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1. 011b is a hexadecimal representation of the polynomial $m(X) = X^8 + X^4 + X^3 + X + 1$ (so you should never regard it as an integer). This polynomial has coefficients in the finite field $\mathrm{GF}(2)$, which is just the math-y way to say that its coefficients are in $\{0,1\}$:

    hex | 0    1    1    b
    bin | 0000 0001 0001 1011
    x^n |         8 7654 3210
   

The 1-bits indicate which power has a coefficient of $1$. When you take any polynomial and divide it by $m(x)$, which has degree $8$, the remainder will have degree smaller than $8$, so using the same representation scheme as above, it can be represented on one byte.

2. 57 and 83 are likewise representations of polynomials using that same method, I will leave it to you to check that when you multiply those two polynomials, you do indeed find the asserted result.

3. Your understanding is correct, those two operations are the same. But again, remember that the modulus is a polynomial, not an integer.

EDIT: It's important to understand the difference between the objects AES manipulates and those you are used to. When you think of a data type which takes only one byte, you think of integers in $\{0,\dots,255\}$ with the usual addition and multiplication modulo $256$. The problem with those numbers is that while they have all the nice properties we are used to for addition, they fail catastrophically for multiplication: you can take two non-zero numbers (which ones?), multiply them, and obtain $0$!

This is not acceptable for AES: it requires a "nice" multiplication operation, meaning that when you multiply two non-zero elements, you want to obtain a non-zero element. This weird system of polynomials modulo $m(X)$ does exactly this.

1. 011b is a hexadecimal representation of the polynomial $m(X) = X^8 + X^4 + X^3 + X + 1$ (so you should never regard it as an integer). This polynomial has coefficients in the finite field $\mathrm{GF}(2)$, which is just the math-y way to say that its coefficients are in $\{0,1\}$:

    hex | 0    1    1    b
    bin | 0000 0001 0001 1011
    x^n |         8 7654 3210
   

The 1-bits indicate which power has a coefficient of $1$. When you take any polynomial and divide it by $m(x)$, which has degree $8$, the remainder will have degree smaller than $8$, so using the same representation scheme as above, it can be represented on one byte.

2. 57 and 83 are likewise representations of polynomials using that same method, I will leave it to you to check that when you multiply those two polynomials, you do indeed find the asserted result.

3. Your understanding is correct, those two operations are the same. But again, remember that you are working on polynomials, not integers.

EDIT: It's important to understand the difference between the objects AES manipulates and those you are used to. When you think of a data type which takes only one byte, you think of integers in $\{0,\dots,255\}$ with the usual addition and multiplication modulo $256$. The problem with those numbers is that while they have all the nice properties we are used to for addition, they fail catastrophically for multiplication: you can take two non-zero numbers (which ones?), multiply them, and obtain $0$!

This is not acceptable for AES: it requires a "nice" multiplication operation, meaning that when you multiply two non-zero elements, you want to obtain a non-zero element. This weird system of polynomials modulo $m(X)$ does exactly this.

1. 011b is a hexadecimal representation of the polynomial $m(X) = X^8 + X^4 + X^3 + X + 1$ (so you should never regard it as an integer). This polynomial has coefficients in the finite field $\mathrm{GF}(2)$, which is just the math-y way to say that its coefficients are in $\{0,1\}$:

    hex | 0    1    1    b
    bin | 0000 0001 0001 1011
    x^n |         8 7654 3210
   

The 1-bits indicate which power has a coefficient of $1$. When you take any polynomial and divide it by $m(x)$, which has degree $8$, the remainder will have degree smaller than $8$, so using the same representation scheme as above, it can be represented on one byte.

2. 57 and 83 are likewise representations of polynomials using that same method, I will leave it to you to check that when you multiply those two polynomials, you do indeed find the asserted result.

3. Your understanding is correct, those two operations are the same. But again, remember that the modulus is a polynomial, not an integer.

1. 011b is a hexadecimal representation of the polynomial $m(X) = X^8 + X^4 + X^3 + X + 1$ (so you should never regard it as an integer). This polynomial has coefficients in the finite field $\mathrm{GF}(2)$, which is just the math-y way to say that its coefficients are in $\{0,1\}$:

    hex | 0    1    1    b
    bin | 0000 0001 0001 1011
    x^n |         8 7654 3210
   

The 1-bits indicate which power has a coefficient of $1$. When you take any polynomial and divide it by $m(x)$, which has degree $8$, the remainder will have degree smaller than $8$, so using the same representation scheme as above, it can be represented on one byte.

2. 57 and 83 are likewise representations of polynomials using that same method, I will leave it to you to check that when you multiply those two polynomials, you do indeed find the asserted result.

3. Your understanding is correct, those two operations are the same. But again, remember that the modulus is a polynomial, not an integer.

EDIT: It's important to understand the difference between the objects AES manipulates and those you are used to. When you think of a data type which takes only one byte, you think of integers in $\{0,\dots,255\}$ with the usual addition and multiplication modulo $256$. The problem with those numbers is that while they have all the nice properties we are used to for addition, they fail catastrophically for multiplication: you can take two non-zero numbers (which ones?), multiply them, and obtain $0$!

This is not acceptable for AES: it requires a "nice" multiplication operation, meaning that when you multiply two non-zero elements, you want to obtain a non-zero element. This weird system of polynomials modulo $m(X)$ does exactly this.

011b is a hexadecimal representation of the polynomial $m(X) = X^8 + X^4 + X^3 + X + 1$ (so you should never regard it as an integer). This polynomial has coefficients in the finite field $\mathrm{GF}(2)$, which is just the math-y way to say that its coefficients are in $\{0,1\}$:

hex | 0    1    1    b
bin | 0000 0001 0001 1011
x^n |         8 7654 3210

The 1-bits indicate which power has a coefficient of $1$. When you take any polynomial and divide it by $m(x)$, which has degree $8$, the remainder will have degree smaller than $8$, so using the same representation scheme as above, it can be represented on one byte.

57 and 83 are likewise representations of polynomial using that same method, I will leave it to you to check that when you multiply those two polynomials, you do indeed find the asserted result.

1. 011b is a hexadecimal representation of the polynomial $m(X) = X^8 + X^4 + X^3 + X + 1$ (so you should never regard it as an integer). This polynomial has coefficients in the finite field $\mathrm{GF}(2)$, which is just the math-y way to say that its coefficients are in $\{0,1\}$:

    hex | 0    1    1    b
    bin | 0000 0001 0001 1011
    x^n |         8 7654 3210
   

The 1-bits indicate which power has a coefficient of $1$. When you take any polynomial and divide it by $m(x)$, which has degree $8$, the remainder will have degree smaller than $8$, so using the same representation scheme as above, it can be represented on one byte.

2. 57 and 83 are likewise representations of polynomials using that same method, I will leave it to you to check that when you multiply those two polynomials, you do indeed find the asserted result.

3. Your understanding is correct, those two operations are the same. But again, remember that the modulus is a polynomial, not an integer.