to include 2i, 3i, (4/91)i, etc. in the field. The inverses have to be included as well, both additive (-i, -2i, -3i . . .) and mutiplicative (i/2, i/7, i/(-193/16), etc.) and closed under addition (1+9i, -4+6i, 2-17i). This procedure is repeated with all of the new elements. Eventually, this leads to a definition of complex numbers: the set of all numbers in the form A+Bi where A and B are real numbers. Thus the complex numbers are an extension field of the real numbers.

This question/answer is a fancy way to say that the every root of every polynomial with real coefficients is a complex number. The Fundamental Theorem of Algebra guarantees that these roots do exist. More over, the number of roots equals the degree of the polynomial (i.e. the largest power of X). Interestingly enough, the complex numbers are an algebraically closed field which means a polynomial with complex coefficients has all of their roots in the complex field. This is why we don't go beyond complex numbers in high school. They form a kind of cul-de-sac in exploring polynomials.

An extension field is made by taking a field adding one or more elements and creating a new field using the properties discussed in Question One. The complex numbers start out with the field of real numbers and the square root of -1 (i). Since the field has to be closed under multiplication, we have to include 2i, 3i, (4/91)i, etc. in the field. The inverses have to be included as well, both additive (-i, -2i, -3i . . .) and mutiplicative (i/2, i/7, i/(-193/16), etc.) and closed under addition (1+9i, -4+6i, 2-17i). This procedure is repeated with all of the new elements. Eventually, this leads to a definition of complex numbers: the set of all numbers in the form A+Bi where A and B are real numbers. Thus the complex numbers are an extension field of the real numbers.

This question/answer is a fancy way to say that the every root of every polynomial with real coefficients is a complex number. The Fundamental Theorem of Algebra guarantees that these roots do exist. More over, the number of roots equals the degree of the polynomial (i.e. the largest power of X). Interestingly enough, the complex numbers are an algebraically closed field which means a polynomial with complex coefficients has all of their roots in the complex field. This is why we don't go beyond complex numbers in high school. They form a kind of cul-de-sac in exploring polynomials.

Answer:
**An integral domain**

An example of a ring that is not an integral domain are the integers modulo 6 (the remainder when a number is divided by 6) abbreviated as Z[6]. This is not an integral domain because 2x3=6 which is 0 (mod 6) but neither 2 nor 3 are 0 (mod 6). The ring Z[p] with p being a prime number is always an integral domain because the only factors of a prime number are 1 and itself.

Based on this logic, it should be pretty easy for you to show that Z[c] with c being a composite number is NEVER an integral domain.

An example of a ring that is not an integral domain are the integers modulo 6 (the remainder when a number is divided by 6) abbreviated as Z[6]. This is not an integral domain because 2x3=6 which is 0 (mod 6) but neither 2 nor 3 are 0 (mod 6). The ring Z[p] with p being a prime number is always an integral domain because the only factors of a prime number are 1 and itself.

Based on this logic, it should be pretty easy for you to show that Z[c] with c being a composite number is NEVER an integral domain.

Answer:
**The even numbers**

An ideal is a subring that "absorbs" elements of a ring through multiplication. The even numbers are a subring (i.e. have all of the properties of a ring when by themselves) and an even number times any integer gives an even number, hence the even numbers are an ideal. The same cannot be said of odd numbers or positive numbers or negative numbers. None of these set are subrings since the odd numbers are not closed under addition (odd+odd=even), the positive numbers do not have a positive additive inverse (e.g. there is no positive number you can add to 8 to make 0) and the negative numbers do not have a negative additive inverse (e.g. there is no negative number you can add to -19 to make 0).

The mathematician Kummel thought that ideals could be used to solve Fermat's Last Theorem (FLT) and he did make some major advancements on the problem. Ultimately, this idea did not solve FLT, but the usefulness of ideals still remain.

An ideal is a subring that "absorbs" elements of a ring through multiplication. The even numbers are a subring (i.e. have all of the properties of a ring when by themselves) and an even number times any integer gives an even number, hence the even numbers are an ideal. The same cannot be said of odd numbers or positive numbers or negative numbers. None of these set are subrings since the odd numbers are not closed under addition (odd+odd=even), the positive numbers do not have a positive additive inverse (e.g. there is no positive number you can add to 8 to make 0) and the negative numbers do not have a negative additive inverse (e.g. there is no negative number you can add to -19 to make 0).

The mathematician Kummel thought that ideals could be used to solve Fermat's Last Theorem (FLT) and he did make some major advancements on the problem. Ultimately, this idea did not solve FLT, but the usefulness of ideals still remain.

Answer:
**6**

N.B.: sqrt stands for "square root of". cbrt stands for "cube root of".

The splitting field is an extension field that contains all of the roots the polynomial. For example, The polynomial 5X+1 has a rational number as its root, so the splitting field is the field of rational numbers (abbreviated as Q). For X^2-2, the roots are +/- sqrt(2) so we need to take the rationals and generate an extension field by that contains sqrt(2). This means the splitting field is Q[sqrt(2)].

Degree is just that - the degree of the polynomial that has an element of the field as a root. For example, what is the degree of Q(cbrt(2)) over Q? The polynomial that has the smallest degree with rational coefficients with cbrt(2) as a root is X^3-2. This polynomial had degree 3 so the degree of Q(cbrt(2)) over Q (written as [Q(cbrt(2)):Q] is 3.

The splitting field contains all three roots of X^3 - 2, specifically cbrt(2), w[cbrt(2)], and w^2[cbrt(2)] where w is [-sqrt(3)]/2 + (1/2)i. We note that Q(cbrt(2), w) as an extension field contains all 3 roots. As stated above, E=Q[cbrt(2)] is degree 3 over Q but cbrt(2) is real and cannot account for the needed complex numbers. The smallest polynomial with elements of Q[cbrt(2)] for coefficients AND w as a root is X^2 + X + 1 so F=E(W) is degree 2 over E. Putting everything together, [Q(cbrt(2)),w:Q] = [Q(cbrt(2)),w:Q(cbrt(2))] x [Q(cbrt(2)):Q] = 2 x 3 = 6.

N.B.: sqrt stands for "square root of". cbrt stands for "cube root of".

The splitting field is an extension field that contains all of the roots the polynomial. For example, The polynomial 5X+1 has a rational number as its root, so the splitting field is the field of rational numbers (abbreviated as Q). For X^2-2, the roots are +/- sqrt(2) so we need to take the rationals and generate an extension field by that contains sqrt(2). This means the splitting field is Q[sqrt(2)].

Degree is just that - the degree of the polynomial that has an element of the field as a root. For example, what is the degree of Q(cbrt(2)) over Q? The polynomial that has the smallest degree with rational coefficients with cbrt(2) as a root is X^3-2. This polynomial had degree 3 so the degree of Q(cbrt(2)) over Q (written as [Q(cbrt(2)):Q] is 3.

The splitting field contains all three roots of X^3 - 2, specifically cbrt(2), w[cbrt(2)], and w^2[cbrt(2)] where w is [-sqrt(3)]/2 + (1/2)i. We note that Q(cbrt(2), w) as an extension field contains all 3 roots. As stated above, E=Q[cbrt(2)] is degree 3 over Q but cbrt(2) is real and cannot account for the needed complex numbers. The smallest polynomial with elements of Q[cbrt(2)] for coefficients AND w as a root is X^2 + X + 1 so F=E(W) is degree 2 over E. Putting everything together, [Q(cbrt(2)),w:Q] = [Q(cbrt(2)),w:Q(cbrt(2))] x [Q(cbrt(2)):Q] = 2 x 3 = 6.

Answer:
**C2 x C2**

The Galois group is the group of automorphisms of a field within an extention field. For example, take the real numbers (R) and the extension field of the complex numbers (C). Map each complex number to its conjugate (i.e. change the sign of the imaginary part) so for example 2+8i becomes 2-8i. Notice that the real numbers don't change since there is no imaginary part (-5.6 maps to -5.6) so this mapping would be an automorphism of the reals in the complex field. It is to Evariste Galois credit that he recognized that the structure of this group can answer many (previously) unsolved problems in mathematics.

The trick is to show that Q[sqrt(2)+ sqrt(3)] = Q[sqrt(2), sqrt(3)]. Clearly Q[sqrt(2), sqrt(3)] has sqrt(2)+sqrt(3) as an element, so Q[sqrt(2)+ sqrt(3)] is a subfield of Q[sqrt(2), sqrt(3)]. To go the other way, [sqrt(2)+ sqrt(3)]^3 = 11[sqrt(2)] + 9[sqrt(3)]. Subtract 9[sqrt(2)+ sqrt(3)] so 2[sqrt(2)] (hence sqrt(2)) is an element of Q[sqrt(2)+ sqrt(3)] and therefore so is sqrt(3). Thus Q[sqrt(2), sqrt(3)] is a subfield of Q[sqrt(2) + sqrt(3)]. By double inclusion, the fields are the same.

Whew! Now the easy part. By other theorems in Galois theory we can establish that in this problem, sqrt(2)is mapped to +/- sqrt(2) and sqrt(3) is mapped to +/- sqrt(3). This means the Galois group is C2xC2.

The Galois group is the group of automorphisms of a field within an extention field. For example, take the real numbers (R) and the extension field of the complex numbers (C). Map each complex number to its conjugate (i.e. change the sign of the imaginary part) so for example 2+8i becomes 2-8i. Notice that the real numbers don't change since there is no imaginary part (-5.6 maps to -5.6) so this mapping would be an automorphism of the reals in the complex field. It is to Evariste Galois credit that he recognized that the structure of this group can answer many (previously) unsolved problems in mathematics.

The trick is to show that Q[sqrt(2)+ sqrt(3)] = Q[sqrt(2), sqrt(3)]. Clearly Q[sqrt(2), sqrt(3)] has sqrt(2)+sqrt(3) as an element, so Q[sqrt(2)+ sqrt(3)] is a subfield of Q[sqrt(2), sqrt(3)]. To go the other way, [sqrt(2)+ sqrt(3)]^3 = 11[sqrt(2)] + 9[sqrt(3)]. Subtract 9[sqrt(2)+ sqrt(3)] so 2[sqrt(2)] (hence sqrt(2)) is an element of Q[sqrt(2)+ sqrt(3)] and therefore so is sqrt(3). Thus Q[sqrt(2), sqrt(3)] is a subfield of Q[sqrt(2) + sqrt(3)]. By double inclusion, the fields are the same.

Whew! Now the easy part. By other theorems in Galois theory we can establish that in this problem, sqrt(2)is mapped to +/- sqrt(2) and sqrt(3) is mapped to +/- sqrt(3). This means the Galois group is C2xC2.

Answer:
**solvable.**

Start with a group G. Find a subgroup G1 such that the coset G/G1 (roughly equivalent to division in any ring) is Abelian. Now find a subgroup of G1 called G2 such that G1/G2 is Abelian. Keep going and if you can eventually get down the identity element (e), the group is solvable. That is why the general quintic (5th degree) polynomial is not solvable - the group S5 is not solvable.

Start with a group G. Find a subgroup G1 such that the coset G/G1 (roughly equivalent to division in any ring) is Abelian. Now find a subgroup of G1 called G2 such that G1/G2 is Abelian. Keep going and if you can eventually get down the identity element (e), the group is solvable. That is why the general quintic (5th degree) polynomial is not solvable - the group S5 is not solvable.

This quiz was reviewed by FunTrivia editor crisw before going online.

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